Class 10: Equation of a Line

$\displaystyle \textbf{Question 1: }\text{Point }P\text{ divides the line segment joining }A(8,0)\text{ and }B(16,-8)\text{ in the ratio }3:5.\hfill$
$\displaystyle \text{Find the coordinates of }P.\text{ Also, find the equation of the line through }P\text{ parallel to }3x+5y=7.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }P\text{ divides }AB\text{ in the ratio }3:5,$
$\displaystyle x=\frac{3(16)+5(8)}{3+5}=\frac{88}{8}=11$
$\displaystyle y=\frac{3(-8)+5(0)}{3+5}=\frac{-24}{8}=-3$
$\displaystyle \therefore P=(11,-3)$
$\displaystyle \text{Given line }3x+5y=7$
$\displaystyle 5y=-3x+7$
$\displaystyle \therefore \text{Slope}=-\frac{3}{5}$
$\displaystyle \text{A parallel line has the same slope.}$
$\displaystyle y-(-3)=-\frac{3}{5}(x-11)$
$\displaystyle 5y+15=-3x+33$
$\displaystyle 3x+5y=18$
$\displaystyle \therefore \text{The required equation is }3x+5y=18.$
$\\$

$\displaystyle \textbf{Question 2: }\text{The line segment joining }A(3,-4)\text{ and }B(-2,1)\text{ is divided internally}$
$\displaystyle \text{in the ratio }1:3\text{ at }P.\text{ Find the coordinates of }P.\text{ Also, find the equation of the line}$
$\displaystyle \text{through }P\text{ perpendicular to }5x-3y=4.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }P\text{ divides }AB\text{ in the ratio }1:3,$
$\displaystyle x=\frac{1(-2)+3(3)}{1+3}=\frac{7}{4}$
$\displaystyle y=\frac{1(1)+3(-4)}{1+3}=-\frac{11}{4}$
$\displaystyle \therefore P=\left(\frac{7}{4},-\frac{11}{4}\right)$
$\displaystyle \text{Given line }5x-3y=4$
$\displaystyle 3y=5x-4$
$\displaystyle \therefore \text{Slope}=\frac{5}{3}$
$\displaystyle \text{Slope of the perpendicular line}=-\frac{3}{5}$
$\displaystyle y+\frac{11}{4}=-\frac{3}{5}\left(x-\frac{7}{4}\right)$
$\displaystyle 20y+55=-12x+21$
$\displaystyle 12x+20y+34=0$
$\displaystyle 6x+10y+17=0$
$\displaystyle \therefore \text{The required equation is }6x+10y+17=0.$
$\\$

$\displaystyle \textbf{Question 3: }\text{A line }5x+3y+15=0\text{ meets the }y\text{-axis at }P.\text{ Find the} \\ \text{coordinates of }P.\text{ Also, find the equation of the line through }P\text{ and perpendicular} \\ \text{to }x-3y+4=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{On the }y\text{-axis, }x=0$
$\displaystyle 5(0)+3y+15=0$
$\displaystyle 3y=-15$
$\displaystyle y=-5$
$\displaystyle \therefore P=(0,-5)$
$\displaystyle \text{Given line }x-3y+4=0$
$\displaystyle 3y=x+4$
$\displaystyle \therefore \text{Slope}=\frac{1}{3}$
$\displaystyle \text{Slope of the perpendicular line}=-3$
$\displaystyle y-(-5)=-3(x-0)$
$\displaystyle y+5=-3x$
$\displaystyle 3x+y+5=0$
$\displaystyle \therefore \text{The required equation is }3x+y+5=0.$
$\\$

$\displaystyle \textbf{Question 4: }\text{Find the value of }k\text{ for which the lines }kx-5y+4=0\text{ and } \\ 5x-2y+5=0\text{ are perpendicular to each other.}\hfill\text{[ICSE 2003]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Slope of }kx-5y+4=0=\frac{k}{5}$
$\displaystyle \text{Slope of }5x-2y+5=0=\frac{5}{2}$
$\displaystyle \text{Since the lines are perpendicular,}$
$\displaystyle \frac{k}{5}\times\frac{5}{2}=-1$
$\displaystyle \frac{k}{2}=-1$
$\displaystyle k=-2$
$\displaystyle \therefore \text{The required value of }k\text{ is }-2.$
$\\$

$\displaystyle \textbf{Question 5: }\text{A straight line passes through }P(-1,4)\text{ and }Q(5,-2).\text{ It intersects} \\ \text{the coordinate axes at }A\text{ and }B.\text{ }M\text{ is the mid-point of }AB.\text{ Find} \\ \text{the equation of the line, coordinates of }A,\;B\text{ and }M.\hfill\text{[ICSE 2003]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given points are }P(-1,4)\text{ and }Q(5,-2).$
$\displaystyle \text{Slope}=\frac{-2-4}{5-(-1)}=\frac{-6}{6}=-1$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-4=-1(x+1)$
$\displaystyle y-4=-x-1$
$\displaystyle x+y=3$
$\displaystyle \therefore \text{The equation of the line is }x+y=3.$
$\displaystyle \text{For }x\text{-intercept, put }y=0.$
$\displaystyle x=3$
$\displaystyle \therefore A=(3,0)$
$\displaystyle \text{For }y\text{-intercept, put }x=0.$
$\displaystyle y=3$
$\displaystyle \therefore B=(0,3)$
$\displaystyle M=\left(\frac{3+0}{2},\frac{0+3}{2}\right)=\left(\frac{3}{2},\frac{3}{2}\right)$
$\displaystyle \therefore \text{The coordinates of }M\text{ are }\left(\frac{3}{2},\frac{3}{2}\right).$
$\\$

$\displaystyle \textbf{Question 6: }(1,5)\text{ and }(-3,-1)\text{ are the coordinates of vertices }A\text{ and } \\ C\text{ respectively of rhombus }ABCD.\text{ Find the equations of the diagonals } \\ AC\text{ and }BD.$
$\displaystyle \text{Answer:}$
$\displaystyle A=(1,5)\text{ and }C=(-3,-1)$
$\displaystyle \text{Mid-point of }AC=\left(\frac{1+(-3)}{2},\frac{5+(-1)}{2}\right)=(-1,2)$
$\displaystyle \text{Slope of }AC=\frac{-1-5}{-3-1}=\frac{-6}{-4}=\frac{3}{2}$
$\displaystyle \text{Equation of }AC:$
$\displaystyle y-5=\frac{3}{2}(x-1)$
$\displaystyle 2y-10=3x-3$
$\displaystyle 3x-2y+7=0$
$\displaystyle \text{In a rhombus, diagonals are perpendicular to each other.}$
$\displaystyle \therefore \text{Slope of }BD=-\frac{2}{3}$
$\displaystyle \text{Since }BD\text{ passes through }(-1,2),$
$\displaystyle y-2=-\frac{2}{3}(x+1)$
$\displaystyle 3y-6=-2x-2$
$\displaystyle 2x+3y-4=0$
$\displaystyle \therefore \text{The equations of the diagonals are }3x-2y+7=0\text{ and }2x+3y-4=0.$
$\\$

$\displaystyle \textbf{Question 7: }\text{Show that }A(3,2),\;B(6,-2)\text{ and }C(2,-5)\text{ can be vertices} \\ \text{of a square. Find the fourth vertex }D,\text{ if }ABCD\text{ is a square. Without using }D, \\ \text{ find the equation of }AD\text{ and }BD.$
$\displaystyle \text{Answer:}$
$\displaystyle AB^2=(6-3)^2+(-2-2)^2=9+16=25$
$\displaystyle BC^2=(2-6)^2+(-5+2)^2=16+9=25$
$\displaystyle AC^2=(2-3)^2+(-5-2)^2=1+49=50$
$\displaystyle \text{Since }AB^2+BC^2=AC^2\text{ and }AB=BC,\text{ the given points can be vertices of a square.}$
$\displaystyle \text{Mid-point of }AC=\left(\frac{3+2}{2},\frac{2+(-5)}{2}\right)=\left(\frac{5}{2},-\frac{3}{2}\right)$
$\displaystyle \text{Let }D=(x,y).$
$\displaystyle \text{In a square, diagonals bisect each other.}$
$\displaystyle \frac{x+6}{2}=\frac{5}{2}$
$\displaystyle x=-1$
$\displaystyle \frac{y+(-2)}{2}=-\frac{3}{2}$
$\displaystyle y=-1$
$\displaystyle \therefore D=(-1,-1)$
$\displaystyle \text{Slope of }AB=\frac{-2-2}{6-3}=-\frac{4}{3}$
$\displaystyle \text{Since }AB\perp AD,\text{ slope of }AD=\frac{3}{4}$
$\displaystyle y-2=\frac{3}{4}(x-3)$
$\displaystyle 4y-8=3x-9$
$\displaystyle 3x-4y-1=0$
$\displaystyle \therefore \text{Equation of }AD\text{ is }3x-4y-1=0.$
$\displaystyle \text{Mid-point of }AC=\left(\frac{5}{2},-\frac{3}{2}\right)$
$\displaystyle \text{Slope of }BD=\frac{-\frac{3}{2}-(-2)}{\frac{5}{2}-6}=-\frac{1}{7}$
$\displaystyle y+2=-\frac{1}{7}(x-6)$
$\displaystyle 7y+14=-x+6$
$\displaystyle x+7y+8=0$
$\displaystyle \therefore \text{Equation of }BD\text{ is }x+7y+8=0.$
$\\$

$\displaystyle \textbf{Question 8: }\text{A line through the origin meets the line }x=3y+2\text{ at right} \\ \text{angles at }X.\text{ Find the coordinates of }X.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given line is }x=3y+2$
$\displaystyle y=\frac{x-2}{3}$
$\displaystyle \therefore \text{Slope of the given line}=\frac{1}{3}$
$\displaystyle \text{Slope of the perpendicular line}=-3$
$\displaystyle \text{Equation of the line through origin is }y=-3x$
$\displaystyle \text{Solving }x=3y+2\text{ and }y=-3x,$
$\displaystyle x=3(-3x)+2$
$\displaystyle x=-9x+2$
$\displaystyle 10x=2$
$\displaystyle x=\frac{1}{5}$
$\displaystyle y=-3\left(\frac{1}{5}\right)=-\frac{3}{5}$
$\displaystyle \therefore X=\left(\frac{1}{5},-\frac{3}{5}\right)$
$\\$

$\displaystyle \textbf{Question 9: }\text{A straight line passes through }(3,2)\text{ and the portion of this line} \\ \text{intercepted between the positive axes is bisected at this point. Find the equation of the line. }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the }x\text{-intercept be }(a,0)\text{ and the }y\text{-intercept be }(0,b).$
$\displaystyle \text{Given }(3,2)\text{ is the mid-point of }(a,0)\text{ and }(0,b).$
$\displaystyle \frac{a+0}{2}=3$
$\displaystyle a=6$
$\displaystyle \frac{0+b}{2}=2$
$\displaystyle b=4$
$\displaystyle \therefore \text{The intercepts are }(6,0)\text{ and }(0,4).$
$\displaystyle \text{Slope of the line}=\frac{0-4}{6-0}=-\frac{2}{3}$
$\displaystyle y-0=-\frac{2}{3}(x-6)$
$\displaystyle 3y=-2x+12$
$\displaystyle 2x+3y=12$
$\displaystyle \therefore \text{The required equation is }2x+3y=12.$
$\\$

$\displaystyle \textbf{Question 10: }\text{Find the equation of the line passing through the point of intersection} \\ \text{of }7x+6y=71\text{ and }5x-8y=-23,\text{ and perpendicular to }4x-2y=1.$
$\displaystyle \text{Answer:}$
$\displaystyle 7x+6y=71\qquad \text{...(i)}$
$\displaystyle 5x-8y=-23\qquad \text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }4\text{ and (ii) by }3,$
$\displaystyle 28x+24y=284$
$\displaystyle 15x-24y=-69$
$\displaystyle \text{Adding,}$
$\displaystyle 43x=215$
$\displaystyle x=5$
$\displaystyle \text{Substituting }x=5\text{ in (i),}$
$\displaystyle 7(5)+6y=71$
$\displaystyle 35+6y=71$
$\displaystyle y=6$
$\displaystyle \therefore \text{Point of intersection is }(5,6).$
$\displaystyle \text{Given line is }4x-2y=1$
$\displaystyle 2y=4x-1$
$\displaystyle y=2x-\frac{1}{2}$
$\displaystyle \therefore \text{Slope of the given line}=2$
$\displaystyle \text{Slope of the perpendicular line}=-\frac{1}{2}$
$\displaystyle y-6=-\frac{1}{2}(x-5)$
$\displaystyle 2y-12=-x+5$
$\displaystyle x+2y=17$
$\displaystyle \therefore \text{The required equation is }x+2y=17.$
$\\$

$\displaystyle \textbf{Question 11: }\text{Find the equation of the line which is perpendicular to the line } \\ \frac{x}{a}-\frac{y}{b}=1\text{ at the point where this line meets the }y\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x}{a}-\frac{y}{b}=1$
$\displaystyle \frac{y}{b}=\frac{x}{a}-1$
$\displaystyle y=\frac{b}{a}x-b$
$\displaystyle \therefore \text{Slope of the given line}=\frac{b}{a}$
$\displaystyle \text{Slope of the perpendicular line}=-\frac{a}{b}$
$\displaystyle \text{At the }y\text{-axis, }x=0$
$\displaystyle y=-b$
$\displaystyle \therefore \text{The line passes through }(0,-b).$
$\displaystyle y-(-b)=-\frac{a}{b}(x-0)$
$\displaystyle y+b=-\frac{a}{b}x$
$\displaystyle by+b^2=-ax$
$\displaystyle ax+by+b^2=0$
$\displaystyle \therefore \text{The required equation is }ax+by+b^2=0.$
$\\$

$\displaystyle \textbf{Question 12: }O(0,0),\;A(3,5)\text{ and }B(-5,-3)\text{ are the vertices of }\triangle OAB.\text{ Find:}$
$\displaystyle \text{i) The equation of the median through }O\qquad \text{ii) The equation of the altitude} \\ \text{through }B$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Mid-point of }AB=\left(\frac{3+(-5)}{2},\frac{5+(-3)}{2}\right)=(-1,1)$
$\displaystyle \text{The median through }O\text{ passes through }(0,0)\text{ and }(-1,1).$
$\displaystyle \text{Slope}=\frac{1-0}{-1-0}=-1$
$\displaystyle y-0=-1(x-0)$
$\displaystyle x+y=0$
$\displaystyle \therefore \text{The equation of the median is }x+y=0.$
$\displaystyle \text{ii) Slope of }OA=\frac{5-0}{3-0}=\frac{5}{3}$
$\displaystyle \text{Slope of the altitude through }B=-\frac{3}{5}$
$\displaystyle y-(-3)=-\frac{3}{5}\{x-(-5)\}$
$\displaystyle y+3=-\frac{3}{5}(x+5)$
$\displaystyle 5y+15=-3x-15$
$\displaystyle 3x+5y+30=0$
$\displaystyle \therefore \text{The equation of the altitude is }3x+5y+30=0.$
$\\$

$\displaystyle \textbf{Question 13: }\text{Determine whether the line through }(-2,3)\text{ and }(4,1)\text{ is perpendicular} \\ \text{to }3x=y+1.\text{ Does }3x=y+1\text{ bisect the line segment joining the two given} \\ \text{points?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Slope of the line through }(-2,3)\text{ and }(4,1)=\frac{1-3}{4-(-2)}=-\frac{2}{6}=-\frac{1}{3}$
$\displaystyle 3x=y+1$
$\displaystyle y=3x-1$
$\displaystyle \therefore \text{Slope of }3x=y+1\text{ is }3.$
$\displaystyle \text{Since }-\frac{1}{3}\times3=-1,$
$\displaystyle \therefore \text{the two lines are perpendicular.}$
$\displaystyle \text{Mid-point of }(-2,3)\text{ and }(4,1)=\left(\frac{-2+4}{2},\frac{3+1}{2}\right)=(1,2)$
$\displaystyle \text{Substituting }(1,2)\text{ in }3x=y+1,$
$\displaystyle 3(1)=2+1$
$\displaystyle 3=3$
$\displaystyle \therefore (1,2)\text{ lies on the line }3x=y+1.$
$\displaystyle \therefore 3x=y+1\text{ bisects the line segment joining }(-2,3)\text{ and }(4,1).$
$\\$

$\displaystyle \textbf{Question 14: }\text{Given the straight line }x\cos30^\circ+y\sin30^\circ=2.\text{ Determine} \\ \text{the equation of the other line which is parallel to it and passes through }(4,3).$
$\displaystyle \text{Answer:}$
$\displaystyle x\cos30^\circ+y\sin30^\circ=2$
$\displaystyle \frac{\sqrt{3}}{2}x+\frac{1}{2}y=2$
$\displaystyle \sqrt{3}x+y=4$
$\displaystyle y=-\sqrt{3}x+4$
$\displaystyle \therefore \text{Slope of the given line}=-\sqrt{3}$
$\displaystyle \text{A parallel line has the same slope.}$
$\displaystyle y-3=-\sqrt{3}(x-4)$
$\displaystyle y-3=-\sqrt{3}x+4\sqrt{3}$
$\displaystyle \sqrt{3}x+y=4\sqrt{3}+3$
$\displaystyle \therefore \text{The required equation is }\sqrt{3}x+y=4\sqrt{3}+3.$
$\\$

$\displaystyle \textbf{Question 15: }\text{Find the value of }k\text{ such that the line }(k-2)x+(k+3)y-5=0\text{ is:}$
$\displaystyle \text{i) Perpendicular to the line }2x-y+7=0\qquad \text{ii) Parallel to it. }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given line is }(k-2)x+(k+3)y-5=0$
$\displaystyle \therefore \text{Slope}=-\frac{k-2}{k+3}$
$\displaystyle \text{For }2x-y+7=0,$
$\displaystyle y=2x+7$
$\displaystyle \therefore \text{Slope}=2$
$\displaystyle \text{i) For the lines to be perpendicular,}$
$\displaystyle -\frac{k-2}{k+3}=-\frac{1}{2}$
$\displaystyle 2(k-2)=k+3$
$\displaystyle 2k-4=k+3$
$\displaystyle k=7$
$\displaystyle \therefore \text{Required value of }k=7.$
$\displaystyle \text{ii) For the lines to be parallel,}$
$\displaystyle -\frac{k-2}{k+3}=2$
$\displaystyle -(k-2)=2(k+3)$
$\displaystyle -k+2=2k+6$
$\displaystyle -3k=4$
$\displaystyle k=-\frac{4}{3}$
$\displaystyle \therefore \text{Required value of }k=-\frac{4}{3}.$
$\\$

$\displaystyle \textbf{Question 16: }\text{The vertices of }\triangle ABC\text{ are }A(0,5),\;B(-1,-2)\text{ and } \\ C(11,7).\text{ Write down the equation of }BC.\text{ Find:}$
$\displaystyle \text{i) The equation of the line through }A\text{ and perpendicular to }BC.$
$\displaystyle \text{ii) The coordinates of }P,\text{ where the perpendicular through }A\text{ meets }BC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Slope of }BC=\frac{7-(-2)}{11-(-1)}=\frac{9}{12}=\frac{3}{4}$
$\displaystyle \text{Equation of }BC:$
$\displaystyle y-7=\frac{3}{4}(x-11)$
$\displaystyle 4y-28=3x-33$
$\displaystyle 3x-4y=5$
$\displaystyle \text{i) Slope of the line perpendicular to }BC=-\frac{4}{3}$
$\displaystyle \text{Equation of the line through }A(0,5):$
$\displaystyle y-5=-\frac{4}{3}(x-0)$
$\displaystyle 3y-15=-4x$
$\displaystyle 4x+3y=15$
$\displaystyle \therefore \text{The required perpendicular line is }4x+3y=15.$
$\displaystyle \text{ii) Solving }3x-4y=5\text{ and }4x+3y=15,$
$\displaystyle x=3,\qquad y=1$
$\displaystyle \therefore P=(3,1)$
$\\$

$\displaystyle \textbf{Question 17: }\text{From the given figure, find:}$
$\displaystyle \text{i) The coordinates of }A,\;B\text{ and }C\qquad \text{ii) The equation of the line through } \\ A\text{ and parallel to }BC.\hfill\text{[ICSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) From the figure, }A=(2,3),\;B=(-1,2)\text{ and }C=(3,0).$
$\displaystyle \text{ii) Slope of }BC=\frac{0-2}{3-(-1)}=\frac{-2}{4}=-\frac{1}{2}$
$\displaystyle \text{A line parallel to }BC\text{ has the same slope.}$
$\displaystyle \text{Equation of the line through }A(2,3)\text{ is:}$
$\displaystyle y-3=-\frac{1}{2}(x-2)$
$\displaystyle 2y-6=-x+2$
$\displaystyle x+2y=8$
$\displaystyle \therefore \text{The required equation is }x+2y=8.$
$\\$

$\displaystyle \textbf{Question 18: }P(3,4),\;Q(7,-2)\text{ and }R(-2,-1)\text{ are the vertices of }\triangle PQR. \\ \text{ Write down the equation of the median through }R.\hfill\text{[ICSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Mid-point of }PQ=\left(\frac{3+7}{2},\frac{4+(-2)}{2}\right)=(5,1)$
$\displaystyle \text{The median through }R\text{ passes through }R(-2,-1)\text{ and }(5,1).$
$\displaystyle \text{Slope}=\frac{1-(-1)}{5-(-2)}=\frac{2}{7}$
$\displaystyle y-1=\frac{2}{7}(x-5)$
$\displaystyle 7y-7=2x-10$
$\displaystyle 2x-7y-3=0$
$\displaystyle \therefore \text{The equation of the median through }R\text{ is }2x-7y-3=0.$
$\\$

$\displaystyle \textbf{Question 19: }A(8,-6),\;B(-4,2)\text{ and }C(0,-10)\text{ are vertices of }\triangle ABC. \\ \text{ If }P\text{ is the mid-point of }AB\text{ and }Q\text{ is the mid-point of }AC,\text{ use coordinate} \\ \text{geometry to show that }PQ\parallel BC.\text{ Give a special name to quadrilateral }PBCQ.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Coordinates of }P=\left(\frac{8+(-4)}{2},\frac{-6+2}{2}\right)=(2,-2)$
$\displaystyle \text{Coordinates of }Q=\left(\frac{8+0}{2},\frac{-6+(-10)}{2}\right)=(4,-8)$
$\displaystyle \text{Slope of }PQ=\frac{-8-(-2)}{4-2}=\frac{-6}{2}=-3$
$\displaystyle \text{Slope of }BC=\frac{-10-2}{0-(-4)}=\frac{-12}{4}=-3$
$\displaystyle \text{Since slope of }PQ=\text{slope of }BC,$
$\displaystyle \therefore PQ\parallel BC.$
$\displaystyle \text{Since one pair of opposite sides is parallel, }PBCQ\text{ is a trapezium.}$
$\\$

$\displaystyle \textbf{Question 20: }\text{A line }AB\text{ meets the }x\text{-axis at }A\text{ and the }y \\ \text{-axis at }B.\text{ The point }P(-4,-2)\text{ divides }AB\text{ internally in the ratio } \\ AP:PB=1:2.\text{ Find:}$
$\displaystyle \text{i) The coordinates of }A\text{ and }B\qquad \text{ii) The equation of the line through } \\ P\text{ and perpendicular to }AB.$
$\displaystyle \text{Answer:}$
$\displaystyle AP:PB=1:2$
$\displaystyle \text{Let }A=(x,0)\text{ and }B=(0,y).$
$\displaystyle \text{Since }P(-4,-2)\text{ divides }AB\text{ in the ratio }1:2,$
$\displaystyle -4=\frac{1(0)+2x}{1+2}$
$\displaystyle -4=\frac{2x}{3}$
$\displaystyle x=-6$
$\displaystyle -2=\frac{1(y)+2(0)}{1+2}$
$\displaystyle -2=\frac{y}{3}$
$\displaystyle y=-6$
$\displaystyle \therefore A=(-6,0)\text{ and }B=(0,-6).$
$\displaystyle \text{Slope of }AB=\frac{-6-0}{0-(-6)}=-1$
$\displaystyle \text{Slope of the perpendicular line}=1$
$\displaystyle y-(-2)=1\{x-(-4)\}$
$\displaystyle y+2=x+4$
$\displaystyle y=x+2$
$\displaystyle \therefore \text{The required equation is }y=x+2.$
$\\$

$\displaystyle \textbf{Question 21: }\text{A line intersects the }x\text{-axis at }(-2,0)\text{ and cuts off an} \\ \text{intercept of }3\text{ units on the positive }y\text{-axis. Find the equation of the line.}\hfill\text{[ICSE 1992]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The }x\text{-intercept is }(-2,0).$
$\displaystyle \text{The }y\text{-intercept is }(0,3).$
$\displaystyle \text{Slope of the line}=\frac{3-0}{0-(-2)}=\frac{3}{2}$
$\displaystyle y-3=\frac{3}{2}(x-0)$
$\displaystyle 2y-6=3x$
$\displaystyle 3x-2y+6=0$
$\displaystyle \therefore \text{The required equation is }3x-2y+6=0.$
$\\$

$\displaystyle \textbf{Question 22: }\text{Find the equation of a line passing through }(2,3)\text{ and having} \\ \text{the }x\text{-intercept of }4\text{ units.}\hfill\text{[ICSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The }x\text{-intercept is }(4,0).$
$\displaystyle \text{Slope of the line}=\frac{0-3}{4-2}=-\frac{3}{2}$
$\displaystyle y-0=-\frac{3}{2}(x-4)$
$\displaystyle 2y=-3x+12$
$\displaystyle 3x+2y-12=0$
$\displaystyle \therefore \text{The required equation is }3x+2y-12=0.$
$\\$

$\displaystyle \textbf{Question 23: }\text{The given figure (not drawn to scale) shows two straight lines }AB \\ \text{ and }CD.\text{ If the equation of }AB\text{ is }y=x+1\text{ and the equation of }CD\text{ is } \\ y=\sqrt{3}x-1,\text{ write down the inclinations of }AB\text{ and }CD,\text{ and find the} \\ \text{angle between them.}\hfill\text{[ICSE 1989]}$
$\displaystyle \text{Answer:}$
$\displaystyle AB:\;y=x+1$
$\displaystyle \therefore \text{Slope of }AB=1$
$\displaystyle \tan\alpha_1=1$
$\displaystyle \therefore \alpha_1=45^\circ$
$\displaystyle CD:\;y=\sqrt{3}x-1$
$\displaystyle \therefore \text{Slope of }CD=\sqrt{3}$
$\displaystyle \tan\alpha_2=\sqrt{3}$
$\displaystyle \therefore \alpha_2=60^\circ$
$\displaystyle \text{Angle between the two lines}=60^\circ-45^\circ=15^\circ$
$\displaystyle \therefore \text{Inclination of }AB=45^\circ,\;\text{inclination of }CD=60^\circ,\;\text{and the angle between them is }15^\circ.$
$\\$

$\displaystyle \textbf{Question 24: }\text{Write down the equation of the line whose gradient is }\frac{2}{3} \\ \text{ and which passes through }P,\text{ where }P\text{ divides the line segment joining }A(-2,6) \\ \text{ and }B(3,-4)\text{ in the ratio }2:3.\hfill\text{[ICSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }P\text{ divides }AB\text{ in the ratio }2:3,$
$\displaystyle x=\frac{2(3)+3(-2)}{2+3}=\frac{6-6}{5}=0$
$\displaystyle y=\frac{2(-4)+3(6)}{2+3}=\frac{-8+18}{5}=2$
$\displaystyle \therefore P=(0,2)$
$\displaystyle \text{Given gradient}=\frac{2}{3}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-2=\frac{2}{3}(x-0)$
$\displaystyle 3y-6=2x$
$\displaystyle 2x-3y+6=0$
$\displaystyle \therefore \text{The required equation is }2x-3y+6=0.$
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$\displaystyle \textbf{Question 25: }\text{The ordinate of a point lying on the line joining }(6,4)\text{ and } \\ (7,-5)\text{ is }-23.\text{ Find the coordinates of that point.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required point be }(x,-23).$
$\displaystyle \text{Slope of the line joining }(6,4)\text{ and }(7,-5)=\frac{-5-4}{7-6}=-9$
$\displaystyle \text{Equation of the line:}$
$\displaystyle y-4=-9(x-6)$
$\displaystyle y-4=-9x+54$
$\displaystyle 9x+y=58$
$\displaystyle \text{Substituting }y=-23,$
$\displaystyle 9x-23=58$
$\displaystyle 9x=81$
$\displaystyle x=9$
$\displaystyle \therefore \text{The required point is }(9,-23).$
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$\displaystyle \textbf{Question 26: }\text{Points }A\text{ and }B\text{ have coordinates }(7,-3)\text{ and } \\ (1,9)\text{ respectively. Find:}$
$\displaystyle \text{i) The slope of }AB\qquad \text{ii) The equation of the perpendicular bisector of }AB\qquad \\ \text{iii) The value of }p\text{ if }(-2,p)\text{ lies on it.}\hfill\text{[ICSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Slope of }AB=\frac{9-(-3)}{1-7}=\frac{12}{-6}=-2$
$\displaystyle \text{ii) Mid-point of }AB=\left(\frac{7+1}{2},\frac{-3+9}{2}\right)=(4,3)$
$\displaystyle \text{Slope of the perpendicular bisector}=\frac{1}{2}$
$\displaystyle \text{Equation of the perpendicular bisector:}$
$\displaystyle y-3=\frac{1}{2}(x-4)$
$\displaystyle 2y-6=x-4$
$\displaystyle x-2y+2=0$
$\displaystyle \text{iii) Since }(-2,p)\text{ lies on }x-2y+2=0,$
$\displaystyle -2-2p+2=0$
$\displaystyle -2p=0$
$\displaystyle p=0$
$\displaystyle \therefore \text{The slope of }AB=-2,\text{ the perpendicular bisector is }x-2y+2=0\text{ and }p=0.$
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$\displaystyle \textbf{Question 27: }A\text{ and }B\text{ are two points on the }x\text{-axis and }y\text{-axis} \\ \text{respectively. }P(2,-3)\text{ is the mid-point of }AB.\text{ Find:}$
$\displaystyle \text{i) The coordinates of }A\text{ and }B\qquad \text{ii) The slope of }AB\qquad \\ \text{iii) The equation of }AB.\hfill\text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=(x,0)\text{ and }B=(0,y).$
$\displaystyle \text{Since }P(2,-3)\text{ is the mid-point of }AB,$
$\displaystyle 2=\frac{x+0}{2}$
$\displaystyle x=4$
$\displaystyle -3=\frac{0+y}{2}$
$\displaystyle y=-6$
$\displaystyle \therefore A=(4,0)\text{ and }B=(0,-6)$
$\displaystyle \text{ii) Slope of }AB=\frac{-6-0}{0-4}=\frac{-6}{-4}=\frac{3}{2}$
$\displaystyle \text{iii) Using point-slope form through }A(4,0),$
$\displaystyle y-0=\frac{3}{2}(x-4)$
$\displaystyle 2y=3x-12$
$\displaystyle 3x-2y-12=0$
$\displaystyle \therefore \text{The required equation of }AB\text{ is }3x-2y-12=0.$
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$\displaystyle \textbf{Question 28: }\text{The equation of a line is }3x+4y-7=0.\text{ Find:}$
$\displaystyle \text{i) The slope of the line.}\qquad \text{ii) The equation of a line perpendicular to the given} \\ \text{line and passing through the intersection of }x-y+2=0\text{ and }3x+y-10=0.\hfill\text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }3x+4y-7=0$
$\displaystyle 4y=-3x+7$
$\displaystyle y=-\frac{3}{4}x+\frac{7}{4}$
$\displaystyle \therefore \text{Slope of the line}=-\frac{3}{4}$
$\displaystyle \text{ii) Slope of the perpendicular line}=\frac{4}{3}$
$\displaystyle \text{Point of intersection of }x-y+2=0\text{ and }3x+y-10=0:$
$\displaystyle x-y+2=0\qquad \text{...(i)}$
$\displaystyle 3x+y-10=0\qquad \text{...(ii)}$
$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle 4x-8=0$
$\displaystyle x=2$
$\displaystyle \text{Substituting in (i),}$
$\displaystyle 2-y+2=0$
$\displaystyle y=4$
$\displaystyle \therefore \text{Point of intersection is }(2,4)$
$\displaystyle \text{Equation of the required line:}$
$\displaystyle y-4=\frac{4}{3}(x-2)$
$\displaystyle 3y-12=4x-8$
$\displaystyle 4x-3y+4=0$
$\displaystyle \therefore \text{The required equation is }4x-3y+4=0.$
$\\$

$\displaystyle \textbf{Question 29: }ABCD\text{ is a parallelogram where }A(x,y),\;B(5,8),\;C(4,7)\text{ and } \\ D(2,-4).\text{ Find: i) Coordinates of }A\text{ ii) Equation of diagonal }BD.\hfill\text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Mid-point of }BD=\left(\frac{5+2}{2},\frac{8+(-4)}{2}\right)=\left(\frac{7}{2},2\right)$
$\displaystyle \text{Let the mid-point of the diagonals be }O=\left(\frac{7}{2},2\right).$
$\displaystyle \text{In a parallelogram, diagonals bisect each other.}$
$\displaystyle \therefore O\text{ is also the mid-point of }AC.$
$\displaystyle \frac{x+4}{2}=\frac{7}{2}$
$\displaystyle x+4=7$
$\displaystyle x=3$
$\displaystyle \frac{y+7}{2}=2$
$\displaystyle y+7=4$
$\displaystyle y=-3$
$\displaystyle \therefore A=(3,-3)$
$\displaystyle \text{ii) Slope of }BD=\frac{-4-8}{2-5}=\frac{-12}{-3}=4$
$\displaystyle \text{Equation of }BD:$
$\displaystyle y-8=4(x-5)$
$\displaystyle y-8=4x-20$
$\displaystyle 4x-y-12=0$
$\displaystyle \therefore \text{The equation of diagonal }BD\text{ is }4x-y-12=0.$
$\\$

$\displaystyle \textbf{Question 30: }\text{Given the equation of line }L_1\text{ is }y=4.$
$\displaystyle \text{i) Write the slope of line }L_2,\text{ if }L_2\text{ is the bisector of }\angle O.$
$\displaystyle \text{ii) Write the coordinates of point }P.\qquad \text{iii) Find the equation of }L_2.$
$\displaystyle \text{Answer:}$
$\displaystyle L_1:\;y=4$
$\displaystyle \text{i) Since }L_2\text{ bisects the right angle at }O,$
$\displaystyle \text{its inclination is }45^\circ$
$\displaystyle \therefore \text{Slope of }L_2=\tan45^\circ=1$
$\displaystyle \text{ii) }L_2\text{ passes through the origin and }L_1\text{ is }y=4.$
$\displaystyle \therefore P=(4,4)$
$\displaystyle \text{iii) Equation of }L_2:$
$\displaystyle y-0=1(x-0)$
$\displaystyle y=x$
$\displaystyle \therefore \text{The required equation of }L_2\text{ is }y=x.$
$\\$