\displaystyle \textbf{Note:}
\displaystyle \text{If }ax+by+c=0\text{ is a line, then the equation of any line perpendicular}
\displaystyle \text{to it is }bx-ay+\lambda=0,\text{ where }\lambda\text{ is a constant.}
\displaystyle \text{If }ax+by+c=0\text{ is a line, then the equation of any line parallel}
\displaystyle \text{to it is }ax+by+\lambda=0,\text{ where }\lambda\text{ is a constant.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find the equation of a line passing through the point }(2,3)
\displaystyle \text{and parallel to the line }3x-4y+5=0.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }3x-4y+5=0.
\displaystyle \text{A line parallel to the given line is}
\displaystyle 3x-4y+\lambda=0,\text{ where }\lambda\text{ is a constant.}
\displaystyle \text{Since it passes through }(2,3),
\displaystyle 3(2)-4(3)+\lambda=0.
\displaystyle \Rightarrow 6-12+\lambda=0.
\displaystyle \Rightarrow \lambda=6.
\displaystyle \therefore \text{The required equation is }3x-4y+6=0.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the equation of a line passing through }(3,-2)\text{ and perpendicular}
\displaystyle \text{to the line }x-3y+5=0.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }x-3y+5=0.
\displaystyle \text{Comparing it with }ax+by+c=0,\text{ we get }a=1\text{ and }b=-3.
\displaystyle \text{Therefore, a line perpendicular to the given line is}
\displaystyle -3x-y+\lambda=0.
\displaystyle \text{Equivalently, }3x+y-\lambda=0.
\displaystyle \text{Since it passes through }(3,-2),
\displaystyle 3(3)+(-2)-\lambda=0.
\displaystyle \Rightarrow 7-\lambda=0.
\displaystyle \Rightarrow \lambda=7.
\displaystyle \therefore \text{The required equation is }3x+y-7=0.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the equation of the perpendicular bisector of the line joining}
\displaystyle \text{the points }(1,3)\text{ and }(3,1).
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }A(1,3)\text{ and }B(3,1).
\displaystyle \text{The midpoint of }AB\text{ is}
\displaystyle C=\left(\frac{1+3}{2},\frac{3+1}{2}\right)=(2,2).
\displaystyle \text{The equation of }AB\text{ is}
\displaystyle \frac{x-1}{3-1}=\frac{y-3}{1-3}.
\displaystyle \Rightarrow x+y-4=0.
\displaystyle \text{Comparing with }ax+by+c=0,\text{ we get }a=1,\ b=1.
\displaystyle \text{Hence, the perpendicular bisector is of the form}
\displaystyle x-y+\lambda=0.
\displaystyle \text{Since it passes through }(2,2),
\displaystyle 2-2+\lambda=0.
\displaystyle \Rightarrow \lambda=0.
\displaystyle \therefore \text{The equation of the perpendicular bisector is }x-y=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the equations of the altitudes of }\triangle ABC\text{ whose vertices are}
\displaystyle A(1,4),\quad B(-3,2)\text{ and }C(-5,-3).
\displaystyle \textbf{Answer:}
\displaystyle \text{Please refer to the adjoining figure.}
\displaystyle \text{First, we find the equation of }BC.
\displaystyle \frac{x+3}{-5+3}=\frac{y-2}{-3-2}
\displaystyle \Rightarrow \frac{x+3}{-2}=\frac{y-2}{-5}
\displaystyle \Rightarrow 5(x+3)=2(y-2)
\displaystyle \Rightarrow 5x-2y+19=0.
\displaystyle \text{A line perpendicular to }5x-2y+19=0\text{ is of the form}
\displaystyle -2x-5y+\lambda=0.
\displaystyle \text{Since the altitude from }A\text{ passes through }A(1,4),
\displaystyle -2(1)-5(4)+\lambda=0.
\displaystyle \Rightarrow \lambda=22.
\displaystyle \therefore AD:-2x-5y+22=0.
\displaystyle \text{Equivalently, }AD:2x+5y-22=0.
\displaystyle \text{Next, we find the equation of }AC.
\displaystyle \frac{x-1}{-5-1}=\frac{y-4}{-3-4}
\displaystyle \Rightarrow \frac{x-1}{-6}=\frac{y-4}{-7}
\displaystyle \Rightarrow 7(x-1)=6(y-4)
\displaystyle \Rightarrow 7x-6y+17=0.
\displaystyle \text{A line perpendicular to }7x-6y+17=0\text{ is of the form}
\displaystyle -6x-7y+\lambda=0.
\displaystyle \text{Since the altitude from }B\text{ passes through }B(-3,2),
\displaystyle -6(-3)-7(2)+\lambda=0.
\displaystyle \Rightarrow 18-14+\lambda=0.
\displaystyle \Rightarrow \lambda=-4.
\displaystyle \therefore BE:-6x-7y-4=0.
\displaystyle \text{Equivalently, }BE:6x+7y+4=0.
\displaystyle \text{Finally, we find the equation of }AB.
\displaystyle \frac{x-1}{-3-1}=\frac{y-4}{2-4}
\displaystyle \Rightarrow \frac{x-1}{-4}=\frac{y-4}{-2}
\displaystyle \Rightarrow 2(x-1)=4(y-4)
\displaystyle \Rightarrow x-2y+7=0.
\displaystyle \text{A line perpendicular to }x-2y+7=0\text{ is of the form}
\displaystyle -2x-y+\lambda=0.
\displaystyle \text{Since the altitude from }C\text{ passes through }C(-5,-3),
\displaystyle -2(-5)-(-3)+\lambda=0.
\displaystyle \Rightarrow 13+\lambda=0.
\displaystyle \Rightarrow \lambda=-13.
\displaystyle \therefore CF:-2x-y-13=0.
\displaystyle \text{Equivalently, }CF:2x+y+13=0.
\displaystyle \therefore \text{The equations of the altitudes are}
\displaystyle AD:2x+5y-22=0,\quad BE:6x+7y+4=0
\displaystyle \text{and }CF:2x+y+13=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the equation of a line which is perpendicular to the line}
\displaystyle \sqrt{3}x-y+5=0\text{ and which cuts off an intercept of }4\text{ units on the}
\displaystyle \text{negative direction of the }y\text{-axis.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }\sqrt{3}x-y+5=0.
\displaystyle \text{Comparing it with }ax+by+c=0,\text{ we get }a=\sqrt{3}\text{ and }b=-1.
\displaystyle \text{Therefore, a line perpendicular to the given line is}
\displaystyle -x-\sqrt{3}y+\lambda=0.
\displaystyle \text{Since it cuts off an intercept of }4\text{ units on the negative }y\text{-axis,}
\displaystyle \text{it passes through }(0,-4).
\displaystyle -0-\sqrt{3}(-4)+\lambda=0.
\displaystyle \Rightarrow 4\sqrt{3}+\lambda=0.
\displaystyle \Rightarrow \lambda=-4\sqrt{3}.
\displaystyle \therefore -x-\sqrt{3}y-4\sqrt{3}=0.
\displaystyle \text{Equivalently, }x+\sqrt{3}y+4\sqrt{3}=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If the image of the point }(2,1)\text{ with respect to a line mirror is}
\displaystyle (5,2),\text{ find the equation of the mirror.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }A(2,1)\text{ and }A'(5,2).
\displaystyle \text{The midpoint of }AA'\text{ is}
\displaystyle M=\left(\frac{2+5}{2},\frac{1+2}{2}\right)=\left(\frac{7}{2},\frac{3}{2}\right).
\displaystyle \text{The equation of the line }AA'\text{ is}
\displaystyle \frac{x-2}{5-2}=\frac{y-1}{2-1}.
\displaystyle \Rightarrow x-3y+1=0.
\displaystyle \text{A line perpendicular to }x-3y+1=0\text{ is of the form}
\displaystyle 3x+y+\lambda=0.
\displaystyle \text{Since the mirror passes through }\left(\frac{7}{2},\frac{3}{2}\right),
\displaystyle 3\left(\frac{7}{2}\right)+\frac{3}{2}+\lambda=0.
\displaystyle \Rightarrow 12+\lambda=0.
\displaystyle \Rightarrow \lambda=-12.
\displaystyle \therefore \text{The equation of the mirror is }3x+y-12=0.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the equation of the straight line through the point }(\alpha,\beta)
\displaystyle \text{and perpendicular to the line }lx+my+n=0.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }lx+my+n=0.
\displaystyle \text{Comparing it with }ax+by+c=0,\text{ we get }a=l\text{ and }b=m.
\displaystyle \text{Therefore, a line perpendicular to the given line is}
\displaystyle mx-ly+\lambda=0.
\displaystyle \text{Since it passes through }(\alpha,\beta),
\displaystyle m\alpha-l\beta+\lambda=0.
\displaystyle \Rightarrow \lambda=l\beta-m\alpha.
\displaystyle \therefore mx-ly+l\beta-m\alpha=0.
\displaystyle \therefore m(x-\alpha)=l(y-\beta).
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the equation of the straight line perpendicular to }2x-3y=5
\displaystyle \text{and cutting off an intercept of }1\text{ on the positive direction of the }x\text{-axis.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }2x-3y=5.
\displaystyle \text{Comparing it with }ax+by+c=0,\text{ we get }a=2\text{ and }b=-3.
\displaystyle \text{Therefore, a line perpendicular to the given line is}
\displaystyle -3x-2y+\lambda=0.
\displaystyle \text{Since it cuts off an intercept of }1\text{ on the positive }x\text{-axis,}
\displaystyle \text{it passes through }(1,0).
\displaystyle -3(1)-2(0)+\lambda=0.
\displaystyle \Rightarrow \lambda=3.
\displaystyle \therefore -3x-2y+3=0.
\displaystyle \text{Equivalently, }3x+2y-3=0.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the equation of the straight line perpendicular to }5x-2y=8
\displaystyle \text{and which passes through the mid-point of the line segment joining}
\displaystyle (2,3)\text{ and }(4,5).
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }5x-2y=8.
\displaystyle \text{Comparing it with }ax+by+c=0,\text{ we get }a=5\text{ and }b=-2.
\displaystyle \text{Therefore, a line perpendicular to the given line is}
\displaystyle -2x-5y+\lambda=0.
\displaystyle \text{Let }A(2,3)\text{ and }B(4,5).
\displaystyle \text{The midpoint of }AB\text{ is}
\displaystyle C=\left(\frac{2+4}{2},\frac{3+5}{2}\right)=(3,4).
\displaystyle \text{Since the required line passes through }(3,4),
\displaystyle -2(3)-5(4)+\lambda=0.
\displaystyle \Rightarrow -6-20+\lambda=0.
\displaystyle \Rightarrow \lambda=26.
\displaystyle \therefore -2x-5y+26=0.
\displaystyle \text{Equivalently, }2x+5y-26=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the equation of the straight line which has }y\text{-intercept}
\displaystyle \text{equal to }\frac{4}{3}\text{ and is perpendicular to }3x-4y+11=0.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }3x-4y+11=0.
\displaystyle \text{Comparing it with }ax+by+c=0,\text{ we get }a=3\text{ and }b=-4.
\displaystyle \text{Therefore, a line perpendicular to the given line is}
\displaystyle -4x-3y+\lambda=0.
\displaystyle \text{Since its }y\text{-intercept is }\frac{4}{3},\text{ it passes through }\left(0,\frac{4}{3}\right).
\displaystyle -4(0)-3\left(\frac{4}{3}\right)+\lambda=0.
\displaystyle \Rightarrow \lambda=4.
\displaystyle \therefore -4x-3y+4=0.
\displaystyle \text{Equivalently, }4x+3y-4=0.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the equation of the right bisector of the line segment joining}
\displaystyle \text{the points }(a,b)\text{ and }(a_1,b_1).
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }A(a,b)\text{ and }B(a_1,b_1).
\displaystyle \text{The midpoint of }AB\text{ is}
\displaystyle C\left(\frac{a+a_1}{2},\frac{b+b_1}{2}\right).
\displaystyle \text{The equation of }AB\text{ is}
\displaystyle (b_1-b)x-(a_1-a)y+ab_1-a_1b=0.
\displaystyle \text{Hence, a line perpendicular to }AB\text{ is of the form}
\displaystyle (a_1-a)x+(b_1-b)y+\lambda=0.
\displaystyle \text{Since it passes through }C\left(\frac{a+a_1}{2},\frac{b+b_1}{2}\right),
\displaystyle \frac{(a_1-a)(a+a_1)}{2}+\frac{(b_1-b)(b+b_1)}{2}+\lambda=0.
\displaystyle \Rightarrow \lambda=-\frac{a_1^2-a^2+b_1^2-b^2}{2}.
\displaystyle \therefore 2(a_1-a)x+2(b_1-b)y
\displaystyle \qquad +(a^2+b^2)-(a_1^2+b_1^2)=0.
\displaystyle \therefore \text{The required right bisector is}
\displaystyle 2(a_1-a)x+2(b_1-b)y+(a^2+b^2)-(a_1^2+b_1^2)=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the image of the point }(2,1)\text{ with respect to the line}
\displaystyle x+y-5=0.
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the image of }A(2,1)\text{ be }B(a,b).
\displaystyle \text{The midpoint of }AB\text{ is}
\displaystyle C\left(\frac{2+a}{2},\frac{1+b}{2}\right).
\displaystyle \text{Since }C\text{ lies on }x+y-5=0,
\displaystyle \frac{2+a}{2}+\frac{1+b}{2}-5=0.
\displaystyle \Rightarrow a+b=7.\qquad\ldots\ldots\text{i)}
\displaystyle \text{The line }AB\text{ is perpendicular to the mirror }x+y-5=0.
\displaystyle \text{Hence }AB\text{ is of the form }x-y+\lambda=0.
\displaystyle \text{Since }A(2,1)\text{ lies on }AB,
\displaystyle 2-1+\lambda=0.
\displaystyle \Rightarrow \lambda=-1.
\displaystyle \therefore AB:x-y-1=0.
\displaystyle \text{Since }B(a,b)\text{ lies on }AB,
\displaystyle a-b=1.\qquad\ldots\ldots\text{ii)}
\displaystyle \text{From (i) and (ii),}
\displaystyle a+b=7,\qquad a-b=1.
\displaystyle \Rightarrow 2a=8\Rightarrow a=4.
\displaystyle \Rightarrow b=3.
\displaystyle \therefore \text{The image of }(2,1)\text{ is }(4,3).
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If the image of the point }(2,1)\text{ with respect to a line mirror is}
\displaystyle (5,2),\text{ find the equation of the mirror.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }A(2,1)\text{ and }A'(5,2).
\displaystyle \text{The midpoint of }AA'\text{ is}
\displaystyle M=\left(\frac{2+5}{2},\frac{1+2}{2}\right)=\left(\frac{7}{2},\frac{3}{2}\right).
\displaystyle \text{The equation of }AA'\text{ is}
\displaystyle \frac{x-2}{5-2}=\frac{y-1}{2-1}.
\displaystyle \Rightarrow x-3y+1=0.
\displaystyle \text{Since the mirror is perpendicular to }AA',\text{ its equation is of the form}
\displaystyle 3x+y+\lambda=0.
\displaystyle \text{Since the mirror passes through }M\left(\frac{7}{2},\frac{3}{2}\right),
\displaystyle 3\left(\frac{7}{2}\right)+\frac{3}{2}+\lambda=0.
\displaystyle \Rightarrow 12+\lambda=0.
\displaystyle \Rightarrow \lambda=-12.
\displaystyle \therefore \text{The equation of the mirror is }3x+y-12=0.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the equation of the straight line parallel to }3x-4y+6=0
\displaystyle \text{and passing through the midpoint of the line segment joining }(2,3)\text{ and }(4,-1).
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }3x-4y+6=0.
\displaystyle \text{Therefore, a line parallel to the given line is}
\displaystyle 3x-4y+\lambda=0.
\displaystyle \text{Let }A(2,3)\text{ and }B(4,-1).
\displaystyle \text{The midpoint of }AB\text{ is}
\displaystyle C=\left(\frac{2+4}{2},\frac{3+(-1)}{2}\right)=(3,1).
\displaystyle \text{Since the required line passes through }(3,1),
\displaystyle 3(3)-4(1)+\lambda=0.
\displaystyle \Rightarrow 9-4+\lambda=0.
\displaystyle \Rightarrow \lambda=-5.
\displaystyle \therefore \text{The equation of the required line is }3x-4y-5=0.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Prove that the lines }2x-3y+1=0,\quad x+y=3,
\displaystyle 2x-3y=2\text{ and }x+y=4\text{ form a parallelogram.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The given lines are}
\displaystyle L_1:2x-3y+1=0,\qquad L_2:x+y-3=0,
\displaystyle L_3:2x-3y-2=0,\qquad L_4:x+y-4=0.
\displaystyle \text{The lines }L_1\text{ and }L_3\text{ have the same coefficients of }x\text{ and }y
\displaystyle \text{but different constant terms. Therefore, }L_1\parallel L_3.
\displaystyle \text{Similarly, }L_2\text{ and }L_4\text{ have the same coefficients of }x\text{ and }y
\displaystyle \text{but different constant terms. Therefore, }L_2\parallel L_4.
\displaystyle \text{Also, the coefficients }2,-3\text{ are not proportional to }1,1.
\displaystyle \text{Hence, a line from the first pair intersects each line of the second pair.}
\displaystyle \text{Thus, both pairs of opposite sides are parallel.}
\displaystyle \therefore \text{The four given lines form a parallelogram.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Find the equation of a line drawn perpendicular to the line}
\displaystyle \frac{x}{4}+\frac{y}{6}=1\text{ through the point where it meets the }y\text{-axis.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }\frac{x}{4}+\frac{y}{6}=1.
\displaystyle \Rightarrow 6x+4y=24.
\displaystyle \text{The point where the line meets the }y\text{-axis is obtained by putting }x=0.
\displaystyle \Rightarrow 4y=24.
\displaystyle \Rightarrow y=6.
\displaystyle \therefore \text{The point of intersection is }(0,6).
\displaystyle \text{Comparing }6x+4y-24=0\text{ with }ax+by+c=0,
\displaystyle \text{we get }a=6\text{ and }b=4.
\displaystyle \text{Therefore, a line perpendicular to the given line is}
\displaystyle 4x-6y+\lambda=0.
\displaystyle \text{Since it passes through }(0,6),
\displaystyle 4(0)-6(6)+\lambda=0.
\displaystyle \Rightarrow \lambda=36.
\displaystyle \therefore 4x-6y+36=0.
\displaystyle \text{Equivalently, }2x-3y+18=0.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The perpendicular from the origin to the line }y=mx+c
\displaystyle \text{meets it at the point }(-1,2).\text{ Find the values of }m\text{ and }c.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }y=mx+c.\qquad\ldots\ldots\text{i)}
\displaystyle \text{The slope of the given line is }m.
\displaystyle \text{Hence, the slope of the perpendicular from the origin is }-\frac{1}{m}.
\displaystyle \text{Therefore, its equation is}
\displaystyle y-0=-\frac{1}{m}(x-0).
\displaystyle \Rightarrow x+my=0.\qquad\ldots\ldots\text{ii)}
\displaystyle \text{The lines (i) and (ii) intersect at }(-1,2).
\displaystyle \text{Substituting }x=-1,\;y=2\text{ in (ii),}
\displaystyle -1+2m=0.
\displaystyle \Rightarrow m=\frac{1}{2}.
\displaystyle \text{Substituting }(-1,2)\text{ in (i),}
\displaystyle 2=\frac{1}{2}(-1)+c.
\displaystyle \Rightarrow c=\frac{5}{2}.
\displaystyle \therefore m=\frac{1}{2}\text{ and }c=\frac{5}{2}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Find the equation of the right bisector of the line segment joining}
\displaystyle \text{the points }(3,4)\text{ and }(-1,2).
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }A(3,4)\text{ and }B(-1,2).
\displaystyle \text{The midpoint of }AB\text{ is}
\displaystyle C=\left(\frac{3+(-1)}{2},\frac{4+2}{2}\right)=(1,3).
\displaystyle \text{The equation of }AB\text{ is}
\displaystyle \frac{x-3}{-1-3}=\frac{y-4}{2-4}.
\displaystyle \Rightarrow x-2y+5=0.
\displaystyle \text{Hence, a line perpendicular to }AB\text{ is of the form}
\displaystyle 2x+y+\lambda=0.
\displaystyle \text{Since it passes through }(1,3),
\displaystyle 2(1)+3+\lambda=0.
\displaystyle \Rightarrow \lambda=-5.
\displaystyle \therefore \text{The equation of the right bisector is }2x+y-5=0.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The line through }(h,3)\text{ and }(4,1)\text{ intersects the line}
\displaystyle 7x-9y-19=0\text{ at right angles. Find the value of }h.
\displaystyle \textbf{Answer:}
\displaystyle \text{The equation of the line joining }(h,3)\text{ and }(4,1)\text{ is}
\displaystyle (1-3)x-(4-h)y+\lambda=0.
\displaystyle \Rightarrow -2x+(h-4)y+\lambda=0.
\displaystyle \text{Since this line is perpendicular to }7x-9y-19=0,
\displaystyle \frac{-2}{\,h-4\,}=\frac{-9}{-7}=\frac{9}{7}.
\displaystyle \Rightarrow -14=9(h-4).
\displaystyle \Rightarrow -14=9h-36.
\displaystyle \Rightarrow 9h=22.
\displaystyle \therefore h=\frac{22}{9}.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Find the image of the point }(3,8)\text{ with respect to the line}
\displaystyle x+3y=7,\text{ assuming the line to be a plane mirror.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }A(3,8)\text{ and let its image be }A'(a,b).
\displaystyle \text{Given mirror: }x+3y-7=0.\qquad\ldots\ldots\text{i)}
\displaystyle \text{A line perpendicular to the mirror is of the form}
\displaystyle 3x-y+\lambda=0.
\displaystyle \text{Since it passes through }A(3,8),
\displaystyle 3(3)-8+\lambda=0.
\displaystyle \Rightarrow \lambda=-1.
\displaystyle \therefore AA':3x-y-1=0.\qquad\ldots\ldots\text{ii)}
\displaystyle \text{Let }M\text{ be the point where }AA'\text{ meets the mirror.}
\displaystyle \text{Solving (i) and (ii),}
\displaystyle x+3y=7,\qquad 3x-y=1.
\displaystyle \text{From }3x-y=1,\quad y=3x-1.
\displaystyle \Rightarrow x+3(3x-1)=7.
\displaystyle \Rightarrow 10x=10.
\displaystyle \Rightarrow x=1,\qquad y=2.
\displaystyle \therefore M=(1,2).
\displaystyle \text{Since the mirror bisects }AA',
\displaystyle \left(\frac{3+a}{2},\frac{8+b}{2}\right)=(1,2).
\displaystyle \Rightarrow 3+a=2,\qquad 8+b=4.
\displaystyle \Rightarrow a=-1,\qquad b=-4.
\displaystyle \therefore \text{The image of }(3,8)\text{ is }(-1,-4).
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Find the image of the point }(3,8)\text{ with respect to the line}
\displaystyle x+3y=7,\text{ assuming the line to be a plane mirror.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }A(3,8)\text{ and let its image be }A'(a,b).
\displaystyle \text{Given mirror: }x+3y-7=0.\qquad\ldots\ldots\text{i)}
\displaystyle \text{A line perpendicular to the mirror is of the form}
\displaystyle 3x-y+\lambda=0.
\displaystyle \text{Since it passes through }A(3,8),
\displaystyle 3(3)-8+\lambda=0.
\displaystyle \Rightarrow \lambda=-1.
\displaystyle \therefore AA':3x-y-1=0.\qquad\ldots\ldots\text{ii)}
\displaystyle \text{Let }M\text{ be the point where }AA'\text{ meets the mirror.}
\displaystyle \text{Solving (i) and (ii),}
\displaystyle x+3y=7,\qquad 3x-y=1.
\displaystyle \text{From }3x-y=1,\quad y=3x-1.
\displaystyle \Rightarrow x+3(3x-1)=7.
\displaystyle \Rightarrow 10x=10.
\displaystyle \Rightarrow x=1,\qquad y=2.
\displaystyle \therefore M=(1,2).
\displaystyle \text{Since the mirror bisects }AA',
\displaystyle \left(\frac{3+a}{2},\frac{8+b}{2}\right)=(1,2).
\displaystyle \Rightarrow 3+a=2,\qquad 8+b=4.
\displaystyle \Rightarrow a=-1,\qquad b=-4.
\displaystyle \therefore \text{The image of }(3,8)\text{ is }(-1,-4).
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Find the coordinates of the foot of the perpendicular from the point}
\displaystyle (-1,3)\text{ to the line }3x-4y-16=0.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }3x-4y-16=0.\qquad\ldots\ldots\text{i)}
\displaystyle \text{A line perpendicular to the given line is of the form}
\displaystyle -4x-3y+\lambda=0.
\displaystyle \text{Equivalently, }4x+3y+\lambda=0.
\displaystyle \text{Since it passes through }(-1,3),
\displaystyle 4(-1)+3(3)+\lambda=0.
\displaystyle \Rightarrow -4+9+\lambda=0.
\displaystyle \Rightarrow \lambda=-5.
\displaystyle \therefore 4x+3y-5=0.\qquad\ldots\ldots\text{ii)}
\displaystyle \text{Solving (i) and (ii),}
\displaystyle 3x-4y=16,\qquad 4x+3y=5.
\displaystyle \text{Multiplying the first equation by }3\text{ and the second by }4,
\displaystyle 9x-12y=48,\qquad 16x+12y=20.
\displaystyle \Rightarrow 25x=68.
\displaystyle \Rightarrow x=\frac{68}{25}.
\displaystyle \text{Substituting in }4x+3y=5,
\displaystyle 4\left(\frac{68}{25}\right)+3y=5.
\displaystyle \Rightarrow 3y=-\frac{147}{25}.
\displaystyle \Rightarrow y=-\frac{49}{25}.
\displaystyle \therefore \text{The foot of the perpendicular is }\left(\frac{68}{25},-\frac{49}{25}\right).
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Find the projection of the point }(1,0)\text{ on the line joining}
\displaystyle \text{the points }(-1,2)\text{ and }(5,4).
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }A(-1,2)\text{ and }B(5,4).
\displaystyle \text{The equation of the line }AB\text{ is}
\displaystyle \frac{x+1}{5+1}=\frac{y-2}{4-2}.
\displaystyle \Rightarrow \frac{x+1}{6}=\frac{y-2}{2}.
\displaystyle \Rightarrow x-3y+7=0.\qquad\ldots\ldots\text{i)}
\displaystyle \text{A line perpendicular to }AB\text{ is of the form}
\displaystyle 3x+y+\lambda=0.
\displaystyle \text{Since it passes through }(1,0),
\displaystyle 3(1)+0+\lambda=0.
\displaystyle \Rightarrow \lambda=-3.
\displaystyle \therefore 3x+y-3=0.\qquad\ldots\ldots\text{ii)}
\displaystyle \text{Solving (i) and (ii),}
\displaystyle x-3y+7=0,\qquad 3x+y-3=0.
\displaystyle \text{From (ii), }y=3-3x.
\displaystyle \text{Substituting in (i),}
\displaystyle x-3(3-3x)+7=0.
\displaystyle \Rightarrow 10x-2=0.
\displaystyle \Rightarrow x=\frac{1}{5}.
\displaystyle \Rightarrow y=3-3\left(\frac{1}{5}\right)=\frac{12}{5}.
\displaystyle \therefore \text{The projection of }(1,0)\text{ is }\left(\frac{1}{5},\frac{12}{5}\right).
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Find the equation of a line perpendicular to the line}
\displaystyle \sqrt{3}x-y+5=0\text{ and at a distance of }3\text{ units from the origin.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }\sqrt{3}x-y+5=0.
\displaystyle \text{Comparing it with }ax+by+c=0,\text{ we get }a=\sqrt{3}\text{ and }b=-1.
\displaystyle \text{Therefore, a line perpendicular to the given line is}
\displaystyle -x-\sqrt{3}y+\lambda=0.
\displaystyle \text{Equivalently, }x+\sqrt{3}y-\lambda=0.\qquad\ldots\ldots\text{i)}
\displaystyle \text{The perpendicular distance of (i) from the origin is }3.
\displaystyle \therefore \frac{|-\lambda|}{\sqrt{1+3}}=3.
\displaystyle \Rightarrow \frac{|\lambda|}{2}=3.
\displaystyle \Rightarrow |\lambda|=6.
\displaystyle \Rightarrow \lambda=\pm6.
\displaystyle \therefore \text{The required lines are}
\displaystyle x+\sqrt{3}y-6=0\text{ and }x+\sqrt{3}y+6=0.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The line }2x+3y=12\text{ meets the }x\text{-axis at }A\text{ and the }y\text{-axis at }B.
\displaystyle \text{The line through }(5,5)\text{ perpendicular to }AB\text{ meets the }x\text{-axis and }AB
\displaystyle \text{at }C\text{ and }E\text{ respectively. If }O\text{ is the origin, find the area of }OCEB.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given line: }2x+3y=12.\qquad\ldots\ldots\text{i)}
\displaystyle \text{The }x\text{-intercept and }y\text{-intercept are }A(6,0)\text{ and }B(0,4)\text{ respectively.}
\displaystyle \text{A line perpendicular to (i) is of the form}
\displaystyle 3x-2y+\lambda=0.
\displaystyle \text{Since it passes through }(5,5),
\displaystyle 3(5)-2(5)+\lambda=0.
\displaystyle \Rightarrow \lambda=-5.
\displaystyle \therefore 3x-2y-5=0.\qquad\ldots\ldots\text{ii)}
\displaystyle \text{To find }C,\text{ put }y=0\text{ in (ii).}
\displaystyle 3x-5=0\Rightarrow x=\frac{5}{3}.
\displaystyle \therefore C\left(\frac{5}{3},0\right).
\displaystyle \text{Solving (i) and (ii),}
\displaystyle 2x+3y=12,\qquad 3x-2y=5.
\displaystyle \Rightarrow x=3,\qquad y=2.
\displaystyle \therefore E=(3,2).
\displaystyle \text{Area of }OCEB=\text{Area of }\triangle BCO+\text{Area of }\triangle BCE.
\displaystyle CE=\sqrt{\left(3-\frac{5}{3}\right)^2+(2-0)^2}
\displaystyle =\sqrt{\frac{16}{9}+4}=\frac{2\sqrt{13}}{3}.
\displaystyle BE=\sqrt{(3-0)^2+(2-4)^2}
\displaystyle =\sqrt{9+4}=\sqrt{13}.
\displaystyle \text{Since }CE\perp BE,
\displaystyle \text{Area of }\triangle BCE=\frac{1}{2}\times CE\times BE
\displaystyle =\frac{1}{2}\times\frac{2\sqrt{13}}{3}\times\sqrt{13}=\frac{13}{3}.
\displaystyle \text{Area of }\triangle BCO=\frac{1}{2}\times OC\times OB
\displaystyle =\frac{1}{2}\times\frac{5}{3}\times4=\frac{10}{3}.
\displaystyle \therefore \text{Area of }OCEB=\frac{13}{3}+\frac{10}{3}=\frac{23}{3}\text{ square units.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Find the equation of the straight line which cuts off an intercept}
\displaystyle \text{on the }x\text{-axis twice that on the }y\text{-axis and is at a unit distance from the origin.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the }x\text{-intercept and }y\text{-intercept be }2a\text{ and }a\text{ respectively.}
\displaystyle \text{Therefore, the equation of the line in intercept form is}
\displaystyle \frac{x}{2a}+\frac{y}{a}=1.
\displaystyle \Rightarrow x+2y-2a=0.
\displaystyle \text{The perpendicular distance of this line from the origin is}
\displaystyle \frac{|-2a|}{\sqrt{1^2+2^2}}=\frac{|2a|}{\sqrt{5}}.
\displaystyle \text{Since the distance from the origin is }1,
\displaystyle \frac{|2a|}{\sqrt{5}}=1.
\displaystyle \Rightarrow |2a|=\sqrt{5}.
\displaystyle \Rightarrow 2a=\pm\sqrt{5}.
\displaystyle \therefore x+2y\mp\sqrt{5}=0.
\displaystyle \text{Hence, the required lines are}
\displaystyle x+2y-\sqrt{5}=0\text{ and }x+2y+\sqrt{5}=0.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{The equations of the perpendicular bisectors of the sides }AB\text{ and }AC
\displaystyle \text{of }\triangle ABC\text{ are }x-y+5=0\text{ and }x+2y=0\text{ respectively.}
\displaystyle \text{If }A=(1,-2),\text{ find the equation of the line }BC.
\displaystyle \textbf{Answer:}
\displaystyle \text{The perpendicular bisector of }AB\text{ is}
\displaystyle x-y+5=0.\qquad\ldots\ldots\text{i)}
\displaystyle \text{Therefore, }AB\text{ is perpendicular to (i) and passes through }A(1,-2).
\displaystyle \text{Hence, the equation of }AB\text{ is of the form}
\displaystyle x+y+\lambda=0.
\displaystyle \text{Since it passes through }(1,-2),
\displaystyle 1-2+\lambda=0\Rightarrow\lambda=1.
\displaystyle \therefore AB:x+y+1=0.\qquad\ldots\ldots\text{ii)}
\displaystyle \text{Let }B=(x_1,y_1).
\displaystyle \text{The midpoint of }AB\text{ is}
\displaystyle \left(\frac{1+x_1}{2},\frac{-2+y_1}{2}\right).
\displaystyle \text{Since this midpoint lies on (i),}
\displaystyle \frac{1+x_1}{2}-\frac{-2+y_1}{2}+5=0.
\displaystyle \Rightarrow x_1-y_1+13=0.\qquad\ldots\ldots\text{iii)}
\displaystyle \text{Also, }B\text{ lies on (ii). Therefore,}
\displaystyle x_1+y_1+1=0.\qquad\ldots\ldots\text{iv)}
\displaystyle \text{Solving (iii) and (iv), we get}
\displaystyle x_1=-7,\qquad y_1=6.
\displaystyle \therefore B=(-7,6).
\displaystyle \text{The perpendicular bisector of }AC\text{ is}
\displaystyle x+2y=0.\qquad\ldots\ldots\text{v)}
\displaystyle \text{Therefore, }AC\text{ is perpendicular to (v) and passes through }A(1,-2).
\displaystyle \text{Hence, the equation of }AC\text{ is of the form}
\displaystyle 2x-y+\lambda=0.
\displaystyle \text{Since it passes through }(1,-2),
\displaystyle 2(1)-(-2)+\lambda=0\Rightarrow\lambda=-4.
\displaystyle \therefore AC:2x-y-4=0.\qquad\ldots\ldots\text{vi)}
\displaystyle \text{Let }C=(x_2,y_2).
\displaystyle \text{The midpoint of }AC\text{ is}
\displaystyle \left(\frac{1+x_2}{2},\frac{-2+y_2}{2}\right).
\displaystyle \text{Since this midpoint lies on (v),}
\displaystyle \frac{1+x_2}{2}+2\left(\frac{-2+y_2}{2}\right)=0.
\displaystyle \Rightarrow x_2+2y_2-3=0.\qquad\ldots\ldots\text{vii)}
\displaystyle \text{Also, }C\text{ lies on (vi). Therefore,}
\displaystyle 2x_2-y_2-4=0.\qquad\ldots\ldots\text{viii)}
\displaystyle \text{Solving (vii) and (viii), we get}
\displaystyle x_2=\frac{11}{5},\qquad y_2=\frac{2}{5}.
\displaystyle \therefore C=\left(\frac{11}{5},\frac{2}{5}\right).
\displaystyle \text{The equation of the line through }B(-7,6)\text{ and }C\left(\frac{11}{5},\frac{2}{5}\right)\text{ is}
\displaystyle \frac{y-6}{\frac{2}{5}-6}=\frac{x+7}{\frac{11}{5}+7}.
\displaystyle \Rightarrow \frac{y-6}{-\frac{28}{5}}=\frac{x+7}{\frac{46}{5}}.
\displaystyle \Rightarrow 46(y-6)=-28(x+7).
\displaystyle \Rightarrow 14x+23y-40=0.
\displaystyle \therefore \text{The equation of }BC\text{ is }14x+23y-40=0.
\displaystyle \\

 

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