Class 10: Equation of a Line

$\displaystyle \textbf{Question 1: }\text{Find the slope and y-intercept of the line:}$
$\displaystyle \text{i) }y=4\qquad \text{ii) }ax-by=0\qquad \text{iii) }3x-4y=5$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Given equation is }y=4$
$\displaystyle y=0x+4$
$\displaystyle \therefore \text{Slope }m=0\text{ and y-intercept }=4.$
$\displaystyle \text{ii) Given equation is }ax-by=0$
$\displaystyle by=ax$
$\displaystyle y=\frac{a}{b}x$
$\displaystyle \therefore \text{Slope }m=\frac{a}{b}\text{ and y-intercept }=0.$
$\displaystyle \text{iii) Given equation is }3x-4y=5$
$\displaystyle 4y=3x-5$
$\displaystyle y=\frac{3}{4}x-\frac{5}{4}$
$\displaystyle \therefore \text{Slope }m=\frac{3}{4}\text{ and y-intercept }=-\frac{5}{4}.$
$\\$

$\displaystyle \textbf{Question 2: }\text{The equation of a line is }x-y=4.\text{ Find its slope, y-intercept and inclination.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }x-y=4$
$\displaystyle y=x-4$
$\displaystyle \therefore \text{Slope }m=1\text{ and y-intercept }=-4.$
$\displaystyle m=\tan\theta$
$\displaystyle \tan\theta=1$
$\displaystyle \therefore \theta=45^\circ.$
$\\$

$\displaystyle \textbf{Question 3:}$
$\displaystyle \text{i) Is the line }3x+4y+7=0\text{ perpendicular to the line }28x-21y+50=0?$
$\displaystyle \text{ii) Is the line }x-3y=4\text{ perpendicular to the line }3x-y=7?$
$\displaystyle \text{iii) Is the line }3x+2y=5\text{ parallel to the line }x+2y=1?$
$\displaystyle \text{iv) Determine }x\text{ so that the slope of the line through }(1,4)\text{ and }(x,2)\text{ is }2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }3x+4y+7=0$
$\displaystyle 4y=-3x-7$
$\displaystyle y=-\frac{3}{4}x-\frac{7}{4}$
$\displaystyle \therefore m_1=-\frac{3}{4}$
$\displaystyle 28x-21y+50=0$
$\displaystyle 21y=28x+50$
$\displaystyle y=\frac{4}{3}x+\frac{50}{21}$
$\displaystyle \therefore m_2=\frac{4}{3}$
$\displaystyle m_1m_2=-\frac{3}{4}\times\frac{4}{3}=-1$
$\displaystyle \therefore \text{The two lines are perpendicular.}$
$\displaystyle \text{ii) }x-3y=4$
$\displaystyle 3y=x-4$
$\displaystyle y=\frac{1}{3}x-\frac{4}{3}$
$\displaystyle \therefore m_1=\frac{1}{3}$
$\displaystyle 3x-y=7$
$\displaystyle y=3x-7$
$\displaystyle \therefore m_2=3$
$\displaystyle m_1m_2=\frac{1}{3}\times3=1\neq-1$
$\displaystyle \therefore \text{The two lines are not perpendicular.}$
$\displaystyle \text{iii) }3x+2y=5$
$\displaystyle 2y=-3x+5$
$\displaystyle y=-\frac{3}{2}x+\frac{5}{2}$
$\displaystyle \therefore m_1=-\frac{3}{2}$
$\displaystyle x+2y=1$
$\displaystyle 2y=-x+1$
$\displaystyle y=-\frac{1}{2}x+\frac{1}{2}$
$\displaystyle \therefore m_2=-\frac{1}{2}$
$\displaystyle \text{Since }m_1\neq m_2,\text{ the two lines are not parallel.}$
$\displaystyle \text{iv) Slope}=\frac{2-4}{x-1}=2$
$\displaystyle \frac{-2}{x-1}=2$
$\displaystyle -2=2(x-1)$
$\displaystyle -2=2x-2$
$\displaystyle x=0$
$\displaystyle \therefore \text{The required value of }x\text{ is }0.$
$\\$

$\displaystyle \textbf{Question 4: }\text{Find the slope of the line which is parallel to:}$
$\displaystyle \text{i) }x+2y+3=0\qquad \text{ii) }\frac{x}{2}-\frac{y}{3}-1=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Given equation is }x+2y+3=0$
$\displaystyle 2y=-x-3$
$\displaystyle y=-\frac{1}{2}x-\frac{3}{2}$
$\displaystyle \therefore \text{Slope }m=-\frac{1}{2}$
$\displaystyle \therefore \text{The slope of a line parallel to the given line is }-\frac{1}{2}.$
$\displaystyle \text{ii) Given equation is }\frac{x}{2}-\frac{y}{3}-1=0$
$\displaystyle \frac{y}{3}=\frac{x}{2}-1$
$\displaystyle y=\frac{3}{2}x-3$
$\displaystyle \therefore \text{Slope }m=\frac{3}{2}$
$\displaystyle \therefore \text{The slope of a line parallel to the given line is }\frac{3}{2}.$
$\\$

$\displaystyle \textbf{Question 5: }\text{Find the slope of the line which is perpendicular to:}$
$\displaystyle \text{i) }x-\frac{y}{2}+3=0\qquad \text{ii) }\frac{x}{3}-2y=4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Given equation is }x-\frac{y}{2}+3=0$
$\displaystyle \frac{y}{2}=x+3$
$\displaystyle y=2x+6$
$\displaystyle \therefore \text{Slope }m_1=2$
$\displaystyle m_1m_2=-1$
$\displaystyle m_2=-\frac{1}{2}$
$\displaystyle \therefore \text{The required slope is }-\frac{1}{2}.$
$\displaystyle \text{ii) Given equation is }\frac{x}{3}-2y=4$
$\displaystyle 2y=\frac{x}{3}-4$
$\displaystyle y=\frac{1}{6}x-2$
$\displaystyle \therefore \text{Slope }m_1=\frac{1}{6}$
$\displaystyle m_1m_2=-1$
$\displaystyle m_2=\frac{-1}{\frac{1}{6}}=-6$
$\displaystyle \therefore \text{The required slope is }-6.$
$\\$

$\displaystyle \textbf{Question 6:}$
$\displaystyle \text{i) Lines }2x-by+5=0\text{ and }ax+3y=2\text{ are parallel. Find the relation between }a\text{ and }b.$
$\displaystyle \text{ii) Lines }mx+3y+7=0\text{ and }5x-ny-3=0\text{ are perpendicular. Find the relation between }m\text{ and }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Given equation is }2x-by+5=0$
$\displaystyle by=2x+5$
$\displaystyle y=\frac{2}{b}x+\frac{5}{b}$
$\displaystyle \therefore \text{Slope }m_1=\frac{2}{b}$
$\displaystyle \text{Given equation is }ax+3y=2$
$\displaystyle 3y=-ax+2$
$\displaystyle y=-\frac{a}{3}x+\frac{2}{3}$
$\displaystyle \therefore \text{Slope }m_2=-\frac{a}{3}$
$\displaystyle \text{Since the lines are parallel, }m_1=m_2$
$\displaystyle \frac{2}{b}=-\frac{a}{3}$
$\displaystyle ab=-6$
$\displaystyle \therefore \text{The required relation is }ab=-6.$
$\displaystyle \text{ii) Given equation is }mx+3y+7=0$
$\displaystyle 3y=-mx-7$
$\displaystyle y=-\frac{m}{3}x-\frac{7}{3}$
$\displaystyle \therefore \text{Slope }m_1=-\frac{m}{3}$
$\displaystyle \text{Given equation is }5x-ny-3=0$
$\displaystyle ny=5x-3$
$\displaystyle y=\frac{5}{n}x-\frac{3}{n}$
$\displaystyle \therefore \text{Slope }m_2=\frac{5}{n}$
$\displaystyle \text{Since the lines are perpendicular, }m_1m_2=-1$
$\displaystyle -\frac{m}{3}\times\frac{5}{n}=-1$
$\displaystyle 5m=3n$
$\displaystyle \therefore \text{The required relation is }5m=3n.$
$\\$

$\displaystyle \textbf{Question 7: }\text{Find the value of }p\text{ if the lines }2x-y+5=0\text{ and }px+3y=4\text{ are perpendicular.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }2x-y+5=0$
$\displaystyle y=2x+5$
$\displaystyle \therefore \text{Slope }m_1=2$
$\displaystyle \text{Given equation is }px+3y=4$
$\displaystyle 3y=-px+4$
$\displaystyle y=-\frac{p}{3}x+\frac{4}{3}$
$\displaystyle \therefore \text{Slope }m_2=-\frac{p}{3}$
$\displaystyle \text{Since the lines are perpendicular, }m_1m_2=-1$
$\displaystyle 2\left(-\frac{p}{3}\right)=-1$
$\displaystyle p=\frac{3}{2}$
$\displaystyle \therefore \text{The required value of }p\text{ is }\frac{3}{2}.$
$\\$

$\displaystyle \textbf{Question 8: }\text{The equation of a line }AB\text{ is }2x-2y+3=0.\text{ Find:}$
$\displaystyle \text{i) The slope of }AB\qquad \text{ii) The angle that }AB\text{ makes with the positive direction of the }x\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }2x-2y+3=0$
$\displaystyle 2y=2x+3$
$\displaystyle y=x+\frac{3}{2}$
$\displaystyle \therefore \text{Slope }m=1.$
$\displaystyle m=\tan\theta$
$\displaystyle \tan\theta=1$
$\displaystyle \therefore \theta=45^\circ.$
$\\$

$\displaystyle \textbf{Question 9: }\text{The lines represented by }4x+3y=9\text{ and }px-6y+3=0\text{ are} \\ \text{parallel. Find the value of }p.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }4x+3y=9$
$\displaystyle 3y=-4x+9$
$\displaystyle y=-\frac{4}{3}x+3$
$\displaystyle \therefore \text{Slope }m_1=-\frac{4}{3}$
$\displaystyle \text{Given equation is }px-6y+3=0$
$\displaystyle 6y=px+3$
$\displaystyle y=\frac{p}{6}x+\frac{1}{2}$
$\displaystyle \therefore \text{Slope }m_2=\frac{p}{6}$
$\displaystyle \text{Since the lines are parallel, }m_1=m_2$
$\displaystyle -\frac{4}{3}=\frac{p}{6}$
$\displaystyle p=-8$
$\displaystyle \therefore \text{The required value of }p\text{ is }-8.$
$\\$

$\displaystyle \textbf{Question 10: }\text{If the lines }y=3x+7\text{ and }2y+px=3\text{ are perpendicular, find} \\ \text{the value of }p.\hfill\text{[ICSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }y=3x+7$
$\displaystyle \therefore \text{Slope }m_1=3$
$\displaystyle \text{Given equation is }2y+px=3$
$\displaystyle 2y=-px+3$
$\displaystyle y=-\frac{p}{2}x+\frac{3}{2}$
$\displaystyle \therefore \text{Slope }m_2=-\frac{p}{2}$
$\displaystyle \text{Since the lines are perpendicular, }m_1m_2=-1$
$\displaystyle 3\left(-\frac{p}{2}\right)=-1$
$\displaystyle p=\frac{2}{3}$
$\displaystyle \therefore \text{The required value of }p\text{ is }\frac{2}{3}.$
$\\$

$\displaystyle \textbf{Question 11: }\text{The line through }A(-2,3)\text{ and }B(4,b)\text{ is perpendicular to the} \\ \text{line }2x-4y=5.\text{ Find the value of }b.\hfill\text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Slope of }AB=\frac{b-3}{4-(-2)}=\frac{b-3}{6}$
$\displaystyle \text{Given equation is }2x-4y=5$
$\displaystyle 4y=2x-5$
$\displaystyle y=\frac{1}{2}x-\frac{5}{4}$
$\displaystyle \therefore \text{Slope }m_2=\frac{1}{2}$
$\displaystyle \text{Since the lines are perpendicular, }m_1m_2=-1$
$\displaystyle \frac{b-3}{6}\times\frac{1}{2}=-1$
$\displaystyle b-3=-12$
$\displaystyle b=-9$
$\displaystyle \therefore \text{The required value of }b\text{ is }-9.$
$\\$

$\displaystyle \textbf{Question 12: }\text{Find the equation of the line passing through }(-5,7)\text{ and parallel to:}$
$\displaystyle \text{i) }x\text{-axis}\qquad \text{ii) }y\text{-axis}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) A line parallel to the }x\text{-axis has slope }0.$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-7=0(x+5)$
$\displaystyle y=7$
$\displaystyle \therefore \text{The required equation is }y=7.$
$\displaystyle \text{ii) A line parallel to the }y\text{-axis is a vertical line.}$
$\displaystyle \text{Since it passes through }(-5,7),$
$\displaystyle x=-5$
$\displaystyle \therefore \text{The required equation is }x=-5.$
$\\$

$\displaystyle \textbf{Question 13:}$
$\displaystyle \text{i) Find the equation of the line passing through }(5,-3)\text{ and parallel to }x-3y=4.$
$\displaystyle \text{ii) Find the equation of the line parallel to }3x+2y=8\text{ and passing through }(0,1).\hfill\text{[ICSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Given equation is }x-3y=4$
$\displaystyle 3y=x-4$
$\displaystyle y=\frac{1}{3}x-\frac{4}{3}$
$\displaystyle \therefore \text{Slope }m=\frac{1}{3}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-(-3)=\frac{1}{3}(x-5)$
$\displaystyle 3y+9=x-5$
$\displaystyle x-3y-14=0$
$\displaystyle \therefore \text{The required equation is }x-3y-14=0.$
$\displaystyle \text{ii) Given equation is }3x+2y=8$
$\displaystyle 2y=-3x+8$
$\displaystyle y=-\frac{3}{2}x+4$
$\displaystyle \therefore \text{Slope }m=-\frac{3}{2}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-1=-\frac{3}{2}(x-0)$
$\displaystyle 2y-2=-3x$
$\displaystyle 3x+2y=2$
$\displaystyle \therefore \text{The required equation is }3x+2y=2.$
$\\$

$\displaystyle \textbf{Question 14: }\text{Find the equation of the line passing through }(-2,1)\text{ and} \\ \text{perpendicular to }4x+5y=6.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given point }(-2,1)$
$\displaystyle \text{Given equation is }4x+5y=6$
$\displaystyle 5y=-4x+6$
$\displaystyle y=-\frac{4}{5}x+\frac{6}{5}$
$\displaystyle \therefore \text{Slope of the given line }=-\frac{4}{5}$
$\displaystyle \text{Slope of the required line }=\frac{5}{4}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-1=\frac{5}{4}(x+2)$
$\displaystyle 4y-4=5x+10$
$\displaystyle 4y=5x+14$
$\displaystyle \therefore \text{The required equation is }4y=5x+14.$
$\\$

$\displaystyle \textbf{Question 15: }\text{Find the equation of the perpendicular bisector of the line segment} \\ \text{joining }(6,-3)\text{ and }(0,3).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P\text{ be the midpoint of the line segment joining }(6,-3)\text{ and }(0,3).$
$\displaystyle P=\left(\frac{6+0}{2},\frac{-3+3}{2}\right)=(3,0)$
$\displaystyle \text{Slope of the line joining the two points }=\frac{3-(-3)}{0-6}=\frac{6}{-6}=-1$
$\displaystyle \text{Slope of the perpendicular bisector}=1$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-0=1(x-3)$
$\displaystyle y=x-3$
$\displaystyle \therefore \text{The required equation is }y=x-3.$
$\\$

$\displaystyle \textbf{Question 16: }B(-5,6)\text{ and }D(1,4)\text{ are vertices of rhombus }ABCD.\text{ Find} \\ \text{the equations of diagonals }BD\text{ and }AC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Slope of }BD=\frac{4-6}{1-(-5)}=\frac{-2}{6}=-\frac{1}{3}$
$\displaystyle \text{Equation of }BD:$
$\displaystyle y-6=-\frac{1}{3}(x+5)$
$\displaystyle 3y-18=-x-5$
$\displaystyle x+3y=13$
$\displaystyle \text{Midpoint of }BD=\left(\frac{-5+1}{2},\frac{6+4}{2}\right)=(-2,5)$
$\displaystyle \text{Since diagonals of a rhombus are perpendicular, slope of }AC=3.$
$\displaystyle \text{Equation of }AC:$
$\displaystyle y-5=3(x+2)$
$\displaystyle y=3x+11$
$\displaystyle \therefore \text{The required equations are }x+3y=13\text{ and }y=3x+11.$
$\\$

$\displaystyle \textbf{Question 17: }A(7,-2)\text{ and }C(-1,-6)\text{ are vertices of square }ABCD.\text{ Find} \\ \text{the equations of diagonals }AC\text{ and }BD.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Slope of }AC=\frac{-6-(-2)}{-1-7}=\frac{-4}{-8}=\frac{1}{2}$
$\displaystyle \text{Equation of }AC:$
$\displaystyle y-(-2)=\frac{1}{2}(x-7)$
$\displaystyle 2y+4=x-7$
$\displaystyle x-2y=11$
$\displaystyle \text{Midpoint of }AC=\left(\frac{7+(-1)}{2},\frac{-2+(-6)}{2}\right)=(3,-4)$
$\displaystyle \text{Since diagonals of a square are perpendicular, slope of }BD=-2.$
$\displaystyle \text{Equation of }BD:$
$\displaystyle y-(-4)=-2(x-3)$
$\displaystyle y+4=-2x+6$
$\displaystyle 2x+y=2$
$\displaystyle \therefore \text{The required equations are }x-2y=11\text{ and }2x+y=2.$
$\\$

$\displaystyle \textbf{Question 18: }A(1,-5),\;B(2,2)\text{ and }C(-2,4)\text{ are the vertices of }\triangle ABC. \\ \text{ Find the equation of:}$
$\displaystyle \text{i) The median through }A\qquad \text{ii) The altitude through }B\qquad \text{iii) The line } \\ \text{hrough }C\text{ and parallel to }AB.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Let }D\text{ be the midpoint of }BC.$
$\displaystyle D=\left(\frac{2+(-2)}{2},\frac{2+4}{2}\right)=(0,3)$
$\displaystyle \text{Slope of }AD=\frac{3-(-5)}{0-1}=\frac{8}{-1}=-8$
$\displaystyle \text{Equation of median }AD:$
$\displaystyle y-3=-8(x-0)$
$\displaystyle y+8x=3$
$\displaystyle \therefore \text{The equation of the median through }A\text{ is }y+8x=3.$
$\displaystyle \text{ii) Slope of }AC=\frac{4-(-5)}{-2-1}=\frac{9}{-3}=-3$
$\displaystyle \text{Slope of altitude through }B=-\frac{1}{-3}=\frac{1}{3}$
$\displaystyle \text{Equation of altitude through }B:$
$\displaystyle y-2=\frac{1}{3}(x-2)$
$\displaystyle 3y-6=x-2$
$\displaystyle 3y=x+4$
$\displaystyle \therefore \text{The equation of the altitude through }B\text{ is }3y=x+4.$
$\displaystyle \text{iii) Slope of }AB=\frac{2-(-5)}{2-1}=7$
$\displaystyle \text{The required line is parallel to }AB,\text{ so its slope is }7.$
$\displaystyle \text{Equation of the line through }C(-2,4):$
$\displaystyle y-4=7(x+2)$
$\displaystyle y=7x+18$
$\displaystyle \therefore \text{The required equation is }y=7x+18.$
$\\$

$\displaystyle \textbf{Question 19:}$
$\displaystyle \text{i) Write down the equation of the line }AB\text{ through }(3,2)\text{ and perpendicular} \\ \text{to }2y=3x+5.$
$\displaystyle \text{ii) }AB\text{ meets the }x\text{-axis at }A\text{ and the }y\text{-axis at }B.\text{ Write} \\ \text{down the coordinates of }A\text{ and }B.\text{ Calculate the area of }\triangle OAB,\text{ where } \\ O\text{ is the origin.}\hfill\text{[ICSE 1995]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Given equation is }2y=3x+5$
$\displaystyle y=\frac{3}{2}x+\frac{5}{2}$
$\displaystyle \therefore \text{Slope of the given line }=\frac{3}{2}$
$\displaystyle \text{Slope of the required perpendicular line }=-\frac{2}{3}$
$\displaystyle \text{The required line passes through }(3,2).$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-2=-\frac{2}{3}(x-3)$
$\displaystyle 3y-6=-2x+6$
$\displaystyle 2x+3y=12$
$\displaystyle \therefore \text{The equation of }AB\text{ is }2x+3y=12.$
$\displaystyle \text{ii) For the }x\text{-intercept, put }y=0.$
$\displaystyle 2x+3(0)=12$
$\displaystyle x=6$
$\displaystyle \therefore A=(6,0)$
$\displaystyle \text{For the }y\text{-intercept, put }x=0.$
$\displaystyle 2(0)+3y=12$
$\displaystyle y=4$
$\displaystyle \therefore B=(0,4)$
$\displaystyle \text{Area of }\triangle OAB=\frac{1}{2}\times OA\times OB$
$\displaystyle =\frac{1}{2}\times6\times4=12\text{ sq. units}$
$\displaystyle \therefore \text{Area of }\triangle OAB=12\text{ sq. units.}$
$\\$

$\displaystyle \textbf{Question 20: }\text{The line }4x-3y+12=0\text{ meets the }x\text{-axis at }A.\text{ Write} \\ \text{the coordinates of }A.\text{ Determine the equation of the line through }A\text{ and} \\ \text{perpendicular to }4x-3y+12=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }4x-3y+12=0$
$\displaystyle \text{On the }x\text{-axis, }y=0$
$\displaystyle 4x+12=0$
$\displaystyle x=-3$
$\displaystyle \therefore A=(-3,0)$
$\displaystyle \text{Also, }4x-3y+12=0$
$\displaystyle 3y=4x+12$
$\displaystyle y=\frac{4}{3}x+4$
$\displaystyle \therefore \text{Slope of the given line }=\frac{4}{3}$
$\displaystyle \text{Slope of the required perpendicular line }=-\frac{3}{4}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-0=-\frac{3}{4}(x+3)$
$\displaystyle 4y=-3(x+3)$
$\displaystyle 3x+4y+9=0$
$\displaystyle \therefore \text{The required equation is }3x+4y+9=0.$
$\\$

$\displaystyle \textbf{Question 21: }\text{The point }P\text{ is the foot of the perpendicular from } \\ A(-5,7)\text{ to the line }2x-3y+18=0.\text{ Determine:}$
$\displaystyle \text{i) The equation of }AP\qquad \text{ii) The coordinates of }P.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Given equation is }2x-3y+18=0$
$\displaystyle 3y=2x+18$
$\displaystyle y=\frac{2}{3}x+6$
$\displaystyle \therefore \text{Slope of the given line }=\frac{2}{3}$
$\displaystyle \text{Slope of }AP=-\frac{3}{2}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-7=-\frac{3}{2}(x+5)$
$\displaystyle 2y-14=-3x-15$
$\displaystyle 3x+2y+1=0$
$\displaystyle \therefore \text{The equation of }AP\text{ is }3x+2y+1=0.$
$\displaystyle \text{ii) }P\text{ is the point of intersection of }2x-3y+18=0\text{ and }3x+2y+1=0.$
$\displaystyle \text{Solving the two equations,}$
$\displaystyle x=-3,\qquad y=4$
$\displaystyle \therefore P=(-3,4).$
$\\$

$\displaystyle \textbf{Question 22: }\text{The points }A,\;B\text{ and }C\text{ are }(4,0),\;(2,2)\text{ and } \\ (0,6)\text{ respectively. Find the equations of }AB\text{ and }BC.\text{ If }AB\text{ cuts the }y\text{-axis at } \\ P\text{ and }BC\text{ cuts the }x\text{-axis at }Q,\text{ find the coordinates of }P\text{ and }Q.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given points }A(4,0),\;B(2,2)\text{ and }C(0,6).$
$\displaystyle \text{Slope of }AB=\frac{2-0}{2-4}=\frac{2}{-2}=-1$
$\displaystyle \text{Slope of }BC=\frac{6-2}{0-2}=\frac{4}{-2}=-2$
$\displaystyle \text{Equation of }AB:$
$\displaystyle y-2=-1(x-2)$
$\displaystyle y-2=-x+2$
$\displaystyle x+y=4$
$\displaystyle \text{Equation of }BC:$
$\displaystyle y-6=-2(x-0)$
$\displaystyle y+2x=6$
$\displaystyle \text{On the }y\text{-axis, }x=0.$
$\displaystyle y=4$
$\displaystyle \therefore P=(0,4)$
$\displaystyle \text{On the }x\text{-axis, }y=0.$
$\displaystyle 2x=6$
$\displaystyle x=3$
$\displaystyle \therefore Q=(3,0)$
$\\$

$\displaystyle \textbf{Question 23: }\text{Find the value of }a\text{ for which }A(a,3),\;B(2,1)\text{ and }C(5,a) \\ \text{ are collinear. Hence, find the equation of the line.}\hfill\text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given points are }A(a,3),\;B(2,1)\text{ and }C(5,a).$
$\displaystyle \text{Slope of }AB=\frac{1-3}{2-a}=\frac{-2}{2-a}$
$\displaystyle \text{Slope of }BC=\frac{a-1}{5-2}=\frac{a-1}{3}$
$\displaystyle \text{Since }A,\;B\text{ and }C\text{ are collinear,}$
$\displaystyle \frac{-2}{2-a}=\frac{a-1}{3}$
$\displaystyle -6=(2-a)(a-1)$
$\displaystyle -6=2a-2-a^2+a$
$\displaystyle -6=3a-a^2-2$
$\displaystyle a^2-3a-4=0$
$\displaystyle (a-4)(a+1)=0$
$\displaystyle a=4\text{ or }a=-1$
$\displaystyle \text{When }a=4,\text{ the points are }A(4,3),\;B(2,1)\text{ and }C(5,4).$
$\displaystyle \text{Slope }m=\frac{1-3}{2-4}=1$
$\displaystyle y-1=1(x-2)$
$\displaystyle y=x-1$
$\displaystyle \text{When }a=-1,\text{ the points are }A(-1,3),\;B(2,1)\text{ and }C(5,-1).$
$\displaystyle \text{Slope }m=\frac{1-3}{2-(-1)}=-\frac{2}{3}$
$\displaystyle y-1=-\frac{2}{3}(x-2)$
$\displaystyle 3y-3=-2x+4$
$\displaystyle 2x+3y=7$
$\displaystyle \therefore \text{The possible equations are }y=x-1\text{ and }2x+3y=7.$
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