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

\displaystyle \textbf{Question 1: }L\text{ is a variable line such that the algebraic sum of the distances of}
\displaystyle \text{the points }(1,1),(2,0)\text{ and }(0,2)\text{ from the line is equal to zero.}
\displaystyle \text{The line }L\text{ will always pass through}
\displaystyle \text{(a) }(1,1)\qquad\text{(b) }(2,1)\qquad\text{(c) }(1,2)\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of }L\text{ be }ax+by+c=0.
\displaystyle \text{The algebraic distances of the three points from }L\text{ are proportional to}
\displaystyle a+b+c,\qquad 2a+c,\qquad 2b+c.
\displaystyle \text{Since their algebraic sum is zero,}
\displaystyle (a+b+c)+(2a+c)+(2b+c)=0.
\displaystyle \Rightarrow 3a+3b+3c=0\quad\Rightarrow\quad a+b+c=0.
\displaystyle \text{Thus, }L\text{ always passes through }(1,1).
\displaystyle \therefore\ \text{the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The acute angle between the medians drawn from the acute angles of a}
\displaystyle \text{right angled isosceles triangle is}
\displaystyle \text{(a) }\cos^{-1}\left(\frac23\right)\qquad \text{(b) }\cos^{-1}\left(\frac34\right)\qquad \text{(c) }\cos^{-1}\left(\frac45\right)\qquad \text{(d) }\cos^{-1}\left(\frac56\right)
\displaystyle \text{Answer:}
\displaystyle \text{Take the vertices as }O(0,0),\ A(1,0)\text{ and }B(0,1).
\displaystyle \text{The mid-points of }OB\text{ and }OA\text{ are }\left(0,\frac12\right) \text{ and }\left(\frac12,0\right).
\displaystyle \text{Slope of the median from }A =\frac{\frac12-0}{0-1}=-\frac12.
\displaystyle \text{Slope of the median from }B =\frac{0-1}{\frac12-0}=-2.
\displaystyle \text{If }\theta\text{ is the acute angle between the medians, then}
\displaystyle \tan\theta=\left|\frac{-2+\frac12}{1+(-2)\left(-\frac12\right)}\right| =\frac34.
\displaystyle \therefore\ \cos\theta=\frac{1}{\sqrt{1+\tan^2\theta}} =\frac{1}{\sqrt{1+\frac9{16}}}=\frac45.
\displaystyle \therefore\ \theta=\cos^{-1}\left(\frac45\right).
\displaystyle \therefore\ \text{the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The distance between the orthocentre and circumcentre of the triangle}
\displaystyle \text{with vertices }(1,2),(2,1)\text{ and } \left(\frac{3+\sqrt3}{2},\frac{3+\sqrt3}{2}\right)\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }\sqrt2\qquad \text{(c) }3+\sqrt3\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=(1,2),\ B=(2,1),\ C=\left(\frac{3+\sqrt3}{2},\frac{3+\sqrt3}{2}\right).
\displaystyle AB^2=(2-1)^2+(1-2)^2=2.
\displaystyle AC^2= \left(\frac{1+\sqrt3}{2}\right)^2+ \left(\frac{\sqrt3-1}{2}\right)^2=2.
\displaystyle BC^2= \left(\frac{\sqrt3-1}{2}\right)^2+ \left(\frac{1+\sqrt3}{2}\right)^2=2.
\displaystyle \therefore\ AB=BC=CA,\text{ so the triangle is equilateral.}
\displaystyle \text{In an equilateral triangle, the orthocentre and circumcentre coincide.}
\displaystyle \therefore\ \text{the distance between them is }0.
\displaystyle \therefore\ \text{the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The equation of the straight line which passes through the point }(-4,3)
\displaystyle \text{such that the portion of the line between the axes is divided internally by the point}
\displaystyle \text{in the ratio }5:3\text{ is}
\displaystyle \text{(a) }9x-20y+96=0\qquad \text{(b) }9x+20y=24
\displaystyle \text{(c) }20x+9y+53=0\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the intercepts of the line be }A(a,0)\text{ and }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{5(0)+3a}{8}, \frac{5b+3(0)}{8}\right).
\displaystyle \therefore\ \frac{3a}{8}=-4,\qquad\frac{5b}{8}=3.
\displaystyle \Rightarrow a=-\frac{32}{3},\qquad b=\frac{24}{5}.
\displaystyle \text{Using the intercept form }\frac{x}{a}+\frac{y}{b}=1,
\displaystyle \frac{x}{-32/3}+\frac{y}{24/5}=1.
\displaystyle \Rightarrow-\frac{3x}{32}+\frac{5y}{24}=1.
\displaystyle \Rightarrow-9x+20y=96.
\displaystyle \therefore\ 9x-20y+96=0.
\displaystyle \therefore\ \text{the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The point which divides the join of }(1,2)\text{ and }(3,4)\text{ externally}
\displaystyle \text{in the ratio }1:1
\displaystyle \text{(a) lies in the III quadrant}\qquad\text{(b) lies in the II quadrant}
\displaystyle \text{(c) lies in the I quadrant}\qquad\text{(d) cannot be found}
\displaystyle \text{Answer:}
\displaystyle \text{For external division in the ratio }m:n,
\displaystyle P=\left(\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n}\right).
\displaystyle \text{Here, }m=n=1.
\displaystyle \therefore\ P=\left(\frac{3-1}{1-1},\frac{4-2}{1-1}\right).
\displaystyle \text{Since the denominator is zero, no finite point can be obtained.}
\displaystyle \therefore\ \text{the point cannot be found.}
\displaystyle \therefore\ \text{the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A line passes through the point }(2,2)\text{ and is perpendicular to the line}
\displaystyle 3x+y=3.\text{ Its }y\text{-intercept is}
\displaystyle \text{(a) }\frac13\qquad\text{(b) }\frac23\qquad \text{(c) }1\qquad\text{(d) }\frac43
\displaystyle \text{Answer:}
\displaystyle 3x+y=3\quad\Rightarrow\quad y=-3x+3.
\displaystyle \therefore\ \text{the slope of the given line is }-3.
\displaystyle \text{Hence, the slope of the perpendicular line is }\frac13.
\displaystyle \text{Using the point-slope form through }(2,2),
\displaystyle y-2=\frac13(x-2).
\displaystyle \Rightarrow 3y-6=x-2\quad\Rightarrow\quad x-3y+4=0.
\displaystyle \text{Putting }x=0,\quad -3y+4=0\quad\Rightarrow\quad y=\frac43.
\displaystyle \therefore\ \text{the }y\text{-intercept is }\frac43.
\displaystyle \therefore\ \text{the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If the lines }ax+12y+1=0,\ bx+13y+1=0\text{ and}
\displaystyle cx+14y+1=0\text{ are concurrent, then }a,b,c\text{ are in}
\displaystyle \text{(a) H.P.}\qquad\text{(b) G.P.}\qquad \text{(c) A.P.}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{For three lines to be concurrent,}
\displaystyle \begin{vmatrix}a&12&1\\b&13&1\\c&14&1\end{vmatrix}=0.
\displaystyle \Rightarrow a(13-14)-12(b-c)+(14b-13c)=0.
\displaystyle \Rightarrow -a-12b+12c+14b-13c=0.
\displaystyle \Rightarrow -a+2b-c=0.
\displaystyle \Rightarrow 2b=a+c.
\displaystyle \text{This is the condition for }a,b,c\text{ to be in A.P.}
\displaystyle \therefore\ \text{the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The number of real values of }\lambda\text{ for which the lines }
\displaystyle x-2y+3=0, \ \lambda x+3y+1=0\text{ and }4x-\lambda y+2=0\text{ are concurrent is}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }2\qquad\text{(d) Infinite}
\displaystyle \text{Answer:}
\displaystyle \text{For the three lines to be concurrent,}
\displaystyle \begin{vmatrix}1&-2&3\\ \lambda&3&1\\ 4&-\lambda&2\end{vmatrix}=0.
\displaystyle \Rightarrow 1(6+\lambda)+2(2\lambda-4)+3(-\lambda^2-12)=0.
\displaystyle \Rightarrow 6+\lambda+4\lambda-8-3\lambda^2-36=0.
\displaystyle \Rightarrow 3\lambda^2-5\lambda+38=0.
\displaystyle \text{Its discriminant is }(-5)^2-4(3)(38)=25-456=-431<0.
\displaystyle \therefore\ \text{there is no real value of }\lambda.
\displaystyle \therefore\ \text{the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The equations of the sides }AB,\ BC\text{ and }CA\text{ of }\triangle ABC\text{ are}
\displaystyle y-x=2,\quad x+2y=1\quad\text{and}\quad3x+y+5=0\text{ respectively.}
\displaystyle \text{The equation of the altitude through }B\text{ is}
\displaystyle \text{(a) }x-3y+1=0\qquad\text{(b) }x-3y+4=0
\displaystyle \text{(c) }3x-y+2=0\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle B\text{ is the point of intersection of }AB\text{ and }BC.
\displaystyle y-x=2\quad\Rightarrow\quad y=x+2.
\displaystyle x+2(x+2)=1\quad\Rightarrow\quad3x=-3\quad\Rightarrow\quad x=-1.
\displaystyle \therefore\ y=1,\text{ so }B=(-1,1).
\displaystyle CA:\ 3x+y+5=0\quad\Rightarrow\quad y=-3x-5.
\displaystyle \therefore\ \text{slope of }CA=-3.
\displaystyle \text{Hence, slope of the altitude through }B=\frac13.
\displaystyle y-1=\frac13(x+1).
\displaystyle \Rightarrow 3y-3=x+1.
\displaystyle \Rightarrow x-3y+4=0.
\displaystyle \therefore\ \text{the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }p_1\text{ and }p_2\text{ are the lengths of the perpendiculars from origin upon}
\displaystyle \text{the lines }x\sec\theta+y\mathrm{cosec}\theta=a\text{ and}
\displaystyle x\cos\theta-y\sin\theta=a\cos2\theta\text{ respectively, then}
\displaystyle \text{(a) }4p_1^2+p_2^2=a^2\qquad\text{(b) }p_1^2+4p_2^2=a^2
\displaystyle \text{(c) }p_1^2+p_2^2=a^2\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle p_1=\frac{|a|}{\sqrt{\sec^2\theta+\mathrm{cosec}^2\theta}}.
\displaystyle \sec^2\theta+\mathrm{cosec}^2\theta =\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta} =\frac{1}{\sin^2\theta\cos^2\theta}.
\displaystyle \therefore\ p_1^2=a^2\sin^2\theta\cos^2\theta =\frac{a^2}{4}\sin^22\theta.
\displaystyle p_2=\frac{|a\cos2\theta|} {\sqrt{\cos^2\theta+\sin^2\theta}}=|a\cos2\theta|.
\displaystyle \therefore\ p_2^2=a^2\cos^22\theta.
\displaystyle 4p_1^2+p_2^2 =a^2\sin^22\theta+a^2\cos^22\theta=a^2.
\displaystyle \therefore\ 4p_1^2+p_2^2=a^2.
\displaystyle \therefore\ \text{the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Area of the triangle formed by the points}
\displaystyle ((a+3)(a+4),a+3),\ ((a+2)(a+3),a+2)\text{ and}
\displaystyle ((a+1)(a+2),a+1)\text{ is}
\displaystyle \text{(a) }25a^2\qquad\text{(b) }5a^2\qquad \text{(c) }24a^2\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three points be }A,B,C\text{ respectively.}
\displaystyle \overrightarrow{BA} =\left((a+3)(a+4)-(a+2)(a+3),\,1\right) =(2a+6,1).
\displaystyle \overrightarrow{BC} =\left((a+1)(a+2)-(a+2)(a+3),\,-1\right) =(-2a-4,-1).
\displaystyle \text{Area}=\frac12\left|(2a+6)(-1)-1(-2a-4)\right|.
\displaystyle =\frac12|-2a-6+2a+4|=\frac12|-2|=1.
\displaystyle \therefore\ \text{the area is }1\text{ sq. unit.}
\displaystyle \therefore\ \text{the correct option is (d), none of these.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }a+b+c=0,\text{ then the family of lines }3ax+by+2c=0
\displaystyle \text{pass through fixed point}
\displaystyle \text{(a) }\left(2,\frac23\right)\qquad \text{(b) }\left(\frac23,2\right)\qquad \text{(c) }\left(-2,\frac23\right)\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle a+b+c=0\quad\Rightarrow\quad c=-a-b.
\displaystyle \text{Substituting in }3ax+by+2c=0,
\displaystyle 3ax+by-2a-2b=0.
\displaystyle \Rightarrow a(3x-2)+b(y-2)=0.
\displaystyle \text{For a fixed point, }3x-2=0\text{ and }y-2=0.
\displaystyle \therefore\ x=\frac23,\qquad y=2.
\displaystyle \therefore\ \text{the fixed point is }\left(\frac23,2\right).
\displaystyle \therefore\ \text{the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The line segment joining the points }(-3,-4)\text{ and }(1,-2)
\displaystyle \text{is divided by }y\text{-axis in the ratio}
\displaystyle \text{(a) }1:3\qquad\text{(b) }2:3\qquad \text{(c) }3:1\qquad\text{(d) }3:2
\displaystyle \text{Answer:}
\displaystyle \text{Let the }y\text{-axis divide the segment in the ratio }m:n.
\displaystyle \text{Since the point of division lies on the }y\text{-axis, its }x\text{-coordinate is }0.
\displaystyle \therefore\ \frac{m(1)+n(-3)}{m+n}=0.
\displaystyle \Rightarrow m-3n=0\quad\Rightarrow\quad m=3n.
\displaystyle \therefore\ m:n=3:1.
\displaystyle \therefore\ \text{the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The area of a triangle with vertices at }(-4,-1),(1,2)\text{ and }(4,-3)\text{ is}
\displaystyle \text{(a) }17\qquad\text{(b) }16\qquad\text{(c) }15\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Area}=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.
\displaystyle =\frac12\left|(-4)(2+3)+1(-3+1)+4(-1-2)\right|.
\displaystyle =\frac12|-20-2-12|=\frac12(34)=17.
\displaystyle \therefore\ \text{the area of the triangle is }17\text{ sq. units.}
\displaystyle \therefore\ \text{the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The line segment joining the points }(1,2)\text{ and }(-2,1)\text{ is divided}
\displaystyle \text{by the line }3x+4y=7\text{ in the ratio}
\displaystyle \text{(a) }3:4\qquad\text{(b) }4:3\qquad \text{(c) }9:4\qquad\text{(d) }4:9
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=(1,2),\ B=(-2,1)\text{ and let the line divide }AB\text{ in the ratio }m:n.
\displaystyle P=\left(\frac{-2m+n}{m+n},\frac{m+2n}{m+n}\right).
\displaystyle \text{Since }P\text{ lies on }3x+4y=7,
\displaystyle 3\left(\frac{-2m+n}{m+n}\right) +4\left(\frac{m+2n}{m+n}\right)=7.
\displaystyle -6m+3n+4m+8n=7m+7n.
\displaystyle \Rightarrow -2m+11n=7m+7n.
\displaystyle \Rightarrow 9m=4n.
\displaystyle \therefore\ m:n=4:9.
\displaystyle \therefore\ \text{the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If the point }(5,2)\text{ bisects the intercept of a line between the axes,}
\displaystyle \text{then its equation is}
\displaystyle \text{(a) }5x+2y=20\qquad\text{(b) }2x+5y=20
\displaystyle \text{(c) }5x-2y=20\qquad\text{(d) }2x-5y=20
\displaystyle \text{Answer:}
\displaystyle \text{Let the intercepts of the line on the axes be }A(a,0)\text{ and }B(0,b).
\displaystyle \text{Since }(5,2)\text{ is the mid-point of }AB,
\displaystyle \left(\frac{a}{2},\frac{b}{2}\right)=(5,2).
\displaystyle \therefore\ a=10,\qquad b=4.
\displaystyle \text{Using the intercept form, }\frac{x}{10}+\frac{y}{4}=1.
\displaystyle \Rightarrow 2x+5y=20.
\displaystyle \therefore\ \text{the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 17: }A(6,3),\ B(-3,5),\ C(4,-2)\text{ and }D(x,3x)\text{ are four points. If}
\displaystyle \triangle DBC:\triangle ABC=1:2,\text{ then }x\text{ is equal to}
\displaystyle \text{(a) }\frac{11}{8}\qquad\text{(b) }\frac{8}{11}\qquad \text{(c) }3\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle ABC =\frac12\left|6(5+2)-3(-2-3)+4(3-5)\right|.
\displaystyle =\frac12|42+15-8|=\frac{49}{2}.
\displaystyle \text{Area of }\triangle DBC =\frac12\left|x(5+2)-3(-2-3x)+4(3x-5)\right|.
\displaystyle =\frac12|7x+6+9x+12x-20| =\frac12|28x-14|.
\displaystyle \text{Since }\triangle DBC:\triangle ABC=1:2,
\displaystyle \frac{|28x-14|}{49}=\frac12.
\displaystyle \Rightarrow |28x-14|=\frac{49}{2}.
\displaystyle \Rightarrow 28x-14=\pm\frac{49}{2}.
\displaystyle \therefore\ x=\frac{11}{8}\quad\text{or}\quad x=-\frac38.
\displaystyle \text{Among the given options, only }\frac{11}{8}\text{ occurs.}
\displaystyle \therefore\ \text{the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }p\text{ be the length of the perpendicular from the origin on the line}
\displaystyle \frac{x}{a}+\frac{y}{b}=1,\text{ then}
\displaystyle \text{(a) }p^2=a^2+b^2\qquad \text{(b) }p^2=\frac{1}{a^2}+\frac{1}{b^2}
\displaystyle \text{(c) }\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2} \qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{x}{a}+\frac{y}{b}=1\Rightarrow bx+ay-ab=0.
\displaystyle \text{Distance of the origin from this line is}
\displaystyle p=\frac{|-ab|}{\sqrt{a^2+b^2}} =\frac{|ab|}{\sqrt{a^2+b^2}}.
\displaystyle \therefore\ p^2=\frac{a^2b^2}{a^2+b^2}.
\displaystyle \Rightarrow\frac{1}{p^2} =\frac{a^2+b^2}{a^2b^2} =\frac{1}{a^2}+\frac{1}{b^2}.
\displaystyle \therefore\ \text{the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The equation of the line passing through }(1,5)\text{ and perpendicular}
\displaystyle \text{to the line }3x-5y+7=0\text{ is}
\displaystyle \text{(a) }5x+3y-20=0\qquad \text{(b) }3x-5y+7=0
\displaystyle \text{(c) }3x-5y+6=0\qquad \text{(d) }5x+3y+7=0
\displaystyle \text{Answer:}
\displaystyle 3x-5y+7=0\Rightarrow y=\frac35x+\frac75.
\displaystyle \therefore\ \text{slope of the given line}=\frac35.
\displaystyle \text{Slope of the perpendicular line}=-\frac53.
\displaystyle \therefore\ y-5=-\frac53(x-1).
\displaystyle \Rightarrow 3y-15=-5x+5.
\displaystyle \Rightarrow 5x+3y-20=0.
\displaystyle \text{Verification: }5(1)+3(5)-20=0.
\displaystyle \therefore\ \text{the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The figure formed by the lines }a|x|\pm b|y|\pm c=0\text{ is}
\displaystyle \text{(a) a rectangle}\qquad\text{(b) a square}\qquad \text{(c) a rhombus}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{The four lines are }a|x|+b|y|\pm c=0 \text{ and }a|x|-b|y|\pm c=0.
\displaystyle \text{Equivalently, their sides have slopes }\frac{a}{b} \text{ and }-\frac{a}{b}.
\displaystyle \text{The figure is symmetric about both coordinate axes, and its vertices are}
\displaystyle \left(\frac{c}{a},0\right),\left(0,\frac{c}{b}\right), \left(-\frac{c}{a},0\right),\left(0,-\frac{c}{b}\right).
\displaystyle \text{Each side has length } \sqrt{\frac{c^2}{a^2}+\frac{c^2}{b^2}}.
\displaystyle \therefore\ \text{all four sides are equal, so the figure is a rhombus.}
\displaystyle \therefore\ \text{the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Two vertices of a triangle are }(2,-1)\text{ and }(3,2)\text{ and third vertex lies}
\displaystyle \text{on the line }x+y=5.\text{ If the area of the triangle is }4\text{ square units, then the third vertex is}
\displaystyle \text{(a) }(0,5)\text{ or }(4,1)\qquad \text{(b) }(5,0)\text{ or }(1,4)
\displaystyle \text{(c) }(5,0)\text{ or }(4,1)\qquad \text{(d) }(0,5)\text{ or }(1,4)
\displaystyle \text{Answer:}
\displaystyle \text{Let the third vertex be }C(x,y).
\displaystyle \text{Since }C\text{ lies on }x+y=5,\quad y=5-x.
\displaystyle \text{Area}=\frac12\left|2(2-y)+3(y+1)+x(-1-2)\right|=4.
\displaystyle \Rightarrow |4-2y+3y+3-3x|=8.
\displaystyle \Rightarrow |7+y-3x|=8.
\displaystyle \text{Substituting }y=5-x,
\displaystyle |12-4x|=8.
\displaystyle \Rightarrow 12-4x=\pm8.
\displaystyle \Rightarrow x=1\text{ or }5.
\displaystyle \therefore\ y=4\text{ or }0.
\displaystyle \therefore\ C=(1,4)\text{ or }(5,0).
\displaystyle \text{Verification: both points satisfy }x+y=5 \text{ and give area }4\text{ square units}.
\displaystyle \therefore\ \text{the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The inclination of the straight line passing through the point }(-3,6)\text{ and mid-point}
\displaystyle \text{of the line joining the point }(4,-5)\text{ and }(-2,9)\text{ is}
\displaystyle \text{(a) }\frac{\pi}{4}\qquad \text{(b) }\frac{\pi}{6}\qquad \text{(c) }\frac{\pi}{3}\qquad \text{(d) }\frac{3\pi}{4}
\displaystyle \text{Answer:}
\displaystyle \text{Mid-point of }(4,-5)\text{ and }(-2,9) =\left(\frac{4-2}{2},\frac{-5+9}{2}\right)=(1,2).
\displaystyle \therefore\ \text{slope of the line through }(-3,6)\text{ and }(1,2)
\displaystyle m=\frac{2-6}{1-(-3)}=\frac{-4}{4}=-1.
\displaystyle \text{If }\theta\text{ is the inclination, then }\tan\theta=-1.
\displaystyle \text{Since }0\leq\theta<\pi,\quad\theta=\frac{3\pi}{4}.
\displaystyle \text{Verification: }\tan\frac{3\pi}{4}=-1.
\displaystyle \therefore\ \text{the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Distance between the lines }5x+3y-7=0\text{ and }15x+9y+14=0\text{ is}
\displaystyle \text{(a) }\frac{35}{\sqrt{34}}\qquad \text{(b) }\frac{1}{3\sqrt{34}}\qquad \text{(c) }\frac{35}{3\sqrt{34}}\qquad \text{(d) }\frac{35}{2\sqrt{34}}
\displaystyle \text{Answer:}
\displaystyle 15x+9y+14=0\Rightarrow 5x+3y+\frac{14}{3}=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|-7-\frac{14}{3}\right|}{\sqrt{5^2+3^2}} =\frac{\frac{35}{3}}{\sqrt{34}} =\frac{35}{3\sqrt{34}}.
\displaystyle \text{Verification: the two lines have proportional }x\text{ and }y \text{ coefficients, so they are parallel.}
\displaystyle \therefore\ \text{the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The angle between the lines }2x-y+3=0\text{ and }x+2y+3=0\text{ is}
\displaystyle \text{(a) }90^\circ\qquad\text{(b) }60^\circ\qquad \text{(c) }45^\circ\qquad\text{(d) }30^\circ
\displaystyle \text{Answer:}
\displaystyle \text{For }2x-y+3=0,\quad m_1=2.
\displaystyle \text{For }x+2y+3=0,\quad m_2=-\frac12.
\displaystyle m_1m_2=2\left(-\frac12\right)=-1.
\displaystyle \therefore\ \text{the two lines are perpendicular.}
\displaystyle \therefore\ \theta=90^\circ.
\displaystyle \text{Verification: }m_1m_2=-1\Rightarrow\theta=90^\circ.
\displaystyle \therefore\ \text{the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{The value of }\lambda\text{ for which the lines }3x+4y=5,\ 5x+4y=4
\displaystyle \text{and }\lambda x+4y=6\text{ meet at a point is}
\displaystyle \text{(a) }2\qquad\text{(b) }1\qquad \text{(c) }4\qquad\text{(d) }3
\displaystyle \text{Answer:}
\displaystyle 3x+4y=5\quad\text{and}\quad5x+4y=4.
\displaystyle \text{Subtracting the first equation from the second,}
\displaystyle 2x=-1\Rightarrow x=-\frac12.
\displaystyle 3\left(-\frac12\right)+4y=5 \Rightarrow4y=\frac{13}{2}\Rightarrow y=\frac{13}{8}.
\displaystyle \text{Since }\lambda x+4y=6\text{ passes through this point,}
\displaystyle \lambda\left(-\frac12\right) +4\left(\frac{13}{8}\right)=6.
\displaystyle -\frac{\lambda}{2}+\frac{13}{2}=6.
\displaystyle \Rightarrow-\lambda+13=12\Rightarrow\lambda=1.
\displaystyle \text{Verification: }-\frac12+\frac{13}{2}=6.
\displaystyle \therefore\ \text{the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Three vertices of a parallelogram taken in order are }(-1,-6),(2,-5)\text{ and }(7,2).
\displaystyle \text{The fourth vertex is}
\displaystyle \text{(a) }(1,4)\qquad\text{(b) }(4,1)\qquad \text{(c) }(1,1)\qquad\text{(d) }(4,4)
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=(-1,-6),\ B=(2,-5),\ C=(7,2)\text{ and }D=(x,y).
\displaystyle \text{Since the vertices are taken in order, the diagonals }AC\text{ and }BD\text{ bisect each other.}
\displaystyle \text{Mid-point of }AC=\left(\frac{-1+7}{2},\frac{-6+2}{2}\right)=(3,-2).
\displaystyle \therefore\ \left(\frac{2+x}{2},\frac{-5+y}{2}\right)=(3,-2).
\displaystyle \Rightarrow 2+x=6,\quad -5+y=-4.
\displaystyle \Rightarrow x=4,\quad y=1.
\displaystyle \text{Verification: mid-point of }BD =\left(\frac{2+4}{2},\frac{-5+1}{2}\right)=(3,-2).
\displaystyle \therefore\ \text{the fourth vertex is }(4,1).
\displaystyle \therefore\ \text{the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{The centroid of a triangle is }(2,7)\text{ and two of its vertices are }(4,8)
\displaystyle \text{and }(-2,6).\text{ The third vertex is}
\displaystyle \text{(a) }(0,0)\qquad\text{(b) }(4,7)\qquad \text{(c) }(7,4)\qquad\text{(d) }(7,7)
\displaystyle \text{Answer:}
\displaystyle \text{Let the third vertex be }(x,y).
\displaystyle \text{Using the centroid formula,}
\displaystyle \left(\frac{4-2+x}{3},\frac{8+6+y}{3}\right)=(2,7).
\displaystyle \therefore\ \frac{x+2}{3}=2,\qquad\frac{y+14}{3}=7.
\displaystyle \Rightarrow x=4,\qquad y=7.
\displaystyle \text{Verification: }\left(\frac{4-2+4}{3}, \frac{8+6+7}{3}\right)=(2,7).
\displaystyle \therefore\ \text{the third vertex is }(4,7).
\displaystyle \therefore\ \text{the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If the lines }x+q=0,\ y-2=0\text{ and }3x+2y+5=0\text{ are concurrent,}
\displaystyle \text{then the value of }q\text{ will be}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad \text{(c) }3\qquad\text{(d) }5
\displaystyle \text{Answer:}
\displaystyle x+q=0\Rightarrow x=-q,\qquad y-2=0\Rightarrow y=2.
\displaystyle \text{Since the three lines are concurrent, }(-q,2)\text{ lies on }3x+2y+5=0.
\displaystyle \therefore\ 3(-q)+2(2)+5=0.
\displaystyle \Rightarrow -3q+9=0\Rightarrow q=3.
\displaystyle \text{Verification: }3(-3)+2(2)+5=-9+4+5=0.
\displaystyle \therefore\ \text{the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{The medians }AD\text{ and }BE\text{ of a triangle with vertices }A(0,b),B(0,0)
\displaystyle \text{and }C(a,0)\text{ are perpendicular to each other, if}
\displaystyle \text{(a) }a=\frac{b}{2}\qquad \text{(b) }b=\frac{a}{2}\qquad \text{(c) }ab=1\qquad \text{(d) }a=\pm\sqrt2\,b
\displaystyle \text{Answer:}
\displaystyle D\text{ is the mid-point of }BC,\quad D=\left(\frac{a}{2},0\right).
\displaystyle E\text{ is the mid-point of }AC,\quad E=\left(\frac{a}{2},\frac{b}{2}\right).
\displaystyle \text{Slope of }AD= \frac{0-b}{\frac{a}{2}-0}=-\frac{2b}{a}.
\displaystyle \text{Slope of }BE= \frac{\frac{b}{2}-0}{\frac{a}{2}-0}=\frac{b}{a}.
\displaystyle \text{Since }AD\perp BE,\quad \left(-\frac{2b}{a}\right)\left(\frac{b}{a}\right)=-1.
\displaystyle \Rightarrow \frac{2b^2}{a^2}=1 \Rightarrow a^2=2b^2.
\displaystyle \Rightarrow a=\pm\sqrt2\,b.
\displaystyle \text{Verification: for }a^2=2b^2,\quad m_{AD}m_{BE}=-\frac{2b^2}{a^2}=-1.
\displaystyle \therefore\ \text{the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{The equation of the line with slope }-\frac32\text{ and which is concurrent with the lines}
\displaystyle 4x+3y-7=0\text{ and }8x+5y-1=0\text{ is}
\displaystyle \text{(a) }3x+2y-63=0\qquad \text{(b) }3x+2y-2=0
\displaystyle \text{(c) }2y-3x-2=0\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 4x+3y-7=0,\qquad 8x+5y-1=0.
\displaystyle \text{Multiplying the first equation by }2,
\displaystyle 8x+6y-14=0.
\displaystyle \text{Subtracting }8x+5y-1=0,\quad y-13=0 \Rightarrow y=13.
\displaystyle 4x+3(13)-7=0\Rightarrow4x+32=0\Rightarrow x=-8.
\displaystyle \therefore\ \text{the point of concurrence is }(-8,13).
\displaystyle \text{The required line has slope }-\frac32.
\displaystyle y-13=-\frac32(x+8).
\displaystyle \Rightarrow 2y-26=-3x-24.
\displaystyle \Rightarrow 3x+2y-2=0.
\displaystyle \text{Verification: }3(-8)+2(13)-2=-24+26-2=0.
\displaystyle \therefore\ \text{the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{The vertices of a triangle are }(6,0),(0,6)\text{ and }(6,6).\text{ The distance}
\displaystyle \text{between its circumcentre and centroid is}
\displaystyle \text{(a) }2\sqrt2\qquad\text{(b) }2\qquad \text{(c) }\sqrt2\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{The triangle is right-angled at }(6,6).
\displaystyle \text{Hence, its circumcentre is the mid-point of the hypotenuse joining }(6,0)\text{ and }(0,6).
\displaystyle \therefore\ O=\left(\frac{6+0}{2},\frac{0+6}{2}\right)=(3,3).
\displaystyle \text{The centroid is}
\displaystyle G=\left(\frac{6+0+6}{3},\frac{0+6+6}{3}\right)=(4,4).
\displaystyle OG=\sqrt{(4-3)^2+(4-3)^2}=\sqrt{1+1}=\sqrt2.
\displaystyle \text{Verification: }O=(3,3),\ G=(4,4) \Rightarrow OG=\sqrt2.
\displaystyle \therefore\ \text{the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{A point equidistant from the line }4x+3y+10=0,\ 
\displaystyle 5x-12y+26=0 \ \text{and }7x+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{We verify the distances of }(1,-1)\text{ from the three lines.}
\displaystyle d_1=\frac{|4(1)+3(-1)+10|}{\sqrt{4^2+3^2}} =\frac{11}{5}.
\displaystyle d_2=\frac{|5(1)-12(-1)+26|}{\sqrt{5^2+(-12)^2}} =\frac{43}{13}.
\displaystyle d_3=\frac{|7(1)+24(-1)-50|}{\sqrt{7^2+24^2}} =\frac{67}{25}.
\displaystyle \text{These distances are not equal, so option (a) is not correct.}
\displaystyle \text{Checking }(1,1),
\displaystyle d_1=\frac{|4+3+10|}{5}=\frac{17}{5},\quad d_2=\frac{|5-12+26|}{13}=\frac{19}{13},\quad d_3=\frac{|7+24-50|}{25}=\frac{19}{25}.
\displaystyle \text{Hence, option (b) is also not correct.}
\displaystyle \text{Checking }(0,0),
\displaystyle d_1=\frac{10}{5}=2,\quad d_2=\frac{26}{13}=2,\quad d_3=\frac{50}{25}=2.
\displaystyle \therefore\ (0,0)\text{ is equidistant from all three lines.}
\displaystyle \therefore\ \text{the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{The ratio in which the line }3x+4y+2=0\text{ divides the distance between}
\displaystyle \text{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.}
\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}}=\frac35.
\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}}=\frac75.
\displaystyle \therefore\ d_1:d_2=\frac35:\frac75=3:7.
\displaystyle \text{Verification: }\frac35+\frac75=2, \text{ the distance between the two outer lines.}
\displaystyle \therefore\ \text{the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 34: }\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 x+y-11=0\Rightarrow y=-x+11.
\displaystyle \therefore\ \text{slope of the given line}=-1.
\displaystyle \text{Hence, slope of the perpendicular line}=1.
\displaystyle \text{Equation of the perpendicular through }(2,3)\text{ is}
\displaystyle y-3=x-2\Rightarrow y=x+1.
\displaystyle \text{At the foot of the perpendicular, }x+(x+1)-11=0.
\displaystyle \Rightarrow 2x=10\Rightarrow x=5,\quad y=6.
\displaystyle \text{Verification: }5+6-11=0.
\displaystyle \therefore\ \text{the foot of the perpendicular is }(5,6).
\displaystyle \therefore\ \text{the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{The reflection of the point }(4,-13)\text{ about the line } \\ 5x+y+6=0\text{ is}
\displaystyle \text{(a) }(-1,-14)\qquad\text{(b) }(3,4)\qquad \text{(c) }(0,0)\qquad\text{(d) }(1,2)
\displaystyle \text{Answer:}
\displaystyle \text{Let }P=(4,-13)\text{ and let its reflection be }P'(x',y').
\displaystyle \text{For }ax+by+c=0,\text{ the reflection of }(x_1,y_1)\text{ is given by}
\displaystyle x'=x_1-\frac{2a(ax_1+by_1+c)}{a^2+b^2},\qquad y'=y_1-\frac{2b(ax_1+by_1+c)}{a^2+b^2}.
\displaystyle \text{Here }a=5,\ b=1,\ c=6,\quad ax_1+by_1+c=20-13+6=13.
\displaystyle x'=4-\frac{2(5)(13)}{25+1} =4-\frac{130}{26}=-1.
\displaystyle y'=-13-\frac{2(1)(13)}{26} =-13-1=-14.
\displaystyle \text{Verification: the midpoint of }(4,-13)\text{ and }(-1,-14) \text{ is }\left(\frac32,-\frac{27}{2}\right).
\displaystyle 5\left(\frac32\right)-\frac{27}{2}+6=0, \text{ so the midpoint lies on the given line.}
\displaystyle \therefore\ \text{the reflection is }(-1,-14).
\displaystyle \therefore\ \text{the correct option is (a).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Write an equation representing a pair of lines through the point }(a,b)
\displaystyle \text{and parallel to the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The line through }(a,b)\text{ parallel to the }y\text{-axis is }x=a.
\displaystyle \text{The line through }(a,b)\text{ parallel to the }x\text{-axis is }y=b.
\displaystyle \therefore\ (x-a)(y-b)=0.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the coordinates of the orthocentre of the triangle formed by the lines}
\displaystyle x^2-y^2=0\text{ and }x+6y=18.
\displaystyle \text{Answer:}
\displaystyle x^2-y^2=0\Rightarrow(x-y)(x+y)=0.
\displaystyle \therefore\ x-y=0\text{ and }x+y=0.
\displaystyle \text{These two lines intersect at }O(0,0)\text{ and are perpendicular to each other.}
\displaystyle \therefore\ \text{the triangle is right-angled at }O.
\displaystyle \text{The orthocentre of a right-angled triangle is its right-angled vertex.}
\displaystyle \therefore\ \text{the coordinates of the orthocentre are }(0,0).
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the centroid of a triangle formed by the points }(0,0),
\displaystyle (\cos\theta,\sin\theta)\text{ and }(\sin\theta,-\cos\theta)\text{ lies on the line }y=2x,
\displaystyle \text{then write the value of }\tan\theta.
\displaystyle \text{Answer:}
\displaystyle \text{The coordinates of the centroid are}
\displaystyle \left(\frac{\cos\theta+\sin\theta}{3}, \frac{\sin\theta-\cos\theta}{3}\right).
\displaystyle \text{Since the centroid lies on }y=2x,
\displaystyle \frac{\sin\theta-\cos\theta}{3} =2\left(\frac{\cos\theta+\sin\theta}{3}\right).
\displaystyle \sin\theta-\cos\theta =2\cos\theta+2\sin\theta.
\displaystyle \therefore\ \sin\theta+3\cos\theta=0.
\displaystyle \therefore\ \tan\theta=-3.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Write the value of }\theta\in\left(0,\frac{\pi}{2}\right)\text{ for which area of the triangle}
\displaystyle \text{formed by points }O(0,0),\ A(a\cos\theta,b\sin\theta)\text{ and}
\displaystyle B(a\cos\theta,-b\sin\theta)\text{ is maximum.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle OAB =\frac{1}{2}\left|a\cos\theta(-b\sin\theta)-a\cos\theta(b\sin\theta)\right|.
\displaystyle =ab\sin\theta\cos\theta =\frac{ab}{2}\sin2\theta.
\displaystyle \text{The area is maximum when }\sin2\theta=1.
\displaystyle \therefore\ 2\theta=\frac{\pi}{2} \Rightarrow\theta=\frac{\pi}{4}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Write the distance between the lines }4x+3y-11=0
\displaystyle \text{and }8x+6y-15=0.
\displaystyle \text{Answer:}
\displaystyle 8x+6y-15=0\Rightarrow4x+3y-\frac{15}{2}=0.
\displaystyle \text{The distance between }ax+by+c_1=0\text{ and }ax+by+c_2=0\text{ is}
\displaystyle d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}.
\displaystyle \therefore\ d =\frac{\left|-11+\frac{15}{2}\right|}{\sqrt{4^2+3^2}} =\frac{\frac{7}{2}}{5} =\frac{7}{10}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Write the coordinates of the orthocentre of the triangle formed by the lines}
\displaystyle xy=0\text{ and }x+y=1.
\displaystyle \text{Answer:}
\displaystyle xy=0\Rightarrow x=0\text{ or }y=0.
\displaystyle \text{Thus, the three lines forming the triangle are }x=0,\ y=0\text{ and }x+y=1.
\displaystyle x=0\text{ and }y=0\text{ are perpendicular and intersect at }(0,0).
\displaystyle \therefore\ \text{the triangle is right-angled at }(0,0).
\displaystyle \text{Hence, its orthocentre is }(0,0).
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If the lines }x+ay+a=0,\ bx+y+b=0\text{ and}
\displaystyle cx+cy+1=0\text{ are concurrent, then write the value of }
\displaystyle 2abc-ab-bc-ca.
\displaystyle \text{Answer:}
\displaystyle \text{For three lines }a_1x+b_1y+c_1=0,\ a_2x+b_2y+c_2=0
\displaystyle \text{and }a_3x+b_3y+c_3=0\text{ to be concurrent,}
\displaystyle \begin{vmatrix}1&a&a\\ b&1&b\\ c&c&1\end{vmatrix}=0.
\displaystyle 1(1-bc)-a(b-bc)+a(bc-c)=0.
\displaystyle 1-bc-ab+abc+abc-ac=0.
\displaystyle \therefore\ 1+2abc-ab-bc-ca=0.
\displaystyle \therefore\ 2abc-ab-bc-ca=-1.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write the area of the triangle formed by the coordinate axes and the line}
\displaystyle (\sec\theta-\tan\theta)x+(\sec\theta+\tan\theta)y=2.
\displaystyle \text{Answer:}
\displaystyle \text{Putting }y=0,\text{ the }x\text{-intercept is } \frac{2}{\sec\theta-\tan\theta}.
\displaystyle \text{Putting }x=0,\text{ the }y\text{-intercept is } \frac{2}{\sec\theta+\tan\theta}.
\displaystyle \therefore\ \text{Area} =\frac{1}{2}\left(\frac{2}{\sec\theta-\tan\theta}\right) \left(\frac{2}{\sec\theta+\tan\theta}\right).
\displaystyle =\frac{2}{\sec^2\theta-\tan^2\theta}=2.
\displaystyle \therefore\ \text{the area of the triangle is }2\text{ sq. units}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If the diagonals of the quadrilateral formed by the lines}
\displaystyle l_1x+m_1y+n_1=0,\quad l_2x+m_2y+n_2=0,
\displaystyle l_1x+m_1y+n_1'=0\text{ and }l_2x+m_2y+n_2'=0\text{ are perpendicular,}
\displaystyle \text{then write the value of }l_1^2-l_2^2+m_1^2-m_2^2.
\displaystyle \text{Answer:}
\displaystyle \text{The two pairs of opposite sides are parallel, so the quadrilateral is a parallelogram.}
\displaystyle \text{A parallelogram whose diagonals are perpendicular is a rhombus.}
\displaystyle \therefore\ \text{the perpendicular distances between the two pairs of opposite sides are equal.}
\displaystyle \frac{|n_1-n_1'|}{\sqrt{l_1^2+m_1^2}} =\frac{|n_2-n_2'|}{\sqrt{l_2^2+m_2^2}}.
\displaystyle \text{For the diagonals to be perpendicular, the adjacent side normals have equal magnitudes,}
\displaystyle l_1^2+m_1^2=l_2^2+m_2^2.
\displaystyle \therefore\ l_1^2-l_2^2+m_1^2-m_2^2=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Write the coordinates of the image of the point }(3,8)\text{ in the line}
\displaystyle x+3y-7=0.
\displaystyle \text{Answer:}
\displaystyle \text{For the image of }(x_1,y_1)\text{ in }ax+by+c=0,
\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b} =-\frac{2(ax_1+by_1+c)}{a^2+b^2}.
\displaystyle \text{Here, }a=1,\ b=3,\ c=-7,\ x_1=3,\ y_1=8.
\displaystyle -\frac{2(3+24-7)}{1^2+3^2}=-\frac{40}{10}=-4.
\displaystyle \therefore\ x-3=-4\Rightarrow x=-1,
\displaystyle y-8=3(-4)=-12\Rightarrow y=-4.
\displaystyle \therefore\ \text{the coordinates of the image are }(-1,-4).
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Write the integral values of }m\text{ for which the }x\text{-coordinate of the}
\displaystyle \text{point of intersection of the lines }y=mx+1\text{ and }3x+4y=9\text{ is an integer.}
\displaystyle \text{Answer:}
\displaystyle \text{Substituting }y=mx+1\text{ in }3x+4y=9,
\displaystyle 3x+4(mx+1)=9.
\displaystyle (4m+3)x=5\Rightarrow x=\frac{5}{4m+3}.
\displaystyle \text{For }x\text{ to be an integer, }4m+3\text{ must be an integral divisor of }5.
\displaystyle \therefore\ 4m+3=\pm1,\ \pm5.
\displaystyle 4m+3=1\Rightarrow m=-\frac12,\qquad 4m+3=-1\Rightarrow m=-1.
\displaystyle 4m+3=5\Rightarrow m=\frac12,\qquad 4m+3=-5\Rightarrow m=-2.
\displaystyle \text{Since }m\text{ must be an integer, }m=-1\text{ or }m=-2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }a\ne b\ne c,\text{ write the condition for which the equations}
\displaystyle (b-c)x+(c-a)y+(a-b)=0\text{ and}
\displaystyle (b^3-c^3)x+(c^3-a^3)y+(a^3-b^3)=0\text{ represent the same line.}
\displaystyle \text{Answer:}
\displaystyle \text{For the two equations to represent the same line,}
\displaystyle \frac{b^3-c^3}{b-c}=\frac{c^3-a^3}{c-a} =\frac{a^3-b^3}{a-b}.
\displaystyle \therefore\ b^2+bc+c^2=c^2+ca+a^2=a^2+ab+b^2.
\displaystyle b^2+bc+c^2=c^2+ca+a^2
\displaystyle \Rightarrow b^2-a^2+c(b-a)=0.
\displaystyle \Rightarrow(b-a)(a+b+c)=0.
\displaystyle \text{Since }a\ne b,\quad a+b+c=0.
\displaystyle \therefore\ \text{the required condition is }a+b+c=0.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }a,b,c\text{ are in G.P. write the area of the triangle formed by the line}
\displaystyle ax+by+c=0\text{ with the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle ax+by+c=0.
\displaystyle \text{Putting }y=0,\quad x=-\frac{c}{a}.
\displaystyle \text{Putting }x=0,\quad y=-\frac{c}{b}.
\displaystyle \therefore\ \text{Area} =\frac12\left|\left(-\frac{c}{a}\right)\left(-\frac{c}{b}\right)\right|.
\displaystyle =\left|\frac{c^2}{2ab}\right|.
\displaystyle \text{Since }a,b,c\text{ are in G.P., }b^2=ac,
\displaystyle \text{which is consistent with the required intercepts.}
\displaystyle \therefore\ \text{Area}=\left|\frac{c^2}{2ab}\right|\text{ sq. units}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Write the area of the figure formed by the lines }a|x|+b|y|+c=0.
\displaystyle \text{Answer:}
\displaystyle a|x|+b|y|=-c.
\displaystyle \text{For }y=0,\quad |x|=-\frac{c}{a}.
\displaystyle \text{For }x=0,\quad |y|=-\frac{c}{b}.
\displaystyle \text{Thus, the lengths of the diagonals of the figure are}
\displaystyle 2\left|\frac{c}{a}\right|\text{ and }2\left|\frac{c}{b}\right|.
\displaystyle \therefore\ \text{Area}=\frac12\times2\left|\frac{c}{a}\right| \times2\left|\frac{c}{b}\right|.
\displaystyle \therefore\ \text{Area}=\frac{2c^2}{|ab|}\text{ sq. units}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Write the locus of a point the sum of whose distances from the}
\displaystyle \text{coordinate axes is unity.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ be the required point.}
\displaystyle \text{Its distances from the }x\text{-axis and }y\text{-axis are }|y|\text{ and }|x|.
\displaystyle \therefore\ |x|+|y|=1.
\displaystyle \text{In the four quadrants, this represents the lines}
\displaystyle x+y=1,\quad -x+y=1,\quad -x-y=1,\quad x-y=1.
\displaystyle \text{These lines form a square with vertices}
\displaystyle (1,0),\ (0,1),\ (-1,0),\ (0,-1).
\displaystyle \therefore\ \text{the required locus is a square.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }a,b,c\text{ are in A.P., then the line }ax+by+c=0\text{ passes}
\displaystyle \text{through a fixed point. Write the coordinates of that point.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }a,b,c\text{ are in A.P.,}
\displaystyle 2b=a+c\quad\Rightarrow\quad c=2b-a.
\displaystyle \text{Substituting in }ax+by+c=0,
\displaystyle ax+by+2b-a=0.
\displaystyle \Rightarrow a(x-1)+b(y+2)=0.
\displaystyle \text{For this line to pass through a fixed point,}
\displaystyle x-1=0,\qquad y+2=0.
\displaystyle \therefore\ x=1,\qquad y=-2.
\displaystyle \therefore\ \text{the fixed point is }(1,-2).
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Write the equation of the line passing through the point }(1,-2)
\displaystyle \text{and cutting off equal intercepts from the axes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the equal intercepts on the axes be }k.
\displaystyle \text{Using the intercept form,}
\displaystyle \frac{x}{k}+\frac{y}{k}=1.
\displaystyle \Rightarrow x+y=k.
\displaystyle \text{Since the line passes through }(1,-2),
\displaystyle 1-2=k\quad\Rightarrow\quad k=-1.
\displaystyle \therefore\ x+y=-1.
\displaystyle \therefore\ x+y+1=0.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Find the locus of the mid-points of the portion of the line}
\displaystyle x\sin\theta+y\cos\theta=p\text{ intercepted between the axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }x\sin\theta+y\cos\theta=p.
\displaystyle \text{Its }x\text{-intercept is }p\mathrm{cosec}\theta\text{ and }y\text{-intercept is }p\sec\theta.
\displaystyle \text{Let }(h,k)\text{ be the mid-point of the portion intercepted between the axes.}
\displaystyle h=\frac{p}{2}\mathrm{cosec}\theta,\qquad k=\frac{p}{2}\sec\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 required locus is}
\displaystyle \frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}.
\displaystyle \\

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