\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{The inclination of the line }x-y+3=0\text{ with the positive direction of}
\displaystyle \text{the }x\text{-axis is:}
\displaystyle \text{(A) }45^\circ\qquad\text{(B) }135^\circ\qquad\text{(C) }-45^\circ\qquad\text{(D) }-135^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x-y+3=0.
\displaystyle y=x+3.
\displaystyle \text{Comparing with }y=mx+c,\text{ we get }m=1.
\displaystyle \text{If }\theta\text{ is the inclination of the line, then }m=\tan\theta.
\displaystyle \therefore \tan\theta=1.
\displaystyle \therefore \theta=45^\circ.
\displaystyle \therefore \text{Option (A) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The two lines }ax+by=c\text{ and }a'x+b'y=c'\text{ are perpendicular if:}
\displaystyle \text{(A) }aa'+bb'=0\qquad\text{(B) }ab'=ba'\qquad\text{(C) }ab+a'b'=0
\displaystyle \text{Answer:}
\displaystyle \text{Slope of }ax+by=c\text{ is }m_1=-\frac{a}{b}.
\displaystyle \text{Slope of }a'x+b'y=c'\text{ is }m_2=-\frac{a'}{b'}.
\displaystyle \text{For two lines to be perpendicular, }m_1m_2=-1.
\displaystyle \left(-\frac{a}{b}\right)\left(-\frac{a'}{b'}\right)=-1.
\displaystyle \frac{aa'}{bb'}=-1.
\displaystyle aa'=-bb'.
\displaystyle \therefore aa'+bb'=0.
\displaystyle \therefore \text{Option (A) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The equation of the line passing through }(1,2)\text{ and perpendicular to}
\displaystyle x+y+7=0\text{ is:}
\displaystyle \text{(A) }y-x+1=0\qquad\text{(B) }y-x-1=0
\displaystyle \text{(C) }y-x+2=0\qquad\text{(D) }y-x-2=0
\displaystyle \text{Answer:}
\displaystyle \text{Given line is }x+y+7=0.
\displaystyle y=-x-7.
\displaystyle \therefore \text{Slope of the given line }=-1.
\displaystyle \text{Slope of a line perpendicular to it }=1.
\displaystyle \text{Using the point-slope form through }(1,2),
\displaystyle y-2=1(x-1).
\displaystyle y-2=x-1.
\displaystyle \therefore y-x-1=0.
\displaystyle \therefore \text{Option (B) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The distance of the point }P(1,-3)\text{ from the line }2y-3x=4\text{ is:}
\displaystyle \text{(A) }13\qquad\text{(B) }\frac{7\sqrt{13}}{13}\qquad\text{(C) }\sqrt{13}\qquad\text{(D) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }2y-3x-4=0.
\displaystyle \text{Distance of }(x_1,y_1)\text{ from }Ax+By+C=0\text{ is}
\displaystyle d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.
\displaystyle \therefore d=\frac{|-3(1)+2(-3)-4|}{\sqrt{(-3)^2+2^2}}.
\displaystyle =\frac{|-3-6-4|}{\sqrt{13}}=\frac{13}{\sqrt{13}}=\sqrt{13}.
\displaystyle \therefore \text{Option (C) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The coordinates of the foot of the perpendicular from the point }(2,3)
\displaystyle \text{on the line }x+y-11=0\text{ are:}
\displaystyle \text{(A) }(-6,5)\qquad\text{(B) }(5,6)\qquad\text{(C) }(-5,6)\qquad\text{(D) }(6,5)
\displaystyle \text{Answer:}
\displaystyle \text{Given line is }x+y-11=0.
\displaystyle \text{Its slope is }-1.
\displaystyle \therefore \text{Slope of the perpendicular line }=1.
\displaystyle \text{Equation of the perpendicular line through }(2,3)\text{ is}
\displaystyle y-3=x-2.
\displaystyle \therefore y=x+1.
\displaystyle \text{Substituting }y=x+1\text{ in }x+y-11=0,
\displaystyle x+x+1-11=0.
\displaystyle 2x=10\Rightarrow x=5.
\displaystyle \therefore y=5+1=6.
\displaystyle \therefore \text{The foot of the perpendicular is }(5,6).
\displaystyle \therefore \text{Option (B) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The intercept cut off by a line from the }y\text{-axis is twice that from the}
\displaystyle x\text{-axis, and the line passes through the point }(1,2).\text{ The equation of the line is:}
\displaystyle \text{(A) }2x+y=4\qquad\text{(B) }2x+y+4=0
\displaystyle \text{(C) }2x-y=4\qquad\text{(D) }2x-y+4=0
\displaystyle \text{Answer:}
\displaystyle \text{Let the }x\text{-intercept be }a.
\displaystyle \therefore \text{The }y\text{-intercept is }2a.
\displaystyle \text{Using the intercept form of a line,}
\displaystyle \frac{x}{a}+\frac{y}{2a}=1.
\displaystyle \text{Since the line passes through }(1,2),
\displaystyle \frac{1}{a}+\frac{2}{2a}=1.
\displaystyle \frac{2}{a}=1\Rightarrow a=2.
\displaystyle \therefore \text{The }x\text{-intercept is }2\text{ and the }y\text{-intercept is }4.
\displaystyle \frac{x}{2}+\frac{y}{4}=1.
\displaystyle \therefore 2x+y=4.
\displaystyle \therefore \text{Option (A) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A line cuts off an intercept }-3\text{ from the }y\text{-axis and the tangent of}
\displaystyle \text{its angle of inclination to the }x\text{-axis is }\frac{3}{5}.\text{ Its equation is:}
\displaystyle \text{(a) }5y-3x+15=0\qquad\text{(b) }3x-5y+15=0
\displaystyle \text{(c) }5y-3x-15=0\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{The slope of the line is }m=\tan\theta=\frac{3}{5}.
\displaystyle \text{The }y\text{-intercept is }c=-3.
\displaystyle \text{Using }y=mx+c,
\displaystyle y=\frac{3}{5}x-3.
\displaystyle 5y=3x-15.
\displaystyle \therefore 5y-3x+15=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The slope of a line which cuts off intercepts of equal lengths on the axes is:}
\displaystyle \text{(a) }-1\qquad\text{(b) }0\qquad\text{(c) }2\qquad\text{(d) }\sqrt{3}
\displaystyle \text{Answer:}
\displaystyle \text{Let the equal intercepts on the }x\text{-axis and }y\text{-axis be }a.
\displaystyle \text{Using the intercept form of a line,}
\displaystyle \frac{x}{a}+\frac{y}{a}=1.
\displaystyle x+y=a.
\displaystyle y=-x+a.
\displaystyle \therefore \text{Slope of the line }=-1.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The equation of the straight line passing through the point }(3,2)\text{ and}
\displaystyle \text{perpendicular to the line }y=x\text{ is:}
\displaystyle \text{(a) }x-y=5\qquad\text{(b) }x+y=5\qquad\text{(c) }x+y=1\qquad\text{(d) }x-y=1
\displaystyle \text{Answer:}
\displaystyle \text{The slope of the line }y=x\text{ is }1.
\displaystyle \therefore \text{Slope of a line perpendicular to it is }-1.
\displaystyle \text{Using the point-slope form through }(3,2),
\displaystyle y-2=-1(x-3).
\displaystyle y-2=-x+3.
\displaystyle \therefore x+y=5.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The equation of the line passing through the point }(1,2)\text{ and}
\displaystyle \text{perpendicular to the line }x+y+1=0\text{ is:}
\displaystyle \text{(a) }y-x+1=0\qquad\text{(b) }y-x-1=0
\displaystyle \text{(c) }y-x+2=0\qquad\text{(d) }y-x-2=0
\displaystyle \text{Answer:}
\displaystyle \text{Given line is }x+y+1=0.
\displaystyle y=-x-1.
\displaystyle \therefore \text{Slope of the given line }=-1.
\displaystyle \therefore \text{Slope of a line perpendicular to it }=1.
\displaystyle \text{Using the point-slope form through }(1,2),
\displaystyle y-2=1(x-1).
\displaystyle y-2=x-1.
\displaystyle \therefore y-x-1=0.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If the line }\frac{x}{a}+\frac{y}{b}=1\text{ passes through the points }(2,-3)\text{ and }(4,-5),
\displaystyle \text{then }(a,b)\text{ is:}
\displaystyle \text{(a) }(1,1)\qquad\text{(b) }(-1,1)\qquad\text{(c) }(1,-1)\qquad\text{(d) }(-1,-1)
\displaystyle \text{Answer:}
\displaystyle \text{Since }(2,-3)\text{ lies on the line,}
\displaystyle \frac{2}{a}-\frac{3}{b}=1.\qquad\ldots(1)
\displaystyle \text{Since }(4,-5)\text{ lies on the line,}
\displaystyle \frac{4}{a}-\frac{5}{b}=1.\qquad\ldots(2)
\displaystyle \text{Multiplying (1) by }2,
\displaystyle \frac{4}{a}-\frac{6}{b}=2.\qquad\ldots(3)
\displaystyle \text{Subtracting (3) from (2),}
\displaystyle \frac{1}{b}=-1.
\displaystyle \therefore b=-1.
\displaystyle \text{Substituting }b=-1\text{ in (1),}
\displaystyle \frac{2}{a}+3=1.
\displaystyle \frac{2}{a}=-2.
\displaystyle \therefore a=-1.
\displaystyle \therefore (a,b)=(-1,-1).
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If the coordinates of the midpoint of the portion of a line intercepted}
\displaystyle \text{between the coordinate axes is }(3,2),\text{ then the equation of the line will be:}
\displaystyle \text{(a) }2x+3y=12\qquad\text{(b) }3x+2y=12
\displaystyle \text{(c) }4x-3y=6\qquad\text{(d) }5x-2y=10
\displaystyle \text{Answer:}
\displaystyle \text{Let the }x\text{-intercept and }y\text{-intercept of the line be }a\text{ and }b\text{ respectively.}
\displaystyle \therefore \text{The intercept points are }(a,0)\text{ and }(0,b).
\displaystyle \text{The midpoint of the portion intercepted between the axes is}
\displaystyle \left(\frac{a}{2},\frac{b}{2}\right)=(3,2).
\displaystyle \therefore \frac{a}{2}=3\Rightarrow a=6,\qquad\frac{b}{2}=2\Rightarrow b=4.
\displaystyle \text{Using the intercept form of a line,}
\displaystyle \frac{x}{6}+\frac{y}{4}=1.
\displaystyle \therefore 2x+3y=12.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The equation of the line passing through }(1,2)\text{ and parallel to the line}
\displaystyle y=3x-1\text{ is:}
\displaystyle \text{(a) }y+2=x+1\qquad\text{(b) }y+2=3(x+1)
\displaystyle \text{(c) }y-2=3(x-1)\qquad\text{(d) }y-2=x-1
\displaystyle \text{Answer:}
\displaystyle \text{The slope of the line }y=3x-1\text{ is }3.
\displaystyle \text{Since parallel lines have equal slopes, the slope of the required line is }3.
\displaystyle \text{Using the point-slope form through }(1,2),
\displaystyle y-2=3(x-1).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The point }(4,1)\text{ undergoes the following two successive transformations:}
\displaystyle \text{(i) Reflection about the line }y=x
\displaystyle \text{(ii) Translation through a distance }2\text{ units along the positive }x\text{-axis.}
\displaystyle \text{Then the final coordinates of the point are:}
\displaystyle \text{(a) }(4,3)\qquad\text{(b) }(3,4)\qquad\text{(c) }(1,4)\qquad\text{(d) }\left(\frac{7}{2},\frac{7}{2}\right)
\displaystyle \text{Answer:}
\displaystyle \text{Reflection about the line }y=x\text{ interchanges the }x\text{ and }y\text{ coordinates.}
\displaystyle \therefore (4,1)\rightarrow(1,4).
\displaystyle \text{Translation by }2\text{ units along the positive }x\text{-axis gives}
\displaystyle (1,4)\rightarrow(1+2,4)=(3,4).
\displaystyle \therefore \text{The final coordinates of the point are }(3,4).
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A point equidistant from the lines }4x+3y+10=0,\ 5x-12y+26=0
\displaystyle \text{and }x+24y-50=0\text{ is:}
\displaystyle \text{(a) }(1,-1)\qquad\text{(b) }(1,1)\qquad\text{(c) }(0,0)\qquad\text{(d) }(0,1)
\displaystyle \text{Answer:}
\displaystyle \text{For the point }(0,0),\text{ distance from }4x+3y+10=0\text{ is}
\displaystyle d_1=\frac{|4(0)+3(0)+10|}{\sqrt{4^2+3^2}}=\frac{10}{5}=2.
\displaystyle \text{Distance from }5x-12y+26=0\text{ is}
\displaystyle d_2=\frac{|5(0)-12(0)+26|}{\sqrt{5^2+(-12)^2}}=\frac{26}{13}=2.
\displaystyle \text{Distance from }x+24y-50=0\text{ is}
\displaystyle d_3=\frac{|0+24(0)-50|}{\sqrt{1^2+24^2}}=\frac{50}{25}=2.
\displaystyle \therefore d_1=d_2=d_3=2.
\displaystyle \therefore (0,0)\text{ is equidistant from all three lines.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{A line passes through }(2,2)\text{ and is perpendicular to the line }3x+y=3.
\displaystyle \text{Its }y\text{-intercept is:}
\displaystyle \text{(a) }\frac{1}{3}\qquad\text{(b) }\frac{2}{3}\qquad\text{(c) }1\qquad\text{(d) }\frac{4}{3}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }3x+y=3.
\displaystyle y=-3x+3.
\displaystyle \therefore \text{Slope of the given line }=-3.
\displaystyle \therefore \text{Slope of a line perpendicular to it }=\frac{1}{3}.
\displaystyle \text{Using the point-slope form through }(2,2),
\displaystyle y-2=\frac{1}{3}(x-2).
\displaystyle 3y-6=x-2.
\displaystyle \therefore y=\frac{1}{3}x+\frac{4}{3}.
\displaystyle \therefore \text{The }y\text{-intercept is }\frac{4}{3}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The ratio in which the line }3x+4y+2=0\text{ divides the perpendicular}
\displaystyle \text{distance between the lines }3x+4y+5=0\text{ and }3x+4y-5=0\text{ is:}
\displaystyle \text{(a) }1:2\qquad\text{(b) }3:7\qquad\text{(c) }2:3\qquad\text{(d) }2:5
\displaystyle \text{Answer:}
\displaystyle \text{The three lines are parallel since their coefficients of }x\text{ and }y\text{ are the same.}
\displaystyle \text{Distance between }3x+4y+5=0\text{ and }3x+4y+2=0\text{ is}
\displaystyle d_1=\frac{|5-2|}{\sqrt{3^2+4^2}}=\frac{3}{5}.
\displaystyle \text{Distance between }3x+4y+2=0\text{ and }3x+4y-5=0\text{ is}
\displaystyle d_2=\frac{|2-(-5)|}{\sqrt{3^2+4^2}}=\frac{7}{5}.
\displaystyle \therefore d_1:d_2=\frac{3}{5}:\frac{7}{5}=3:7.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{One vertex of an equilateral triangle with centroid at the origin and}
\displaystyle \text{one side }x+y-2=0\text{ is:}
\displaystyle \text{(a) }(-1,-1)\qquad\text{(b) }(2,2)\qquad\text{(c) }(-2,-2)\qquad\text{(d) }(2,-2)
\displaystyle \text{Answer:}
\displaystyle \text{The centroid is }O(0,0)\text{ and the given side is }x+y-2=0.
\displaystyle \text{Distance of }O\text{ from the given side is}
\displaystyle d=\frac{|0+0-2|}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt2}=\sqrt2.
\displaystyle \text{The perpendicular from }O\text{ to }x+y-2=0\text{ has equation }y=x.
\displaystyle \text{Solving }y=x\text{ and }x+y=2,\text{ the foot of the perpendicular is }(1,1).
\displaystyle \text{In an equilateral triangle, the centroid divides a median in the ratio }2:1.
\displaystyle \therefore \text{The opposite vertex lies twice as far from the centroid as the side,}
\displaystyle \text{and in the direction opposite to }(1,1).
\displaystyle \therefore \text{The required vertex is }(-2,-2).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The equations of the diagonals of the square formed by the lines }x=0,\ y=0,
\displaystyle x=1\text{ and }y=1\text{ are:}
\displaystyle \text{(A) }y=x,\ y+x=1\qquad\text{(B) }y=x,\ x+y=2
\displaystyle \text{(C) }2y=x,\ y+x=\frac{1}{3}\qquad\text{(D) }y=2x,\ y+2x=1
\displaystyle \text{Answer:}
\displaystyle \text{The vertices of the square are }(0,0),(1,0),(1,1)\text{ and }(0,1).
\displaystyle \text{The diagonal joining }(0,0)\text{ and }(1,1)\text{ has equation }y=x.
\displaystyle \text{The other diagonal joins }(0,1)\text{ and }(1,0).
\displaystyle \text{Its equation is }x+y=1.
\displaystyle \therefore \text{The diagonals are }y=x\text{ and }x+y=1.
\displaystyle \therefore \text{Option (A) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{For specifying a straight line, how many geometrical parameters should be}
\displaystyle \text{known?}
\displaystyle \text{(A) }1\qquad\text{(B) }2\qquad\text{(C) }4\qquad\text{(D) }3
\displaystyle \text{Answer:}
\displaystyle \text{A straight line is uniquely determined by two independent geometrical parameters.}
\displaystyle \text{For example, a line can be specified by a point on it and its slope.}
\displaystyle \therefore \text{Two geometrical parameters are required to specify a straight line.}
\displaystyle \therefore \text{Option (B) is correct.}
\displaystyle \\

\displaystyle \text{CASE BASED/SOURCE BASED / PASSAGE BASED}

\displaystyle \textbf{Question 21: }\text{Read the case study given below:}
\displaystyle \text{Neeraj's house is }1\text{ km east of the origin }O(0,0).\text{ While going to school, he first}
\displaystyle \text{takes an auto from his house }A(1,0)\text{ to the hospital }B(4,4).\text{ From the hospital,}
\displaystyle \text{he travels by city bus to the church }C(4,8).\text{ From the church, he takes a metro}
\displaystyle \text{train and reaches the school at }D(-5,8).\text{ All distances are in km.}
\displaystyle \text{Now answer the following questions:}
\displaystyle \text{(i) What is the slope of Neeraj's journey from home to hospital?}
\displaystyle \text{(a) }\frac{3}{4}\qquad\text{(b) }\frac{4}{3}\qquad\text{(c) }\frac{4}{5}\qquad\text{(d) }\frac{5}{4}
\displaystyle \text{Answer:}
\displaystyle \text{Home is at }A(1,0)\text{ and hospital is at }B(4,4).
\displaystyle \text{Slope of }AB=\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{4-1}=\frac{4}{3}.
\displaystyle \therefore \text{Option (b) is correct.}

\displaystyle \text{(ii) What is the distance of the school from the hospital?}
\displaystyle \text{(a) }\sqrt{97}\text{ km}\qquad\text{(b) }10\text{ km}\qquad\text{(c) }\sqrt{145}\text{ km}\qquad\text{(d) }12\text{ km}
\displaystyle \text{Answer:}
\displaystyle \text{Hospital is at }B(4,4)\text{ and school is at }D(-5,8).
\displaystyle BD=\sqrt{(-5-4)^2+(8-4)^2}.
\displaystyle =\sqrt{(-9)^2+4^2}=\sqrt{81+16}=\sqrt{97}\text{ km}.
\displaystyle \therefore \text{Option (a) is correct.}

\displaystyle \text{(iii) What is the equation of the straight line joining the points }A\text{ and }D\text{?}
\displaystyle \text{(a) }4x-3y=4\qquad\text{(b) }5x+4y=10
\displaystyle \text{(c) }4x+3y=4\qquad\text{(d) }6x+7y=-15
\displaystyle \text{Answer:}
\displaystyle \text{The points are }A(1,0)\text{ and }D(-5,8).
\displaystyle \text{Slope of }AD=\frac{8-0}{-5-1}=\frac{8}{-6}=-\frac{4}{3}.
\displaystyle \text{Using the point-slope form through }A(1,0),
\displaystyle y-0=-\frac{4}{3}(x-1).
\displaystyle 3y=-4x+4.
\displaystyle \therefore 4x+3y=4.
\displaystyle \therefore \text{Option (c) is correct.}

\displaystyle \text{(iv) What is the equation of the straight line joining the church and hospital?}
\displaystyle \text{(a) }y=4\qquad\text{(b) }5x+4y=10\qquad\text{(c) }x=-4\qquad\text{(d) }x=4
\displaystyle \text{Answer:}
\displaystyle \text{Church is at }C(4,8)\text{ and hospital is at }B(4,4).
\displaystyle \text{Both points have the same }x\text{-coordinate }4.
\displaystyle \therefore \text{The line joining them is }x=4.
\displaystyle \therefore \text{Option (d) is correct.}

\displaystyle \text{(v) What is the equation of the straight line joining the points }A\text{ and }C\text{?}
\displaystyle \text{(a) }4x-3y=4\qquad\text{(b) }8x-3y=8
\displaystyle \text{(c) }4x+3y=4\qquad\text{(d) }6x+7y=-15
\displaystyle \text{Answer:}
\displaystyle \text{The points are }A(1,0)\text{ and }C(4,8).
\displaystyle \text{Slope of }AC=\frac{8-0}{4-1}=\frac{8}{3}.
\displaystyle \text{Using the point-slope form through }A(1,0),
\displaystyle y-0=\frac{8}{3}(x-1).
\displaystyle 3y=8x-8.
\displaystyle \therefore 8x-3y=8.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Read the case study given below and attempt any four subparts:}
\displaystyle \text{A surveyor was measuring the width of a river. For this, he selected a tree at }Y\text{ on the}
\displaystyle \text{other side of the river. He is standing at the point }O(0,0).\text{ From }O,\text{ he walks }50\text{ m}
\displaystyle \text{along the positive }x\text{-axis to point }B,\text{ where he fixes a stick. From }B,\text{ he walks }20\text{ m}
\displaystyle \text{further along the positive }x\text{-axis to point }C,\text{ where he fixes another stick.}
\displaystyle \text{From }C,\text{ he walks perpendicular to }OC\text{ and fixes a stick at }D,\text{ such that }CD=30\text{ m}.
\displaystyle \text{The points }Y,B\text{ and }D\text{ are approximately collinear. Assume }OC\text{ is the }x\text{-axis}
\displaystyle \text{and }OY\text{ is the }y\text{-axis.}
\displaystyle \text{(i) What are the coordinates of point }D\text{?}
\displaystyle \text{(a) }(50,30)\qquad\text{(b) }(70,-30)\qquad\text{(c) }(50,20)\qquad\text{(d) }(70,30)
\displaystyle \text{Answer:}
\displaystyle \text{Since }OB=50\text{ m and }BC=20\text{ m,}
\displaystyle OC=50+20=70\text{ m}.
\displaystyle \therefore C=(70,0).
\displaystyle \text{Since }CD=30\text{ m perpendicular to the }x\text{-axis in the downward direction,}
\displaystyle D=(70,-30).
\displaystyle \therefore \text{Option (b) is correct.}

\displaystyle \text{(ii) What are the coordinates of point }C\text{?}
\displaystyle \text{Answer:}
\displaystyle OB=50\text{ m and }BC=20\text{ m}.
\displaystyle \therefore OC=50+20=70\text{ m}.
\displaystyle \text{Since }C\text{ lies on the }x\text{-axis, its }y\text{-coordinate is }0.
\displaystyle \therefore C=(70,0).
\displaystyle \therefore \text{None of the printed options is correct.}

\displaystyle \text{(iv) What are the coordinates of point }Y\text{?}
\displaystyle \text{(a) }(0,100)\qquad\text{(b) }(0,-70)\qquad\text{(c) }(75,0)\qquad\text{(d) }(0,75)
\displaystyle \text{Answer:}
\displaystyle B=(50,0)\text{ and }D=(70,-30).
\displaystyle \text{Slope of }BD=\frac{-30-0}{70-50}=-\frac{30}{20}=-\frac{3}{2}.
\displaystyle \text{Using the point-slope form through }B(50,0),
\displaystyle y-0=-\frac{3}{2}(x-50).
\displaystyle 2y=-3x+150.
\displaystyle \therefore 3x+2y=150.
\displaystyle \text{Since }Y\text{ lies on the }y\text{-axis, }x=0.
\displaystyle 2y=150\Rightarrow y=75.
\displaystyle \therefore Y=(0,75).
\displaystyle \therefore \text{Option (d) is correct.}

\displaystyle \text{(v) What is the equation of the straight line }BD\text{?}
\displaystyle \text{(a) }3x+2y=150\qquad\text{(b) }3x+2y=100
\displaystyle \text{(c) }5x+2y=150\qquad\text{(d) }5x-3y=150
\displaystyle \text{Answer:}
\displaystyle B=(50,0)\text{ and }D=(70,-30).
\displaystyle \text{Slope of }BD=\frac{-30-0}{70-50}=-\frac{3}{2}.
\displaystyle \text{Using the point-slope form through }B(50,0),
\displaystyle y=-\frac{3}{2}(x-50).
\displaystyle 2y=-3x+150.
\displaystyle \therefore 3x+2y=150.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Read the case study given below and attempt any four subparts:}
\displaystyle \text{In a colony, as shown in the figure, an electric pole has been installed. The pole is}
\displaystyle \text{supported by a strong wire }PQ,\text{ and some electricians are working on staircase }PS.
\displaystyle \text{On the left and right sides of the pole, two street lights are fixed at heights of }3\text{ m}
\displaystyle \text{and }4\text{ m respectively. These lights are supplied by wires }PN\text{ and }PM.
\displaystyle \text{The height }OP=5\text{ m and }O\text{ is the origin. Answer the following questions:}
\displaystyle \text{(i) What are the coordinates of point }P\text{?}
\displaystyle \text{(a) }(5,0)\qquad\text{(b) }(0,5)\qquad\text{(c) }(0,-5)\qquad\text{(d) }(0,10)
\displaystyle \text{Answer:}
\displaystyle \text{Since }O=(0,0)\text{ and }OP=5\text{ m along the positive }y\text{-axis,}
\displaystyle P=(0,5).
\displaystyle \therefore \text{Option (b) is correct.}

\displaystyle \text{(ii) What is the length of the staircase }PS\text{?}
\displaystyle \text{(a) }12\text{ m}\qquad\text{(b) }15\text{ m}\qquad\text{(c) }13\text{ m}\qquad\text{(d) }20\text{ m}
\displaystyle \text{Answer:}
\displaystyle P=(0,5)\text{ and }S=(-12,0).
\displaystyle PS=\sqrt{(0+12)^2+(5-0)^2}.
\displaystyle =\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{ m}.
\displaystyle \therefore \text{Option (c) is correct.}

\displaystyle \text{(iii) What is the area of }\triangle OPQ\text{?}
\displaystyle \text{(a) }12\text{ m}^2\qquad\text{(b) }15\text{ m}^2\qquad\text{(c) }13\text{ m}^2\qquad\text{(d) }20\text{ m}^2
\displaystyle \text{Answer:}
\displaystyle OP=5\text{ m and }OQ=8\text{ m}.
\displaystyle \text{Since }OP\perp OQ,
\displaystyle \text{Area of }\triangle OPQ=\frac{1}{2}\times OP\times OQ.
\displaystyle =\frac{1}{2}\times5\times8=20\text{ m}^2.
\displaystyle \therefore \text{Option (d) is correct.}

\displaystyle \text{(iv) What is the equation of the line }PN\text{?}
\displaystyle \text{(a) }x+15y=-75\qquad\text{(b) }15x-2y+6=0
\displaystyle \text{(c) }x+5y=50\qquad\text{(d) }x-5y=20
\displaystyle \text{Answer:}
\displaystyle P=(0,5).
\displaystyle \text{Since }OS=12\text{ m and the left light is }3\text{ m further left,}
\displaystyle N=(-15,3).
\displaystyle \text{Slope of }PN=\frac{5-3}{0-(-15)}=\frac{2}{15}.
\displaystyle \text{Using the point-slope form through }P(0,5),
\displaystyle y-5=\frac{2}{15}(x-0).
\displaystyle 15y-75=2x.
\displaystyle \therefore 2x-15y+75=0.
\displaystyle \therefore \text{None of the printed options is correct.}

\displaystyle \text{(v) What is the length of wire }PM\text{?}
\displaystyle \text{(a) }\sqrt{101}\text{ m}\qquad\text{(b) }\sqrt{26}\text{ m}\qquad\text{(c) }26\text{ m}\qquad\text{(d) }25\text{ m}
\displaystyle \text{Answer:}
\displaystyle P=(0,5).
\displaystyle \text{Since }OQ=8\text{ m and }QR=2\text{ m, }R=(10,0).
\displaystyle \text{The right street light is }4\text{ m high, so }M=(10,4).
\displaystyle PM=\sqrt{(10-0)^2+(4-5)^2}.
\displaystyle =\sqrt{100+1}=\sqrt{101}\text{ m}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Read the case study given below and attempt any four subparts:}
\displaystyle \text{In a park, Road 1 and Road 2 of widths }5\text{ m and }4\text{ m respectively cross at the}
\displaystyle \text{origin }O(0,0),\text{ as shown in the figure. The coordinates of trees }A,B,C\text{ and }D
\displaystyle \text{are }(6,8),(12,5),(-5,0)\text{ and }(-3,4)\text{ respectively. Answer the following:}
\displaystyle \text{(i) What is the distance of Tree }C\text{ from the origin?}
\displaystyle \text{(a) }5\text{ m}\qquad\text{(b) }10\text{ m}\qquad\text{(c) }15\text{ m}\qquad\text{(d) }25\text{ m}
\displaystyle \text{Answer:}
\displaystyle O=(0,0)\text{ and }C=(-5,0).
\displaystyle OC=\sqrt{(-5-0)^2+(0-0)^2}=\sqrt{25}=5\text{ m}.
\displaystyle \therefore \text{Option (a) is correct.}

\displaystyle \text{(ii) What is the equation of line }AB\text{?}
\displaystyle \text{(a) }2x+y=22\qquad\text{(b) }x-2y=-6
\displaystyle \text{(c) }x+2y-22=0\qquad\text{(d) }x+2y=6
\displaystyle \text{Answer:}
\displaystyle A=(6,8)\text{ and }B=(12,5).
\displaystyle \text{Slope of }AB=\frac{5-8}{12-6}=-\frac{3}{6}=-\frac{1}{2}.
\displaystyle \text{Using the point-slope form through }A(6,8),
\displaystyle y-8=-\frac{1}{2}(x-6).
\displaystyle 2y-16=-x+6.
\displaystyle \therefore x+2y-22=0.
\displaystyle \therefore \text{Option (c) is correct.}

\displaystyle \text{(iii) What is the slope of line }CD\text{?}
\displaystyle \text{(a) }2\qquad\text{(b) }\frac{1}{2}\qquad\text{(c) }-\frac{1}{2}\qquad\text{(d) }\frac{3}{2}
\displaystyle \text{Answer:}
\displaystyle C=(-5,0)\text{ and }D=(-3,4).
\displaystyle \text{Slope of }CD=\frac{4-0}{-3-(-5)}=\frac{4}{2}=2.
\displaystyle \therefore \text{Option (a) is correct.}

\displaystyle \text{(iv) What is the slope of line }OA\text{?}
\displaystyle \text{(a) }\frac{3}{4}\qquad\text{(b) }1\qquad\text{(c) }\frac{4}{3}\qquad\text{(d) }\frac{6}{8}
\displaystyle \text{Answer:}
\displaystyle O=(0,0)\text{ and }A=(6,8).
\displaystyle \text{Slope of }OA=\frac{8-0}{6-0}=\frac{8}{6}=\frac{4}{3}.
\displaystyle \therefore \text{Option (c) is correct.}

\displaystyle \text{(v) What is the distance of point }B\text{ from the origin?}
\displaystyle \text{(a) }13\text{ m}\qquad\text{(b) }15\text{ m}\qquad\text{(c) }12\text{ m}\qquad\text{(d) }5\text{ m}
\displaystyle \text{Answer:}
\displaystyle O=(0,0)\text{ and }B=(12,5).
\displaystyle OB=\sqrt{(12-0)^2+(5-0)^2}.
\displaystyle =\sqrt{144+25}=\sqrt{169}=13\text{ m}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Read the case study given below and attempt any four subparts:}
\displaystyle \text{The villages of Shanu and Arun are }50\text{ km apart and are situated on the Delhi-Agra}
\displaystyle \text{highway, as shown in the figure. Another highway }YY'\text{ crosses the Delhi-Agra highway}
\displaystyle \text{at }O(0,0).\text{ A small local road }PQ\text{ crosses the two highways at }A\text{ and }B\text{ such that}
\displaystyle OA=10\text{ km and }OB=12\text{ km. The villages of Barun and Jeetu are situated on}
\displaystyle \text{the highway }YY'.\text{ Barun's village is }12\text{ km from }O\text{ and Jeetu's village is }15\text{ km}
\displaystyle \text{from }O.\text{ Now answer the following questions:}
\displaystyle \text{(i) What are the coordinates of }A\text{?}
\displaystyle \text{(a) }(10,0)\qquad\text{(b) }(10,12)\qquad\text{(c) }(0,10)\qquad\text{(d) }(0,15)
\displaystyle \text{Answer:}
\displaystyle O=(0,0)\text{ and }OA=10\text{ km along the positive }x\text{-axis}.
\displaystyle \therefore A=(10,0).
\displaystyle \therefore \text{Option (a) is correct.}

\displaystyle \text{(ii) What is the equation of line }AB\text{?}
\displaystyle \text{(a) }5x+6y=60\qquad\text{(b) }6x+5y=60
\displaystyle \text{(c) }x=10\qquad\text{(d) }y=12
\displaystyle \text{Answer:}
\displaystyle A=(10,0)\text{ and }B=(0,12).
\displaystyle \text{Using the intercept form of a line,}
\displaystyle \frac{x}{10}+\frac{y}{12}=1.
\displaystyle 6x+5y=60.
\displaystyle \therefore \text{Option (b) is correct.}

\displaystyle \text{(iii) What is the perpendicular distance of the line }AB\text{ from }O(0,0)\text{?}
\displaystyle \text{(a) }60\text{ km}\qquad\text{(b) }\frac{60}{\sqrt{61}}\text{ km}\qquad\text{(c) }\sqrt{61}\text{ km}\qquad\text{(d) }6\text{ km}
\displaystyle \text{Answer:}
\displaystyle \text{Equation of }AB\text{ is }6x+5y-60=0.
\displaystyle \text{Distance of }(x_1,y_1)\text{ from }Ax+By+C=0\text{ is}
\displaystyle d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.
\displaystyle \therefore d=\frac{|6(0)+5(0)-60|}{\sqrt{6^2+5^2}}.
\displaystyle =\frac{60}{\sqrt{61}}\text{ km}.
\displaystyle \therefore \text{Option (b) is correct.}

\displaystyle \text{(iv) What is the slope of line }AB\text{?}
\displaystyle \text{(a) }\frac{6}{5}\qquad\text{(b) }\frac{5}{6}\qquad\text{(c) }-\frac{6}{5}\qquad\text{(d) }\frac{10}{12}
\displaystyle \text{Answer:}
\displaystyle A=(10,0)\text{ and }B=(0,12).
\displaystyle \text{Slope of }AB=\frac{12-0}{0-10}=\frac{12}{-10}=-\frac{6}{5}.
\displaystyle \therefore \text{Option (c) is correct.}

\displaystyle \text{(v) What is the length of line segment }AB\text{?}
\displaystyle \text{(a) }\sqrt{61}\text{ km}\qquad\text{(b) }12\text{ km}\qquad\text{(c) }10\text{ km}\qquad\text{(d) }2\sqrt{61}\text{ km}
\displaystyle \text{Answer:}
\displaystyle A=(10,0)\text{ and }B=(0,12).
\displaystyle AB=\sqrt{(0-10)^2+(12-0)^2}.
\displaystyle =\sqrt{100+144}=\sqrt{244}=2\sqrt{61}\text{ km}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \text{ASSERTION REASONING}

\displaystyle \textbf{Question 26: }\text{Assertion (A): The point }(3,0)\text{ is at a distance of }3\text{ units from the}
\displaystyle y\text{-axis, measured along the positive }x\text{-axis, and has zero distance from the }x\text{-axis.}
\displaystyle \text{Reason (R): The point }(3,0)\text{ is at a distance of }3\text{ units from the }x\text{-axis, measured}
\displaystyle \text{along the positive }y\text{-axis, and has zero distance from the }y\text{-axis.}

\displaystyle \text{Select the correct option from the following:}
\displaystyle \text{(a) A is true, R is true; R is the correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not the correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{For the point }(3,0),\text{ distance from the }y\text{-axis}=|3|=3\text{ units.}
\displaystyle \text{Also, distance from the }x\text{-axis}=|0|=0.
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{The Reason states that the point is }3\text{ units from the }x\text{-axis and }0\text{ units from the }y\text{-axis.}
\displaystyle \text{This is incorrect.}
\displaystyle \therefore \text{Reason (R) is false.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Assertion (A): If }x\cos q+y\sin q=2\text{ is perpendicular to the line}
\displaystyle x-y=3,\text{ then one of the values of }q\text{ is }\frac{\pi}{4}.
\displaystyle \text{Reason (R): If two lines }y=m_1x+c_1\text{ and }y=m_2x+c_2\text{ are perpendicular,}
\displaystyle \text{then }m_1=m_2.

\displaystyle \text{Select the correct option from the following:}
\displaystyle \text{(a) A is true, R is true; R is the correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not the correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle x\cos q+y\sin q=2.
\displaystyle y=-\frac{\cos q}{\sin q}x+\frac{2}{\sin q}.
\displaystyle \therefore m_1=-\cot q.
\displaystyle x-y=3\Rightarrow y=x-3.
\displaystyle \therefore m_2=1.
\displaystyle \text{For perpendicular lines, }m_1m_2=-1.
\displaystyle (-\cot q)(1)=-1.
\displaystyle \therefore \cot q=1\Rightarrow\tan q=1.
\displaystyle \therefore q=\frac{\pi}{4}+n\pi,\quad n\in\mathbb{Z}.
\displaystyle \therefore \frac{\pi}{4}\text{ is one of the possible values of }q.
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{For perpendicular lines, }m_1m_2=-1,\text{ not }m_1=m_2.
\displaystyle \therefore \text{Reason (R) is false.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Assertion (A): The slope of the }x\text{-axis is zero and the slope of the}
\displaystyle y\text{-axis is not defined.}
\displaystyle \text{Reason (R): The slope of the }x\text{-axis is not defined and the slope of the }y\text{-axis is zero.}

\displaystyle \text{Select the correct option from the following:}
\displaystyle \text{(a) A is true, R is true; R is the correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not the correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{The }x\text{-axis is a horizontal line.}
\displaystyle \therefore \text{Slope of the }x\text{-axis}=0.
\displaystyle \text{The }y\text{-axis is a vertical line.}
\displaystyle \therefore \text{Slope of the }y\text{-axis is not defined.}
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{The Reason reverses the slopes of the }x\text{-axis and }y\text{-axis.}
\displaystyle \therefore \text{Reason (R) is false.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Assertion (A): The slope of the line }x+7y=0\text{ is }\frac{1}{7}\text{ and its}
\displaystyle y\text{-intercept is }0.
\displaystyle \text{Reason (R): The slope of the line }6x+3y-5=0\text{ is }-2\text{ and its }y\text{-intercept}
\displaystyle \text{is }\frac{5}{3}.

\displaystyle \text{Select the correct option from the following:}
\displaystyle \text{(a) A is true, R is true; R is the correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not the correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle x+7y=0.
\displaystyle y=-\frac{1}{7}x.
\displaystyle \therefore \text{Slope}=-\frac{1}{7}\text{ and }y\text{-intercept}=0.
\displaystyle \therefore \text{Assertion (A) is false.}
\displaystyle 6x+3y-5=0.
\displaystyle 3y=-6x+5.
\displaystyle y=-2x+\frac{5}{3}.
\displaystyle \therefore \text{Slope}=-2\text{ and }y\text{-intercept}=\frac{5}{3}.
\displaystyle \therefore \text{Reason (R) is true.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If }A(-2,-1),\ B(4,0),\ C(3,3)\text{ and }D(-3,2)\text{ are the vertices}
\displaystyle \text{of a parallelogram, then}
\displaystyle \text{Assertion (A): Slope of }AB=\text{Slope of }BC\text{ and slope of }CD=\text{Slope of }AD.
\displaystyle \text{Reason (R): Midpoint of }AC=\text{Midpoint of }BD.

\displaystyle \text{Select the correct option from the following:}
\displaystyle \text{(a) A is true, R is true; R is the correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not the correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{Slope of }AB=\frac{0-(-1)}{4-(-2)}=\frac{1}{6}.
\displaystyle \text{Slope of }BC=\frac{3-0}{3-4}=-3.
\displaystyle \therefore \text{Slope of }AB\ne\text{Slope of }BC.
\displaystyle \therefore \text{Assertion (A) is false.}
\displaystyle \text{Midpoint of }AC=\left(\frac{-2+3}{2},\frac{-1+3}{2}\right)=\left(\frac{1}{2},1\right).
\displaystyle \text{Midpoint of }BD=\left(\frac{4+(-3)}{2},\frac{0+2}{2}\right)=\left(\frac{1}{2},1\right).
\displaystyle \therefore \text{Midpoint of }AC=\text{Midpoint of }BD.
\displaystyle \therefore \text{Reason (R) is true.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 31: }\text{Match the questions given under Column C1 with their appropriate}
\displaystyle \text{answers given under Column C2.}

\displaystyle \begin{array}{|l|l|} \hline \text{\textbf{Column 1}} & \text{\textbf{Column 2}} \\ \hline \text{(a) The coordinates of the points }P\text{ and }Q\text{ on the line } \\ x+5y=13 \text{ which are at a distance of }2\text{ units from the line } \\ 12x-5y+26=0\text{ are} & \text{(i) }(3,1),\,(-7,11) \\ \hline \text{(b) The coordinates of the points on the line }x+y=4 \\ \text{ which are at a unit distance} \text{ from the line }4x+3y-10=0\text{ are} & \text{(ii) }\left(-\frac{1}{3},\frac{11}{3}\right),\, \left(\frac{4}{3},\frac{7}{3}\right) \\ \hline \text{(c) The coordinates of the points }P\text{ and }Q\text{ on the line joining } \\ A(-2,5) \text{ and }B(3,1),\text{ such that }AP=PQ=QB,\text{ are} & \text{(iii) }\left(1,\frac{12}{5}\right),\, \left(-3,\frac{16}{5}\right) \\ \hline \end{array}

\displaystyle \text{Answer:}
\displaystyle \text{For (a), the point }(x,y)\text{ lies on }x+5y=13.
\displaystyle \therefore x=13-5y.
\displaystyle \frac{|12x-5y+26|}{\sqrt{12^2+(-5)^2}}=2.
\displaystyle |12x-5y+26|=26.
\displaystyle |12(13-5y)-5y+26|=26.
\displaystyle |182-65y|=26.
\displaystyle 182-65y=26\quad\text{or}\quad182-65y=-26.
\displaystyle y=\frac{12}{5}\quad\text{or}\quad y=\frac{16}{5}.
\displaystyle \therefore (x,y)=\left(1,\frac{12}{5}\right)\text{ or }\left(-3,\frac{16}{5}\right).
\displaystyle \therefore \text{(a) matches (iii).}

\displaystyle \text{For (b), }x+y=4.
\displaystyle \text{For }(3,1),\quad\frac{|4(3)+3(1)-10|}{\sqrt{4^2+3^2}}=\frac{5}{5}=1.
\displaystyle \text{For }(-7,11),\quad\frac{|4(-7)+3(11)-10|}{5}=\frac{5}{5}=1.
\displaystyle \therefore \text{Both points are at unit distance from }4x+3y-10=0.
\displaystyle \therefore \text{(b) matches (i).}

\displaystyle \text{For (c), }A(-2,5)\text{ and }B(3,1),\text{ with }AP=PQ=QB.
\displaystyle \text{Thus, }P\text{ and }Q\text{ trisect the line segment }AB.
\displaystyle P=\left(\frac{2(-2)+3}{3},\frac{2(5)+1}{3}\right)=\left(-\frac{1}{3},\frac{11}{3}\right).
\displaystyle Q=\left(\frac{-2+2(3)}{3},\frac{5+2(1)}{3}\right)=\left(\frac{4}{3},\frac{7}{3}\right).
\displaystyle \therefore \text{(c) matches (ii).}
\displaystyle \therefore \text{The correct matching is (a)-(iii), (b)-(i), (c)-(ii).}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{The value of }\lambda,\text{ if the lines }(2x+3y+4)+\lambda(6x-y+12)=0
\displaystyle \text{are:}

\displaystyle \begin{array}{|l|l|} \hline \text{\textbf{Column 1}} & \text{\textbf{Column 2}} \\ \hline \text{(a) Parallel to the }y\text{-axis} & \text{(i) }\lambda=-\frac{3}{4} \\ \hline \text{(b) Perpendicular to }7x+y-4=0 & \text{(ii) }\lambda=-\frac{1}{3} \\ \hline \text{(c) Passes through }(1,2) & \text{(iii) }\lambda=-\frac{17}{41} \\ \hline \text{(d) Parallel to the }x\text{-axis} & \text{(iv) }\lambda=3 \\ \hline \end{array}

\displaystyle \text{Answer:}
\displaystyle (2x+3y+4)+\lambda(6x-y+12)=0.
\displaystyle (2+6\lambda)x+(3-\lambda)y+(4+12\lambda)=0.

\displaystyle \text{For (a), the line is parallel to the }y\text{-axis.}
\displaystyle \therefore \text{Coefficient of }y=0.
\displaystyle 3-\lambda=0.
\displaystyle \therefore \lambda=3.
\displaystyle \therefore \text{(a) matches (iv).}

\displaystyle \text{For (b), the line is perpendicular to }7x+y-4=0.
\displaystyle 7x+y-4=0\Rightarrow y=-7x+4.
\displaystyle \therefore \text{Slope of the given line}=-7.
\displaystyle \therefore \text{Slope of the perpendicular line}=\frac{1}{7}.
\displaystyle -\frac{2+6\lambda}{3-\lambda}=\frac{1}{7}.
\displaystyle -14-42\lambda=3-\lambda.
\displaystyle -41\lambda=17.
\displaystyle \therefore \lambda=-\frac{17}{41}.
\displaystyle \therefore \text{(b) matches (iii).}

\displaystyle \text{For (c), the line passes through }(1,2).
\displaystyle (2+6\lambda)(1)+(3-\lambda)(2)+(4+12\lambda)=0.
\displaystyle 2+6\lambda+6-2\lambda+4+12\lambda=0.
\displaystyle 12+16\lambda=0.
\displaystyle \therefore \lambda=-\frac{3}{4}.
\displaystyle \therefore \text{(c) matches (i).}

\displaystyle \text{For (d), the line is parallel to the }x\text{-axis.}
\displaystyle \therefore \text{Coefficient of }x=0.
\displaystyle 2+6\lambda=0.
\displaystyle \therefore \lambda=-\frac{1}{3}.
\displaystyle \therefore \text{(d) matches (ii).}
\displaystyle \therefore \text{The correct matching is (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii).}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{The equation of the line through the intersection of the lines }
\displaystyle 2x-3y=0 \text{and }4x-5y=2\text{ and satisfying the following conditions is:}

\displaystyle \begin{array}{|l|l|} \hline \text{\textbf{Column 1}} & \text{\textbf{Column 2}} \\ \hline \text{(a) Through the point }(2,1) & \text{(i) }2x-y=4 \\ \hline \text{(b) Perpendicular to the line }x+2y+1=0 & \text{(ii) }x+y-5=0 \\ \hline \text{(c) Parallel to the line }3x+4y+5=0 & \text{(iii) }x-y-1=0 \\ \hline \text{(d) Equally inclined to the axes} & \text{(iv) }3x+4y-17=0 \\ \hline \end{array}

\displaystyle \text{Answer:}
\displaystyle 2x-3y=0\qquad\ldots(1)
\displaystyle 4x-5y=2\qquad\ldots(2)
\displaystyle \text{Multiplying (1) by }2,\quad4x-6y=0.
\displaystyle \text{Subtracting this from (2),}\quad y=2.
\displaystyle \therefore x=3.
\displaystyle \therefore \text{The point of intersection is }(3,2).

\displaystyle \text{For (a), the required line passes through }(3,2)\text{ and }(2,1).
\displaystyle \text{Slope}=\frac{2-1}{3-2}=1.
\displaystyle y-2=x-3.
\displaystyle \therefore x-y-1=0.
\displaystyle \therefore \text{(a) matches (iii).}

\displaystyle \text{For (b), }x+2y+1=0.
\displaystyle y=-\frac{1}{2}x-\frac{1}{2}.
\displaystyle \therefore \text{Slope of the given line}=-\frac{1}{2}.
\displaystyle \therefore \text{Slope of the perpendicular line}=2.
\displaystyle y-2=2(x-3).
\displaystyle y=2x-4.
\displaystyle \therefore 2x-y=4.
\displaystyle \therefore \text{(b) matches (i).}

\displaystyle \text{For (c), the required line is parallel to }3x+4y+5=0.
\displaystyle \therefore \text{Its equation is }3x+4y+c=0.
\displaystyle \text{Since it passes through }(3,2),
\displaystyle 3(3)+4(2)+c=0.
\displaystyle 17+c=0\Rightarrow c=-17.
\displaystyle \therefore 3x+4y-17=0.
\displaystyle \therefore \text{(c) matches (iv).}

\displaystyle \text{For (d), a line equally inclined to the axes has slope }1\text{ or }-1.
\displaystyle \text{For slope }1,\quad y-2=x-3\Rightarrow x-y-1=0.
\displaystyle \text{For slope }-1,\quad y-2=-(x-3)\Rightarrow x+y-5=0.
\displaystyle \therefore \text{(d) matches (ii).}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Find the equation of the straight line which passes through the point}
\displaystyle (1,-2)\text{ and has slope }-1.
\displaystyle \text{Answer:}
\displaystyle \text{The given point is }(1,-2)\text{ and slope }m=-1.
\displaystyle \text{Using the point-slope form,}
\displaystyle y-y_1=m(x-x_1).
\displaystyle y-(-2)=-1(x-1).
\displaystyle y+2=-x+1.
\displaystyle \therefore x+y+1=0.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Find the equation of the line passing through the point }(5,2)\text{ and}
\displaystyle \text{perpendicular to the line joining the points }(2,3)\text{ and }(3,-1).
\displaystyle \text{Answer:}
\displaystyle \text{Slope of the line joining }(2,3)\text{ and }(3,-1)\text{ is}
\displaystyle m_1=\frac{-1-3}{3-2}=-4.
\displaystyle \text{Let the slope of the required perpendicular line be }m_2.
\displaystyle \text{For perpendicular lines, }m_1m_2=-1.
\displaystyle (-4)m_2=-1.
\displaystyle \therefore m_2=\frac{1}{4}.
\displaystyle \text{Using the point-slope form through }(5,2),
\displaystyle y-2=\frac{1}{4}(x-5).
\displaystyle 4y-8=x-5.
\displaystyle \therefore x-4y+3=0.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Find the angle between the lines }y=(2-\sqrt{3})(x+5)\text{ and}
\displaystyle y=(2+\sqrt{3})(x-7).
\displaystyle \text{Answer:}
\displaystyle \text{Slope of the first line is }m_1=2-\sqrt{3}.
\displaystyle \text{Slope of the second line is }m_2=2+\sqrt{3}.
\displaystyle \text{If }\theta\text{ is the angle between the two lines, then}
\displaystyle \tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|.
\displaystyle =\left|\frac{(2+\sqrt{3})-(2-\sqrt{3})}{1+(2-\sqrt{3})(2+\sqrt{3})}\right|.
\displaystyle =\frac{2\sqrt{3}}{1+(4-3)}.
\displaystyle =\frac{2\sqrt{3}}{2}=\sqrt{3}.
\displaystyle \therefore \tan\theta=\sqrt{3}.
\displaystyle \therefore \theta=60^\circ.
\displaystyle \therefore \text{The angle between the two lines is }60^\circ.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{Find the equations of the lines which pass through the point }(3,4)
\displaystyle \text{and cut off intercepts from the coordinate axes such that their sum is }14.
\displaystyle \text{Answer:}
\displaystyle \text{Let the }x\text{-intercept and }y\text{-intercept be }a\text{ and }b\text{ respectively.}
\displaystyle \therefore a+b=14.
\displaystyle \therefore b=14-a.
\displaystyle \text{Using the intercept form of a line,}
\displaystyle \frac{x}{a}+\frac{y}{b}=1.
\displaystyle \text{Since the line passes through }(3,4),
\displaystyle \frac{3}{a}+\frac{4}{b}=1.
\displaystyle \frac{3}{a}+\frac{4}{14-a}=1.
\displaystyle 3(14-a)+4a=a(14-a).
\displaystyle 42+a=14a-a^2.
\displaystyle a^2-13a+42=0.
\displaystyle (a-6)(a-7)=0.
\displaystyle \therefore a=6\quad\text{or}\quad a=7.
\displaystyle \text{If }a=6,\text{ then }b=8.
\displaystyle \frac{x}{6}+\frac{y}{8}=1.
\displaystyle \therefore 4x+3y=24.
\displaystyle \text{If }a=7,\text{ then }b=7.
\displaystyle \frac{x}{7}+\frac{y}{7}=1.
\displaystyle \therefore x+y=7.
\displaystyle \therefore \text{The required lines are }4x+3y=24\text{ and }x+y=7.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Find the points on the line }x+y=4\text{ which lie at a unit distance from}
\displaystyle \text{the line }4x+3y=10.
\displaystyle \text{Answer:}
\displaystyle \text{Let }(x,y)\text{ be a point on the line }x+y=4.
\displaystyle \therefore y=4-x.
\displaystyle \text{Its distance from the line }4x+3y-10=0\text{ is }1.
\displaystyle \frac{|4x+3y-10|}{\sqrt{4^2+3^2}}=1.
\displaystyle |4x+3y-10|=5.
\displaystyle \text{Substituting }y=4-x,
\displaystyle |4x+3(4-x)-10|=5.
\displaystyle |x+2|=5.
\displaystyle x+2=5\quad\text{or}\quad x+2=-5.
\displaystyle \therefore x=3\quad\text{or}\quad x=-7.
\displaystyle \text{If }x=3,\text{ then }y=1.
\displaystyle \text{If }x=-7,\text{ then }y=11.
\displaystyle \therefore \text{The required points are }(3,1)\text{ and }(-7,11).
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Find the equations of the lines passing through }(1,2)\text{ and making an}
\displaystyle \text{angle of }30^\circ\text{ with the }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{If a line makes an angle of }30^\circ\text{ with the }y\text{-axis, its inclinations with the}
\displaystyle \text{positive }x\text{-axis are }60^\circ\text{ and }120^\circ.
\displaystyle \therefore m_1=\tan60^\circ=\sqrt{3}.
\displaystyle \therefore m_2=\tan120^\circ=-\sqrt{3}.
\displaystyle \text{Using the point-slope form through }(1,2),
\displaystyle y-2=\sqrt{3}(x-1).
\displaystyle \text{or}
\displaystyle y-2=-\sqrt{3}(x-1).
\displaystyle \therefore \text{The required lines are }y-2=\sqrt{3}(x-1)\text{ and }y-2=-\sqrt{3}(x-1).
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Find the equation of the line passing through the point of intersection}
\displaystyle \text{of }2x+y=5\text{ and }x+3y+8=0,\text{ and parallel to the line }3x+4y=1.
\displaystyle \text{Answer:}
\displaystyle 2x+y=5.\qquad\ldots(1)
\displaystyle x+3y+8=0.\qquad\ldots(2)
\displaystyle \text{From (1), }y=5-2x.
\displaystyle \text{Substituting in (2),}
\displaystyle x+3(5-2x)+8=0.
\displaystyle x+15-6x+8=0.
\displaystyle -5x+23=0.
\displaystyle \therefore x=\frac{23}{5}.
\displaystyle y=5-2\left(\frac{23}{5}\right)=-\frac{21}{5}.
\displaystyle \therefore \text{The point of intersection is }\left(\frac{23}{5},-\frac{21}{5}\right).
\displaystyle \text{A line parallel to }3x+4y=1\text{ is of the form }3x+4y=c.
\displaystyle 3\left(\frac{23}{5}\right)+4\left(-\frac{21}{5}\right)=c.
\displaystyle \frac{69-84}{5}=c.
\displaystyle \therefore c=-3.
\displaystyle \therefore 3x+4y=-3.
\displaystyle \therefore 3x+4y+3=0.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{For what values of }a\text{ and }b\text{ are the intercepts cut off on the coordinate}
\displaystyle \text{axes by the line }ax+by+8=0\text{ equal in length but opposite in sign to those cut off}
\displaystyle \text{by the line }2x-3y+6=0\text{ on the axes?}
\displaystyle \text{Answer:}
\displaystyle \text{For the line }2x-3y+6=0,
\displaystyle \text{putting }y=0,\quad2x+6=0\Rightarrow x=-3.
\displaystyle \therefore \text{The }x\text{-intercept is }-3.
\displaystyle \text{Putting }x=0,\quad-3y+6=0\Rightarrow y=2.
\displaystyle \therefore \text{The }y\text{-intercept is }2.
\displaystyle \text{Hence, the required intercepts are }3\text{ and }-2.
\displaystyle \text{For the line }ax+by+8=0,\text{ the }x\text{-intercept is }-\frac{8}{a}.
\displaystyle -\frac{8}{a}=3.
\displaystyle \therefore a=-\frac{8}{3}.
\displaystyle \text{The }y\text{-intercept is }-\frac{8}{b}.
\displaystyle -\frac{8}{b}=-2.
\displaystyle \therefore b=4.
\displaystyle \therefore a=-\frac{8}{3}\text{ and }b=4.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{If the intercept of a line between the coordinate axes is divided by the point}
\displaystyle (-5,4)\text{ in the ratio }1:2,\text{ find the equation of the line.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the line cut the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{The point }P(-5,4)\text{ divides }AB\text{ internally in the ratio }1:2.
\displaystyle \therefore P=\left(\frac{2a+1(0)}{3},\frac{2(0)+1b}{3}\right).
\displaystyle \therefore \left(\frac{2a}{3},\frac{b}{3}\right)=(-5,4).
\displaystyle \frac{2a}{3}=-5\Rightarrow a=-\frac{15}{2}.
\displaystyle \frac{b}{3}=4\Rightarrow b=12.
\displaystyle \text{Using the intercept form of a line,}
\displaystyle \frac{x}{-\frac{15}{2}}+\frac{y}{12}=1.
\displaystyle -\frac{2x}{15}+\frac{y}{12}=1.
\displaystyle -8x+5y=60.
\displaystyle \therefore 8x-5y+60=0.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{Find the equation of a straight line for which the length of the perpendicular}
\displaystyle \text{from the origin is }4\text{ units and the line makes an angle of }120^\circ\text{ with the positive}
\displaystyle \text{direction of the }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{The inclination of the line is }120^\circ.
\displaystyle \therefore \text{The perpendicular to the line makes an angle of }30^\circ\text{ with the }x\text{-axis.}
\displaystyle \text{Using the normal form }x\cos\alpha+y\sin\alpha=p,
\displaystyle x\cos30^\circ+y\sin30^\circ=4.
\displaystyle \frac{\sqrt{3}}{2}x+\frac{1}{2}y=4.
\displaystyle \therefore \sqrt{3}x+y-8=0.
\displaystyle \text{The parallel line on the opposite side of the origin is}
\displaystyle x\cos30^\circ+y\sin30^\circ=-4.
\displaystyle \frac{\sqrt{3}}{2}x+\frac{1}{2}y=-4.
\displaystyle \therefore \sqrt{3}x+y+8=0.
\displaystyle \therefore \text{The required lines are }\sqrt{3}x+y-8=0\text{ and }\sqrt{3}x+y+8=0.
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{Find the equation of one of the sides of an isosceles right-angled triangle}
\displaystyle \text{whose hypotenuse is given by }3x+4y=4\text{ and the opposite vertex is }(2,2).
\displaystyle \text{Answer:}
\displaystyle 3x+4y=4\Rightarrow y=-\frac{3}{4}x+1.
\displaystyle \therefore \text{Slope of the hypotenuse }m_1=-\frac{3}{4}.
\displaystyle \text{Each side makes an angle of }45^\circ\text{ with the hypotenuse.}
\displaystyle \text{Let }m\text{ be the slope of a side. Then}
\displaystyle \left|\frac{m-m_1}{1+mm_1}\right|=\tan45^\circ=1.
\displaystyle \left|\frac{m+\frac{3}{4}}{1-\frac{3m}{4}}\right|=1.
\displaystyle \frac{4m+3}{4-3m}=\pm1.
\displaystyle \text{For }\frac{4m+3}{4-3m}=1,\quad4m+3=4-3m.
\displaystyle 7m=1\Rightarrow m=\frac{1}{7}.
\displaystyle \text{For }\frac{4m+3}{4-3m}=-1,\quad4m+3=-4+3m.
\displaystyle \therefore m=-7.
\displaystyle \text{Using the point-slope form through }(2,2),
\displaystyle y-2=\frac{1}{7}(x-2).
\displaystyle \therefore x-7y+12=0.
\displaystyle \text{For the other side, }y-2=-7(x-2).
\displaystyle \therefore 7x+y-16=0.
\displaystyle \therefore \text{The required side can be }x-7y+12=0\text{ or }7x+y-16=0.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{If the equation of the base of an equilateral triangle is }
\displaystyle x+y-2=0 \ \text{and the opposite vertex is }(2,-1),\text{ find the length of a side of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{The perpendicular distance of the vertex }(2,-1)\text{ from the base is the altitude.}
\displaystyle h=\frac{|2+(-1)-2|}{\sqrt{1^2+1^2}}.
\displaystyle =\frac{1}{\sqrt{2}}.
\displaystyle \text{If }s\text{ is the side of an equilateral triangle, then }h=\frac{\sqrt{3}}{2}s.
\displaystyle \frac{1}{\sqrt{2}}=\frac{\sqrt{3}}{2}s.
\displaystyle \therefore s=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}.
\displaystyle \therefore \text{The length of a side of the triangle is }\frac{\sqrt{6}}{3} = \sqrt{\frac{2}{3}} \text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{A straight line moves such that the sum of the reciprocals of its intercepts}
\displaystyle \text{made on the coordinate axes is constant. Show that the line passes through a fixed point.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the }x\text{-intercept and }y\text{-intercept of the line be }a\text{ and }b\text{ respectively.}
\displaystyle \text{Using the intercept form of a line,}
\displaystyle \frac{x}{a}+\frac{y}{b}=1.
\displaystyle \text{Given that }\frac{1}{a}+\frac{1}{b}=k,\text{ where }k\text{ is a constant.}
\displaystyle \text{Let }\frac{1}{a}=p.
\displaystyle \therefore \frac{1}{b}=k-p.
\displaystyle \text{Hence, the equation of the moving line becomes}
\displaystyle px+(k-p)y=1.
\displaystyle p(x-y)+ky=1.
\displaystyle \text{For this equation to be independent of }p,\text{ we must have }x-y=0.
\displaystyle \therefore x=y.
\displaystyle \text{Also, }ky=1.
\displaystyle \therefore y=\frac{1}{k}\text{ and }x=\frac{1}{k}.
\displaystyle \therefore \text{The moving line passes through the fixed point }\left(\frac{1}{k},\frac{1}{k}\right).
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{Find the equation of the line which passes through the point }(-4,3)
\displaystyle \text{and the portion of the line intercepted between the axes is divided internally in the}
\displaystyle \text{ratio }5:3\text{ by this point.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the line cut the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{The point }P(-4,3)\text{ divides }AB\text{ internally in the ratio }5:3.
\displaystyle \therefore P=\left(\frac{3a+5(0)}{8},\frac{3(0)+5b}{8}\right).
\displaystyle \therefore \left(\frac{3a}{8},\frac{5b}{8}\right)=(-4,3).
\displaystyle \frac{3a}{8}=-4\Rightarrow a=-\frac{32}{3}.
\displaystyle \frac{5b}{8}=3\Rightarrow b=\frac{24}{5}.
\displaystyle \text{Using the intercept form of a line,}
\displaystyle \frac{x}{-\frac{32}{3}}+\frac{y}{\frac{24}{5}}=1.
\displaystyle -\frac{3x}{32}+\frac{5y}{24}=1.
\displaystyle -9x+20y=96.
\displaystyle \therefore 9x-20y+96=0.
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{If the sum of the distances of a moving point in a plane from the coordinate}
\displaystyle \text{axes is }1,\text{ find the locus of the point.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the moving point be }P(x,y).
\displaystyle \text{Distance of }P\text{ from the }y\text{-axis}=|x|.
\displaystyle \text{Distance of }P\text{ from the }x\text{-axis}=|y|.
\displaystyle \text{Given that the sum of these distances is }1.
\displaystyle \therefore |x|+|y|=1.
\displaystyle \therefore \text{The locus of the point is }|x|+|y|=1.
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Find the equation of a line which passes through the point }(2,3)\text{ and}
\displaystyle \text{makes an angle of }30^\circ\text{ with the positive direction of the }x\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{The inclination of the line is }30^\circ.
\displaystyle \therefore \text{Slope }m=\tan30^\circ=\frac{1}{\sqrt{3}}.
\displaystyle \text{Using the point-slope form through }(2,3),
\displaystyle y-3=\frac{1}{\sqrt{3}}(x-2).
\displaystyle \sqrt{3}(y-3)=x-2.
\displaystyle \therefore x-\sqrt{3}y+3\sqrt{3}-2=0.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{Find the distance between the lines }3x+4y=9\text{ and } \\ 6x+8y=15.
\displaystyle \text{Answer:}
\displaystyle 3x+4y=9\Rightarrow3x+4y-9=0.
\displaystyle 6x+8y=15\Rightarrow3x+4y-\frac{15}{2}=0.
\displaystyle \text{For parallel lines }Ax+By+C_1=0\text{ and }Ax+By+C_2=0,
\displaystyle d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}.
\displaystyle \therefore d=\frac{\left|-9-\left(-\frac{15}{2}\right)\right|}{\sqrt{3^2+4^2}}.
\displaystyle =\frac{\left|-\frac{3}{2}\right|}{5}=\frac{3}{10}.
\displaystyle \therefore \text{The distance between the two lines is }\frac{3}{10}\text{ units.}
\displaystyle \\

\displaystyle \text{SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 51: }\text{Find the equation of the line where the length of the perpendicular}
\displaystyle \text{segment from the origin to the line is }4\text{ and the inclination of the perpendicular}
\displaystyle \text{segment with the positive direction of the }x\text{-axis is }30^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{The normal form of a straight line is}
\displaystyle x\cos\alpha+y\sin\alpha=p.
\displaystyle \text{Here, }p=4\text{ and }\alpha=30^\circ.
\displaystyle \therefore x\cos30^\circ+y\sin30^\circ=4.
\displaystyle \frac{\sqrt{3}}{2}x+\frac{1}{2}y=4.
\displaystyle \therefore \sqrt{3}x+y=8.
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{Find the equation of the straight line passing through }(1,2)\text{ and}
\displaystyle \text{perpendicular to the line }x+y+7=0.
\displaystyle \text{Answer:}
\displaystyle x+y+7=0.
\displaystyle y=-x-7.
\displaystyle \therefore \text{Slope of the given line}=-1.
\displaystyle \therefore \text{Slope of the perpendicular line}=1.
\displaystyle \text{Using the point-slope form through }(1,2),
\displaystyle y-2=1(x-1).
\displaystyle y-2=x-1.
\displaystyle \therefore y-x-1=0.
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{Find the distance between the lines }3x+4y=9\text{ and } \\ 6x+8y=15.
\displaystyle \text{Answer:}
\displaystyle 3x+4y=9\Rightarrow3x+4y-9=0.
\displaystyle 6x+8y=15\Rightarrow3x+4y-\frac{15}{2}=0.
\displaystyle \text{For parallel lines }Ax+By+C_1=0\text{ and }Ax+By+C_2=0,
\displaystyle d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}.
\displaystyle \therefore d=\frac{\left|-9+\frac{15}{2}\right|}{\sqrt{3^2+4^2}}.
\displaystyle =\frac{\frac{3}{2}}{5}=\frac{3}{10}.
\displaystyle \therefore \text{The distance between the two lines is }\frac{3}{10}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{If the slope of a line passing through the point }A(3,2)\text{ is }\frac{3}{4},
\displaystyle \text{find the equation of the line.}
\displaystyle \text{Answer:}
\displaystyle \text{The given point is }A(3,2)\text{ and slope }m=\frac{3}{4}.
\displaystyle \text{Using the point-slope form,}
\displaystyle y-y_1=m(x-x_1).
\displaystyle y-2=\frac{3}{4}(x-3).
\displaystyle 4y-8=3x-9.
\displaystyle \therefore 3x-4y-1=0.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{Find the equation of the straight line passing through }(1,1)\text{ and}
\displaystyle \text{perpendicular to the line }3x-5y+11=0.
\displaystyle \text{Answer:}
\displaystyle 3x-5y+11=0.
\displaystyle y=\frac{3}{5}x+\frac{11}{5}.
\displaystyle \therefore \text{Slope of the given line }m_1=\frac{3}{5}.
\displaystyle \text{Let the slope of the perpendicular line be }m_2.
\displaystyle \text{For perpendicular lines, }m_1m_2=-1.
\displaystyle \frac{3}{5}m_2=-1\Rightarrow m_2=-\frac{5}{3}.
\displaystyle \text{Using the point-slope form through }(1,1),
\displaystyle y-1=-\frac{5}{3}(x-1).
\displaystyle 3y-3=-5x+5.
\displaystyle \therefore 5x+3y-8=0.
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{A ray of light coming from the point }(1,2)\text{ is reflected at a point }A
\displaystyle \text{on the }x\text{-axis and then passes through the point }(5,3).\text{ Find the coordinates of }A.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P=(1,2)\text{ and }Q=(5,3).
\displaystyle \text{The reflection of }Q=(5,3)\text{ in the }x\text{-axis is }Q'=(5,-3).
\displaystyle \text{The point }A\text{ is the point where the line }PQ'\text{ intersects the }x\text{-axis.}
\displaystyle \text{Slope of }PQ'=\frac{-3-2}{5-1}=-\frac{5}{4}.
\displaystyle \text{Equation of }PQ'\text{ through }P(1,2)\text{ is}
\displaystyle y-2=-\frac{5}{4}(x-1).
\displaystyle \text{Since }A\text{ lies on the }x\text{-axis, }y=0.
\displaystyle -2=-\frac{5}{4}(x-1).
\displaystyle 8=5(x-1).
\displaystyle 5x=13\Rightarrow x=\frac{13}{5}.
\displaystyle \therefore A=\left(\frac{13}{5},0\right).
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{If one diagonal of a square is along the line }8x-15y=0\text{ and one of}
\displaystyle \text{its vertices is at }(1,2),\text{ find the equations of the sides of the square passing}
\displaystyle \text{through this vertex.}
\displaystyle \text{Answer:}
\displaystyle 8x-15y=0\Rightarrow y=\frac{8}{15}x.
\displaystyle \therefore \text{Slope of the diagonal }m=\frac{8}{15}.
\displaystyle \text{Each side of a square makes an angle of }45^\circ\text{ with its diagonal.}
\displaystyle \text{Let }m_1\text{ and }m_2\text{ be the slopes of the two sides.}
\displaystyle m_1=\frac{m+\tan45^\circ}{1-m\tan45^\circ}.
\displaystyle =\frac{\frac{8}{15}+1}{1-\frac{8}{15}}=\frac{23}{7}.
\displaystyle m_2=\frac{m-\tan45^\circ}{1+m\tan45^\circ}.
\displaystyle =\frac{\frac{8}{15}-1}{1+\frac{8}{15}}=-\frac{7}{23}.
\displaystyle \text{Using the point-slope form through }(1,2),
\displaystyle y-2=\frac{23}{7}(x-1).
\displaystyle 7y-14=23x-23.
\displaystyle \therefore 23x-7y-9=0.
\displaystyle \text{For the other side,}
\displaystyle y-2=-\frac{7}{23}(x-1).
\displaystyle 23y-46=-7x+7.
\displaystyle \therefore 7x+23y-53=0.
\displaystyle \therefore \text{The required sides are }23x-7y-9=0\text{ and }7x+23y-53=0.
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{Find the angle between the }x\text{-axis and the line joining the points }(3,-1)
\displaystyle \text{and }(4,-2).
\displaystyle \text{Answer:}
\displaystyle \text{Slope of the line joining }(3,-1)\text{ and }(4,-2)\text{ is}
\displaystyle m=\frac{-2-(-1)}{4-3}=-1.
\displaystyle \text{If }\theta\text{ is the inclination of the line, then}
\displaystyle \tan\theta=-1.
\displaystyle \therefore \theta=135^\circ.
\displaystyle \text{Hence, the acute angle between the line and the }x\text{-axis is}
\displaystyle 180^\circ-135^\circ=45^\circ.
\displaystyle \therefore \text{The angle between the line and the }x\text{-axis is }45^\circ.
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{The vertices of }\triangle PQR\text{ are }P(2,1),\ Q(-2,3)\text{ and }R(4,5).
\displaystyle \text{Find the equation of the median through the vertex }R.
\displaystyle \text{Answer:}
\displaystyle \text{Let }M\text{ be the midpoint of }PQ.
\displaystyle M=\left(\frac{2+(-2)}{2},\frac{1+3}{2}\right)=(0,2).
\displaystyle \text{The median through }R\text{ passes through }R(4,5)\text{ and }M(0,2).
\displaystyle \text{Slope of }RM=\frac{5-2}{4-0}=\frac{3}{4}.
\displaystyle \text{Using the point-slope form through }R(4,5),
\displaystyle y-5=\frac{3}{4}(x-4).
\displaystyle 4y-20=3x-12.
\displaystyle \therefore 3x-4y+8=0.
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{Point }R(h,k)\text{ divides a line segment between the coordinate axes}
\displaystyle \text{internally in the ratio }1:2.\text{ Find the equation of the line.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the line cut the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{Given }AR:RB=1:2.
\displaystyle \text{By the section formula,}
\displaystyle R=\left(\frac{2a+1(0)}{3},\frac{2(0)+1b}{3}\right).
\displaystyle \therefore \left(\frac{2a}{3},\frac{b}{3}\right)=(h,k).
\displaystyle \therefore a=\frac{3h}{2},\qquad b=3k.
\displaystyle \text{Using the intercept form of a line,}
\displaystyle \frac{x}{\frac{3h}{2}}+\frac{y}{3k}=1.
\displaystyle \frac{2x}{3h}+\frac{y}{3k}=1.
\displaystyle \therefore 2kx+hy=3hk.
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{Find the angle between the lines }y-\sqrt{3}x-5=0\text{ and}
\displaystyle \sqrt{3}y-x+6=0.
\displaystyle \text{Answer:}
\displaystyle y-\sqrt{3}x-5=0\Rightarrow y=\sqrt{3}x+5.
\displaystyle \therefore m_1=\sqrt{3}.
\displaystyle \sqrt{3}y-x+6=0\Rightarrow y=\frac{1}{\sqrt{3}}x-2\sqrt{3}.
\displaystyle \therefore m_2=\frac{1}{\sqrt{3}}.
\displaystyle \text{If }\theta\text{ is the angle between the two lines, then}
\displaystyle \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|.
\displaystyle =\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)}\right|.
\displaystyle =\frac{\frac{2}{\sqrt{3}}}{2}=\frac{1}{\sqrt{3}}.
\displaystyle \therefore \theta=30^\circ.
\displaystyle \therefore \text{The angle between the two lines is }30^\circ.
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{Find the distance of the point }(-1,1)\text{ from the line } \\ 12(x+6)=5(y-2).
\displaystyle \text{Answer:}
\displaystyle 12(x+6)=5(y-2).
\displaystyle 12x+72=5y-10.
\displaystyle \therefore 12x-5y+82=0.
\displaystyle \text{Distance of }(x_1,y_1)\text{ from }Ax+By+C=0\text{ is}
\displaystyle d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.
\displaystyle \therefore d=\frac{|12(-1)-5(1)+82|}{\sqrt{12^2+(-5)^2}}.
\displaystyle =\frac{|-12-5+82|}{\sqrt{144+25}}.
\displaystyle =\frac{65}{13}=5.
\displaystyle \therefore \text{The distance of the point from the line is }5\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{Two lines passing through the point }(2,3)\text{ intersect each other at}
\displaystyle \text{an angle of }60^\circ.\text{ If the slope of one line is }2,\text{ find the equations of the other lines.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the slope of the other line be }m.
\displaystyle \text{Given, }m_1=2\text{ and }\theta=60^\circ.
\displaystyle \tan\theta=\left|\frac{m-m_1}{1+mm_1}\right|.
\displaystyle \therefore \sqrt{3}=\left|\frac{m-2}{1+2m}\right|.
\displaystyle \therefore \frac{m-2}{1+2m}=\pm\sqrt{3}.
\displaystyle \text{For }\frac{m-2}{1+2m}=\sqrt{3},
\displaystyle m-2=\sqrt{3}(1+2m).
\displaystyle m(1-2\sqrt{3})=2+\sqrt{3}.
\displaystyle \therefore m=-\frac{8+5\sqrt{3}}{11}.
\displaystyle \text{For }\frac{m-2}{1+2m}=-\sqrt{3},
\displaystyle m-2=-\sqrt{3}(1+2m).
\displaystyle m(1+2\sqrt{3})=2-\sqrt{3}.
\displaystyle \therefore m=\frac{5\sqrt{3}-8}{11}.
\displaystyle \text{Using the point-slope form through }(2,3),
\displaystyle y-3=-\frac{8+5\sqrt{3}}{11}(x-2)
\displaystyle \text{or}
\displaystyle y-3=\frac{5\sqrt{3}-8}{11}(x-2).
\displaystyle \therefore \text{These are the equations of the two possible lines.}
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{Find the equations of the lines which cut off intercepts on the axes whose}
\displaystyle \text{sum and product are }1\text{ and }-6,\text{ respectively.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the intercepts on the coordinate axes be }a\text{ and }b.
\displaystyle \therefore a+b=1\text{ and }ab=-6.
\displaystyle \text{Thus, }a\text{ and }b\text{ are the roots of}
\displaystyle t^2-(a+b)t+ab=0.
\displaystyle \therefore t^2-t-6=0.
\displaystyle (t-3)(t+2)=0.
\displaystyle \therefore t=3\text{ or }t=-2.
\displaystyle \therefore \text{The intercepts are }3\text{ and }-2.
\displaystyle \text{Using the intercept form }\frac{x}{a}+\frac{y}{b}=1,
\displaystyle \frac{x}{3}+\frac{y}{-2}=1.
\displaystyle \therefore 2x-3y=6.
\displaystyle \text{Interchanging the intercepts,}
\displaystyle \frac{x}{-2}+\frac{y}{3}=1.
\displaystyle \therefore -3x+2y=6.
\displaystyle \therefore 3x-2y+6=0.
\displaystyle \therefore \text{The required lines are }2x-3y=6\text{ and }3x-2y+6=0.
\displaystyle \\

\displaystyle \textbf{Question 65: }\text{Find the value of }p\text{ so that the three lines }3x+y-2=0,\
\displaystyle px+2y-3=0  \ \text{and }2x-y-3=0\text{ may intersect at one point.}
\displaystyle \text{Answer:}
\displaystyle 3x+y-2=0.\qquad\ldots(1)
\displaystyle 2x-y-3=0.\qquad\ldots(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle 5x-5=0.
\displaystyle \therefore x=1.
\displaystyle \text{Substituting }x=1\text{ in (1),}
\displaystyle 3+y-2=0.
\displaystyle \therefore y=-1.
\displaystyle \therefore \text{The first and third lines intersect at }(1,-1).
\displaystyle \text{For all three lines to intersect at one point, }(1,-1)\text{ must satisfy}
\displaystyle px+2y-3=0.
\displaystyle p(1)+2(-1)-3=0.
\displaystyle p-5=0.
\displaystyle \therefore p=5.
\displaystyle \\

\displaystyle \textbf{Question 66: }\text{Find the equation of the line passing through the point of intersection}
\displaystyle \text{of the lines }4x+7y-3=0\text{ and }2x-3y+1=0\text{ and having equal intercepts}
\displaystyle \text{on the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle 4x+7y-3=0.\qquad\ldots(1)
\displaystyle 2x-3y+1=0.\qquad\ldots(2)
\displaystyle \text{Multiplying (2) by }2,
\displaystyle 4x-6y+2=0.\qquad\ldots(3)
\displaystyle \text{Subtracting (3) from (1),}
\displaystyle 13y-5=0.
\displaystyle \therefore y=\frac{5}{13}.
\displaystyle \text{Substituting }y=\frac{5}{13}\text{ in (2),}
\displaystyle 2x-\frac{15}{13}+1=0.
\displaystyle 2x=\frac{2}{13}.
\displaystyle \therefore x=\frac{1}{13}.
\displaystyle \therefore \text{The point of intersection is }\left(\frac{1}{13},\frac{5}{13}\right).
\displaystyle \text{A line having equal intercepts on the axes is of the form }x+y=a.
\displaystyle \text{Since it passes through }\left(\frac{1}{13},\frac{5}{13}\right),
\displaystyle \frac{1}{13}+\frac{5}{13}=a.
\displaystyle \therefore a=\frac{6}{13}.
\displaystyle \therefore x+y=\frac{6}{13}.
\displaystyle \therefore 13x+13y-6=0.
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{If the lines }y=3x+1\text{ and }2y=x+3\text{ are equally inclined to the line}
\displaystyle y=mx+4,\text{ find the value of }m.
\displaystyle \text{Answer:}
\displaystyle y=3x+1\Rightarrow m_1=3.
\displaystyle 2y=x+3\Rightarrow y=\frac{1}{2}x+\frac{3}{2}.
\displaystyle \therefore m_2=\frac{1}{2}.
\displaystyle \text{Since }y=mx+4\text{ is equally inclined to the two lines,}
\displaystyle \left|\frac{m-3}{1+3m}\right|=\left|\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\right|.
\displaystyle \left|\frac{m-3}{1+3m}\right|=\left|\frac{2m-1}{m+2}\right|.
\displaystyle \therefore \frac{m-3}{1+3m}=\pm\frac{2m-1}{m+2}.
\displaystyle \text{For the positive sign,}
\displaystyle (m-3)(m+2)=(2m-1)(1+3m).
\displaystyle m^2-m-6=6m^2-m-1.
\displaystyle 5m^2+5=0,
\displaystyle \text{which gives no real value of }m.
\displaystyle \text{For the negative sign,}
\displaystyle (m-3)(m+2)=-(2m-1)(1+3m).
\displaystyle m^2-m-6=-6m^2+m+1.
\displaystyle 7m^2-2m-7=0.
\displaystyle m=\frac{2\pm\sqrt{(-2)^2-4(7)(-7)}}{14}.
\displaystyle =\frac{2\pm\sqrt{200}}{14}=\frac{2\pm10\sqrt{2}}{14}.
\displaystyle \therefore m=\frac{1+5\sqrt{2}}{7}\quad\text{or}\quad m=\frac{1-5\sqrt{2}}{7}.
\displaystyle \\

\displaystyle \textbf{Question 68: }\text{Find the equation of the line which is equidistant from the parallel lines}
\displaystyle 9x+6y-7=0\text{ and }3x+2y+6=0.
\displaystyle \text{Answer:}
\displaystyle \text{The given lines are }9x+6y-7=0\text{ and }3x+2y+6=0.
\displaystyle \text{Multiplying the second equation by }3,
\displaystyle 9x+6y+18=0.
\displaystyle \text{The line equidistant from two parallel lines lies midway between them.}
\displaystyle \therefore 9x+6y+\frac{-7+18}{2}=0.
\displaystyle 9x+6y+\frac{11}{2}=0.
\displaystyle \therefore 18x+12y+11=0.
\displaystyle \\

\displaystyle \textbf{Question 69: }\text{The hypotenuse of a right-angled triangle has its ends at the points}
\displaystyle (1,3)\text{ and }(-4,1).\text{ If the legs of the triangle are parallel to the coordinate axes, find}
\displaystyle \text{the equations of the legs of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the endpoints of the hypotenuse be }A(1,3)\text{ and }B(-4,1).
\displaystyle \text{Since the legs are parallel to the coordinate axes, one leg is vertical and the other is horizontal.}
\displaystyle \text{The vertical leg passes through }A(1,3).
\displaystyle \therefore \text{Its equation is }x=1.
\displaystyle \text{The horizontal leg passes through }B(-4,1).
\displaystyle \therefore \text{Its equation is }y=1.
\displaystyle \therefore \text{The equations of the legs are }x=1\text{ and }y=1.
\displaystyle \\

\displaystyle \textbf{Question 70: }\text{Find the equations of the lines through the point }(3,2)\text{ which make an}
\displaystyle \text{angle of }45^\circ\text{ with the line }x-2y=3.
\displaystyle \text{Answer:}
\displaystyle x-2y=3\Rightarrow y=\frac{1}{2}x-\frac{3}{2}.
\displaystyle \therefore \text{Slope of the given line }m_1=\frac{1}{2}.
\displaystyle \text{Let }m\text{ be the slope of a required line.}
\displaystyle \tan45^\circ=\left|\frac{m-m_1}{1+mm_1}\right|.
\displaystyle 1=\left|\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\right|.
\displaystyle \therefore \frac{2m-1}{m+2}=\pm1.
\displaystyle \text{For }\frac{2m-1}{m+2}=1,\quad2m-1=m+2.
\displaystyle \therefore m=3.
\displaystyle \text{For }\frac{2m-1}{m+2}=-1,\quad2m-1=-m-2.
\displaystyle \therefore 3m=-1\Rightarrow m=-\frac{1}{3}.
\displaystyle \text{For }m=3,\quad y-2=3(x-3).
\displaystyle \therefore 3x-y-7=0.
\displaystyle \text{For }m=-\frac{1}{3},\quad y-2=-\frac{1}{3}(x-3).
\displaystyle \therefore x+3y-9=0.
\displaystyle \therefore \text{The required lines are }3x-y-7=0\text{ and }x+3y-9=0.
\displaystyle \\

\displaystyle \text{LONG ANSWER QUESTIONS}

\displaystyle \textbf{Question 71: }\text{A variable line passes through a fixed point }P.\text{ The algebraic sum of}
\displaystyle \text{the perpendiculars drawn from the points }(2,0),(0,2)\text{ and }(1,1)\text{ on the line}
\displaystyle \text{is zero. Find the coordinates of the point }P.
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of the variable line be}
\displaystyle ax+by+c=0.
\displaystyle \text{The algebraic perpendicular distance of }(x_1,y_1)\text{ from this line is}
\displaystyle \frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}.
\displaystyle \text{Therefore, the algebraic sum of the perpendiculars from the three given points is}
\displaystyle \frac{2a+c}{\sqrt{a^2+b^2}}+\frac{2b+c}{\sqrt{a^2+b^2}}+\frac{a+b+c}{\sqrt{a^2+b^2}}=0.
\displaystyle \frac{3a+3b+3c}{\sqrt{a^2+b^2}}=0.
\displaystyle \therefore a+b+c=0.
\displaystyle \text{Hence every such variable line passes through the point }(1,1),\text{ since}
\displaystyle a(1)+b(1)+c=0.
\displaystyle \therefore P=(1,1).
\displaystyle \therefore \text{The coordinates of the fixed point are }(1,1).
\displaystyle \\

\displaystyle \textbf{Question 72: }\text{In what direction should a line be drawn through the point }(1,2)\text{ so}
\displaystyle \text{that its point of intersection with the line }x+y=4\text{ is at a distance }\frac{\sqrt{6}}{3}
\displaystyle \text{from the given point?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P=(1,2)\text{ and let }Q(x,y)\text{ be the point of intersection with }x+y=4.
\displaystyle \therefore x+y=4.
\displaystyle \therefore y=4-x.
\displaystyle \text{Given, }PQ=\frac{\sqrt{6}}{3}.
\displaystyle \therefore (x-1)^2+(y-2)^2=\left(\frac{\sqrt{6}}{3}\right)^2.
\displaystyle (x-1)^2+(2-x)^2=\frac{2}{3}.
\displaystyle 2x^2-6x+5=\frac{2}{3}.
\displaystyle 6x^2-18x+13=0.
\displaystyle x=\frac{18\pm\sqrt{18^2-4(6)(13)}}{12}.
\displaystyle =\frac{18\pm2\sqrt{3}}{12}=\frac{9\pm\sqrt{3}}{6}.
\displaystyle \therefore y=4-x=\frac{15\mp\sqrt{3}}{6}.
\displaystyle \text{The slope of the line }PQ\text{ is}
\displaystyle m=\frac{y-2}{x-1}.
\displaystyle \text{For }x=\frac{9+\sqrt{3}}{6},\ y=\frac{15-\sqrt{3}}{6},
\displaystyle m=\frac{3-\sqrt{3}}{3+\sqrt{3}}=2-\sqrt{3}.
\displaystyle \text{For }x=\frac{9-\sqrt{3}}{6},\ y=\frac{15+\sqrt{3}}{6},
\displaystyle m=\frac{3+\sqrt{3}}{3-\sqrt{3}}=2+\sqrt{3}.
\displaystyle \text{Since }\tan15^\circ=2-\sqrt{3}\text{ and }\tan75^\circ=2+\sqrt{3},
\displaystyle \therefore \text{the required directions are }15^\circ\text{ and }75^\circ\text{ with the positive }x\text{-axis.}
\displaystyle \\

\displaystyle \textbf{Question 73: }\text{Find the equations of the lines through the point of intersection of}
\displaystyle \text{the lines }x-y+1=0\text{ and }2x-3y+5=0,\text{ whose distance from the point }(3,2)
\displaystyle \text{is }\frac{7}{5}.
\displaystyle \text{Answer:}
\displaystyle x-y+1=0.\qquad\ldots(1)
\displaystyle 2x-3y+5=0.\qquad\ldots(2)
\displaystyle \text{From (1), }y=x+1.
\displaystyle \text{Substituting in (2),}
\displaystyle 2x-3(x+1)+5=0.
\displaystyle -x+2=0\Rightarrow x=2.
\displaystyle \therefore y=3.
\displaystyle \therefore \text{The point of intersection is }(2,3).
\displaystyle \text{Let the slope of a required line through }(2,3)\text{ be }m.
\displaystyle y-3=m(x-2).
\displaystyle \therefore mx-y+3-2m=0.
\displaystyle \text{Its distance from }(3,2)\text{ is }\frac{7}{5}.
\displaystyle \therefore \frac{|3m-2+3-2m|}{\sqrt{m^2+1}}=\frac{7}{5}.
\displaystyle \frac{|m+1|}{\sqrt{m^2+1}}=\frac{7}{5}.
\displaystyle 25(m+1)^2=49(m^2+1).
\displaystyle 25m^2+50m+25=49m^2+49.
\displaystyle 12m^2-25m+12=0.
\displaystyle (3m-4)(4m-3)=0.
\displaystyle \therefore m=\frac{4}{3}\quad\text{or}\quad m=\frac{3}{4}.
\displaystyle \text{For }m=\frac{4}{3},
\displaystyle y-3=\frac{4}{3}(x-2).
\displaystyle \therefore 4x-3y+1=0.
\displaystyle \text{For }m=\frac{3}{4},
\displaystyle y-3=\frac{3}{4}(x-2).
\displaystyle \therefore 3x-4y+6=0.
\displaystyle \therefore \text{The required lines are }4x-3y+1=0\text{ and }3x-4y+6=0.
\displaystyle \\

\displaystyle \textbf{Question 74: }\text{The point }(4,1)\text{ undergoes the following two successive transformations:}
\displaystyle \text{(i) Reflection about the line }y=x
\displaystyle \text{(ii) Translation through a distance of }2\text{ units along the positive }x\text{-axis.}
\displaystyle \text{Then the final coordinates of the point are:}
\displaystyle \text{Answer:}
\displaystyle \text{Reflection about the line }y=x\text{ interchanges the }x\text{ and }y\text{ coordinates.}
\displaystyle \therefore (4,1)\rightarrow(1,4).
\displaystyle \text{Translation through }2\text{ units along the positive }x\text{-axis gives}
\displaystyle (1,4)\rightarrow(1+2,4)=(3,4).
\displaystyle \therefore \text{The final coordinates of the point are }(3,4).
\displaystyle \\

\displaystyle \textbf{Question 75: }\text{Find the locus of the midpoints of the portion of the line}
\displaystyle x\sin\theta+y\cos\theta=p\text{ intercepted between the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }x\sin\theta+y\cos\theta=p.
\displaystyle \text{Putting }y=0,\text{ the }x\text{-intercept is }\frac{p}{\sin\theta}.
\displaystyle \text{Putting }x=0,\text{ the }y\text{-intercept is }\frac{p}{\cos\theta}.
\displaystyle \text{Let }(h,k)\text{ be the midpoint of the intercepted portion.}
\displaystyle \therefore h=\frac{p}{2\sin\theta},\qquad k=\frac{p}{2\cos\theta}.
\displaystyle \therefore \sin\theta=\frac{p}{2h},\qquad\cos\theta=\frac{p}{2k}.
\displaystyle \text{Using }\sin^2\theta+\cos^2\theta=1,
\displaystyle \frac{p^2}{4h^2}+\frac{p^2}{4k^2}=1.
\displaystyle \therefore \frac{1}{h^2}+\frac{1}{k^2}=\frac{4}{p^2}.
\displaystyle \text{Replacing }h,k\text{ by }x,y,\text{ the locus is}
\displaystyle \therefore \frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}.
\displaystyle \\

\displaystyle \textbf{Question 76: }\text{If }p\text{ is the length of the perpendicular from the origin to the line whose}
\displaystyle \text{intercepts on the coordinate axes are }a\text{ and }b,\text{ show that }\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}.
\displaystyle \text{Answer:}
\displaystyle \text{The intercept form of the line is}
\displaystyle \frac{x}{a}+\frac{y}{b}=1.
\displaystyle bx+ay-ab=0.
\displaystyle \text{The perpendicular distance of the origin }(0,0)\text{ from this line is }p.
\displaystyle \therefore p=\frac{|b(0)+a(0)-ab|}{\sqrt{a^2+b^2}}.
\displaystyle =\frac{|ab|}{\sqrt{a^2+b^2}}.
\displaystyle \therefore p^2=\frac{a^2b^2}{a^2+b^2}.
\displaystyle \therefore \frac{1}{p^2}=\frac{a^2+b^2}{a^2b^2}.
\displaystyle =\frac{1}{a^2}+\frac{1}{b^2}.
\displaystyle \therefore \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}.
\displaystyle \\

\displaystyle \textbf{Question 77: }\text{Show that the path of a moving point such that its distances from}
\displaystyle \text{the two lines }3x-2y=5\text{ and }3x+2y=5\text{ are equal consists of straight lines.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ be the moving point.}
\displaystyle \text{Distance of }P\text{ from }3x-2y-5=0\text{ is}
\displaystyle d_1=\frac{|3x-2y-5|}{\sqrt{3^2+(-2)^2}}=\frac{|3x-2y-5|}{\sqrt{13}}.
\displaystyle \text{Distance of }P\text{ from }3x+2y-5=0\text{ is}
\displaystyle d_2=\frac{|3x+2y-5|}{\sqrt{3^2+2^2}}=\frac{|3x+2y-5|}{\sqrt{13}}.
\displaystyle \text{Since the distances are equal, }d_1=d_2.
\displaystyle |3x-2y-5|=|3x+2y-5|.
\displaystyle \therefore 3x-2y-5=3x+2y-5
\displaystyle \text{or}
\displaystyle 3x-2y-5=-(3x+2y-5).
\displaystyle \text{From the first equation,}
\displaystyle -2y=2y\Rightarrow y=0.
\displaystyle \text{From the second equation,}
\displaystyle 3x-2y-5=-3x-2y+5.
\displaystyle 6x=10\Rightarrow x=\frac{5}{3}.
\displaystyle \therefore \text{The locus consists of the straight lines }y=0\text{ and }x=\frac{5}{3}.
\displaystyle \\

\displaystyle \textbf{Question 78: }\text{Find the area of the triangle formed by the lines }
\displaystyle y-x=0,\ x+y=0 \ \text{and }x-k=0.
\displaystyle \text{Answer:}
\displaystyle y-x=0\Rightarrow y=x.
\displaystyle x+y=0\Rightarrow y=-x.
\displaystyle \text{The lines }y=x\text{ and }y=-x\text{ intersect at }O(0,0).
\displaystyle \text{For }x=k\text{ and }y=x,\text{ the point of intersection is }A(k,k).
\displaystyle \text{For }x=k\text{ and }y=-x,\text{ the point of intersection is }B(k,-k).
\displaystyle AB=|k-(-k)|=2|k|.
\displaystyle \text{The perpendicular distance of }O(0,0)\text{ from }x=k\text{ is }|k|.
\displaystyle \text{Area of }\triangle OAB=\frac{1}{2}\times AB\times |k|.
\displaystyle =\frac{1}{2}\times2|k|\times|k|=k^2.
\displaystyle \therefore \text{The area of the triangle is }k^2\text{ square units.}
\displaystyle \\

\displaystyle \textbf{Question 79: }\text{If }p\text{ and }q\text{ are the lengths of perpendiculars from the origin to the}
\displaystyle \text{lines }x\cos\theta-y\sin\theta=k\cos2\theta\text{ and }x\sec\theta+y\,\mathrm{cosec}\theta=k,
\displaystyle \text{respectively, prove that }p^2+4q^2=k^2.
\displaystyle \text{Answer:}
\displaystyle \text{For the line }x\cos\theta-y\sin\theta-k\cos2\theta=0,
\displaystyle p=\frac{|-k\cos2\theta|}{\sqrt{\cos^2\theta+\sin^2\theta}}.
\displaystyle \therefore p=|k\cos2\theta|.
\displaystyle \therefore p^2=k^2\cos^22\theta.\qquad\ldots(1)
\displaystyle \text{For the line }x\sec\theta+y\,\mathrm{cosec}\theta-k=0,
\displaystyle q=\frac{|k|}{\sqrt{\sec^2\theta+\mathrm{cosec}^2\theta}}.
\displaystyle \therefore q^2=\frac{k^2}{\sec^2\theta+\mathrm{cosec}^2\theta}.
\displaystyle \sec^2\theta+\mathrm{cosec}^2\theta
\displaystyle =\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}
\displaystyle =\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta\cos^2\theta}
\displaystyle =\frac{1}{\sin^2\theta\cos^2\theta}.
\displaystyle \therefore q^2=k^2\sin^2\theta\cos^2\theta.
\displaystyle =\frac{k^2}{4}\sin^22\theta.
\displaystyle \therefore 4q^2=k^2\sin^22\theta.\qquad\ldots(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle p^2+4q^2=k^2(\cos^22\theta+\sin^22\theta).
\displaystyle \therefore p^2+4q^2=k^2.
\displaystyle \\

\displaystyle \textbf{Question 80: }\text{Find the distance of the line }4x+7y+5=0\text{ from the point }(1,2)
\displaystyle \text{along the line }2x-y=0.
\displaystyle \text{Answer:}
\displaystyle \text{The point }(1,2)\text{ lies on }2x-y=0,\text{ since }2(1)-2=0.
\displaystyle \text{Let }Q\text{ be the point where }2x-y=0\text{ meets }4x+7y+5=0.
\displaystyle 2x-y=0\Rightarrow y=2x.
\displaystyle \text{Substituting }y=2x\text{ in }4x+7y+5=0,
\displaystyle 4x+14x+5=0.
\displaystyle 18x=-5.
\displaystyle \therefore x=-\frac{5}{18}.
\displaystyle \therefore y=2\left(-\frac{5}{18}\right)=-\frac{5}{9}.
\displaystyle \therefore Q=\left(-\frac{5}{18},-\frac{5}{9}\right).
\displaystyle \text{Required distance }=PQ.
\displaystyle PQ=\sqrt{\left(-\frac{5}{18}-1\right)^2+\left(-\frac{5}{9}-2\right)^2}.
\displaystyle =\sqrt{\left(-\frac{23}{18}\right)^2+\left(-\frac{23}{9}\right)^2}.
\displaystyle =\frac{23}{18}\sqrt{1+4}.
\displaystyle \therefore PQ=\frac{23\sqrt5}{18}\text{ units}.
\displaystyle \therefore \text{The required distance is }\frac{23\sqrt5}{18}\text{ units.}
\displaystyle \\

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