Class 10: Equation of a Straight Line

$\displaystyle \textbf{Question 1. }\text{A line intersects }x\text{-axis at point }(-2,0)\text{ and cuts off an} \\ \text{intercept of }3\text{ units from the positive side of }y\text{-axis. Find the equation} \\ \text{of the line. [CBSE 1992]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that the line cuts the }x\text{-axis at }(-2,0)\text{ and the }y\text{-axis at }(0,3).$
$\displaystyle \therefore \text{x-intercept}=(-2,0),\qquad \text{y-intercept}=(0,3)$
$\displaystyle m=\frac{3-0}{0-(-2)}=\frac{3}{2}$
$\displaystyle \text{Using the point-slope form through }(0,3),$
$\displaystyle y-3=\frac{3}{2}(x-0)$
$\displaystyle 2y-6=3x$
$\displaystyle 3x-2y+6=0$
$\displaystyle \text{Hence, the required equation of the line is }3x-2y+6=0.$
$\\$

$\displaystyle \textbf{Question 2. }\text{Find the equation of a line passing through the point }(2,3)\text{ and} \\ \text{having the }x\text{-intercept of }4\text{ units. [CBSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that the line passes through }(2,3)\text{ and has }x\text{-intercept }4\text{ units.}$
$\displaystyle \therefore \text{x-intercept}=(4,0)$
$\displaystyle m=\frac{0-3}{4-2}=-\frac{3}{2}$
$\displaystyle \text{Using the point-slope form through }(4,0),$
$\displaystyle y-0=-\frac{3}{2}(x-4)$
$\displaystyle 2y=-3x+12$
$\displaystyle 3x+2y-12=0$
$\displaystyle \text{Hence, the required equation of the line is }3x+2y-12=0.$
$\\$

$\displaystyle \textbf{Question 3. }\text{The given figure shows two straight lines }AB\text{ and }CD.$
$\displaystyle \text{If equation of the line }AB\text{ is }y=x+1\text{ and equation of }CD\text{ is }y=\sqrt{3}x-1,$
$\displaystyle \text{write down the inclination of lines }AB\text{ and }CD;\text{ also, find the angle between }\\ AB\text{ and }CD.\text{ [ICSE 1989]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For line }AB:y=x+1$
$\displaystyle \text{Slope of }AB=1$
$\displaystyle \tan \alpha_1=1$
$\displaystyle \therefore \alpha_1=45^\circ$
$\displaystyle \text{For line }CD:y=\sqrt{3}x-1$
$\displaystyle \text{Slope of }CD=\sqrt{3}$
$\displaystyle \tan \alpha_2=\sqrt{3}$
$\displaystyle \therefore \alpha_2=60^\circ$
$\displaystyle \text{Angle between the two lines}=60^\circ-45^\circ$
$\displaystyle =15^\circ$
$\displaystyle \text{Hence, the inclinations of }AB\text{ and }CD\text{ are }45^\circ\text{ and }60^\circ\text{ respectively,}$
$\displaystyle \text{and the angle between them is }15^\circ.$
$\\$

$\displaystyle \textbf{Question 4. }\text{Write down the equation of the line whose gradient is }\frac{2}{3} \\ \text{and which passes through }P,\text{ where }P\text{ divides the line segment joining }A(-2,6)\text{ and }\\ B(3,-4)\text{in the ratio }2:3.\text{ [CBSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }P\text{ divides the line joining }A(-2,6)\text{ and }B(3,-4)\text{ in the ratio }2:3.$
$\displaystyle \text{Using the section formula,}$
$\displaystyle x=\frac{2(3)+3(-2)}{2+3}=\frac{6-6}{5}=0$
$\displaystyle y=\frac{2(-4)+3(6)}{2+3}=\frac{-8+18}{5}=2$
$\displaystyle \therefore P=(0,2)$
$\displaystyle \text{Gradient of the line }=\frac{2}{3}$
$\displaystyle \text{Using the point-slope form through }(0,2),$
$\displaystyle y-2=\frac{2}{3}(x-0)$
$\displaystyle 3y-6=2x$
$\displaystyle 2x-3y+6=0$
$\displaystyle \text{Hence, the required equation of the line is }2x-3y+6=0.$
$\\$

$\displaystyle \textbf{Question 5. }\text{Point }A\text{ and }B\text{ have coordinates }(7,-3)\text{ and }(1,9)\text{ respectively. Find:}$
$\displaystyle \text{(i) The slope of }AB$
$\displaystyle \text{(ii) The equation of perpendicular bisector of the line segment }AB$
$\displaystyle \text{(iii) The value of }p\text{ if }(-2,p)\text{ lies on it. [CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Slope of }AB=\frac{9-(-3)}{1-7}=\frac{12}{-6}=-2$
$\displaystyle \text{(ii) Midpoint of }AB=\left(\frac{7+1}{2},\frac{-3+9}{2}\right)=(4,3)$
$\displaystyle \text{Slope of perpendicular bisector}=\frac{1}{2}$
$\displaystyle \text{Equation of line passing through }(4,3)\text{ with slope }\frac{1}{2}\text{ is}$
$\displaystyle y-3=\frac{1}{2}(x-4)$
$\displaystyle 2y-6=x-4$
$\displaystyle 2y=x+2$
$\displaystyle x-2y+2=0$
$\displaystyle \text{(iii) Since }(-2,p)\text{ lies on }2y=x+2,$
$\displaystyle 2p=-2+2$
$\displaystyle 2p=0$
$\displaystyle p=0$
$\displaystyle \text{Hence, the slope of }AB\text{ is }-2,\text{ the perpendicular bisector is }x-2y+2=0,\text{ and }p=0.$
$\\$

$\displaystyle \textbf{Question 6. }\text{ }A\text{ and }B\text{ are two points on the }x\text{-axis and }y \\ \text{-axis respectively. }P(2,-3)\text{ is the midpoint of }AB.\text{ Find the:}$
$\displaystyle \text{(i) Coordinates of }A\text{ and }B$
$\displaystyle \text{(ii) Slope of line }AB$
$\displaystyle \text{(iii) Equation of line }AB\text{. [CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(x,0)\text{ and }B(0,y).$
$\displaystyle \text{Since }P(2,-3)\text{ is the midpoint of }AB,$
$\displaystyle \frac{x+0}{2}=2,\qquad \frac{0+y}{2}=-3$
$\displaystyle x=4,\qquad y=-6$
$\displaystyle \therefore A(4,0)\text{ and }B(0,-6)$
$\displaystyle \text{(ii) Slope of }AB=\frac{-6-0}{0-4}=\frac{-6}{-4}=\frac{3}{2}$
$\displaystyle \text{(iii) Using the point-slope form through }(4,0),$
$\displaystyle y-0=\frac{3}{2}(x-4)$
$\displaystyle 2y=3x-12$
$\displaystyle 3x-2y-12=0$
$\displaystyle \text{Hence, }A=(4,0),\ B=(0,-6),\ \text{slope of }AB=\frac{3}{2},\text{ and the equation of }AB\text{ is }3x-2y-12=0.$
$\\$

$\displaystyle \textbf{Question 7. }\text{The equation of a line is }3x+4y-7=0.\text{ Find:}$
$\displaystyle \text{(i) Slope of the line.}$
$\displaystyle \text{(ii) The equation of a line perpendicular to the given line and passing through the intersection} \\ \text{of the lines }x-y+2=0\text{ and }3x+y-10=0.\text{ [CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given line is }3x+4y-7=0.$
$\displaystyle 4y=-3x+7$
$\displaystyle y=-\frac{3}{4}x+\frac{7}{4}$
$\displaystyle \therefore \text{Slope of the given line}=-\frac{3}{4}$
$\displaystyle \text{Slope of the line perpendicular to it}=\frac{4}{3}$
$\displaystyle \text{Now, }x-y+2=0$
$\displaystyle \Rightarrow y=x+2$
$\displaystyle \text{Substituting }y=x+2\text{ in }3x+y-10=0,$
$\displaystyle 3x+x+2-10=0$
$\displaystyle 4x-8=0$
$\displaystyle x=2$
$\displaystyle y=2+2=4$
$\displaystyle \therefore \text{Point of intersection is }(2,4).$
$\displaystyle \text{Equation of line passing through }(2,4)\text{ with slope }\frac{4}{3}\text{ is}$
$\displaystyle y-4=\frac{4}{3}(x-2)$
$\displaystyle 3y-12=4x-8$
$\displaystyle 3y=4x+4$
$\displaystyle 4x-3y+4=0$
$\displaystyle \text{Hence, the slope of the given line is }-\frac{3}{4}\text{ and the required equation is }4x-3y+4=0.$
$\\$

$\displaystyle \textbf{Question 8. }ABCD\text{ is a parallelogram where }A(x,y),B(5,8),C(4,7)\text{ and }D(2,-4).\text{ Find:}$
$\displaystyle \text{(i) Coordinates of }A$
$\displaystyle \text{(ii) Equation of diagonal }BD.\text{ [CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In a parallelogram, diagonals bisect each other.}$
$\displaystyle \text{Midpoint of }BD=\left(\frac{5+2}{2},\frac{8+(-4)}{2}\right)=\left(\frac{7}{2},2\right)$
$\displaystyle \text{Therefore, midpoint of }AC=\left(\frac{7}{2},2\right)$
$\displaystyle \therefore \frac{x+4}{2}=\frac{7}{2}$
$\displaystyle x+4=7$
$\displaystyle x=3$
$\displaystyle \frac{y+7}{2}=2$
$\displaystyle y+7=4$
$\displaystyle y=-3$
$\displaystyle \therefore A=(3,-3)$
$\displaystyle \text{Now, diagonal }BD\text{ passes through }B(5,8)\text{ and }D(2,-4).$
$\displaystyle \text{Slope of }BD=\frac{-4-8}{2-5}=\frac{-12}{-3}=4$
$\displaystyle y-8=4(x-5)$
$\displaystyle y-8=4x-20$
$\displaystyle y=4x-12$
$\displaystyle 4x-y-12=0$
$\displaystyle \text{Hence, }A=(3,-3)\text{ and the equation of }BD\text{ is }4x-y-12=0.$
$\\$

$\displaystyle \textbf{Question 9. }\text{From the given figure, find:}$
$\displaystyle \text{(i) The coordinates of }A,\ B,\text{ and }C.$
$\displaystyle \text{(ii) The equation of the line through }A\text{ and parallel to }BC.\text{ [CBSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) From the graph, }A=(2,3),\ B=(3,0)\text{ and }C=(-1,2).$
$\displaystyle \text{(ii) Slope of }BC=\frac{0-2}{3-(-1)}=\frac{-2}{4}=-\frac{1}{2}$
$\displaystyle \text{A line parallel to }BC\text{ has the same slope }-\frac{1}{2}.$
$\displaystyle \text{Using the point-slope form through }A(2,3),$
$\displaystyle y-3=-\frac{1}{2}(x-2)$
$\displaystyle 2y-6=-x+2$
$\displaystyle x+2y-8=0$
$\displaystyle \text{Hence, }A=(2,3),\ B=(3,0),\ C=(-1,2)\text{ and the required equation is }x+2y-8=0.$
$\\$

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

$\displaystyle \textbf{Question 11. }\text{Find the value of }k\text{ for which the lines }kx-5y+4=0\text{ and } \\ 5x-2y+5=0\text{ are perpendicular to each other. [CBSE 2003]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }kx-5y+4=0,$
$\displaystyle 5y=kx+4$
$\displaystyle y=\frac{k}{5}x+\frac{4}{5}$
$\displaystyle \therefore m_1=\frac{k}{5}$
$\displaystyle \text{For }5x-2y+5=0,$
$\displaystyle 2y=5x+5$
$\displaystyle y=\frac{5}{2}x+\frac{5}{2}$
$\displaystyle \therefore m_2=\frac{5}{2}$
$\displaystyle \text{Since the two lines are perpendicular, }m_1m_2=-1$
$\displaystyle \frac{k}{5}\cdot\frac{5}{2}=-1$
$\displaystyle \frac{k}{2}=-1$
$\displaystyle k=-2$
$\displaystyle \text{Hence, }k=-2.$
$\\$

$\displaystyle \textbf{Question 12. }\text{A straight line passes through the points }P(-1,4)\text{ and }Q(5,-2). \\ \text{It intersects the coordinate axes at points }A\text{ and }B.\text{ }M\text{ is the midpoint} \\ \text{of the line segment }AB.\text{ Find:}$
$\displaystyle \text{(i) The equation of the line}$
$\displaystyle \text{(ii) The coordinates of }A\text{ and }B$
$\displaystyle \text{(iii) The coordinates of }M.\text{ [CBSE 2003]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given points are }P(-1,4)\text{ and }Q(5,-2).$
$\displaystyle \text{Slope of the line}=\frac{-2-4}{5-(-1)}=\frac{-6}{6}=-1$
$\displaystyle \text{Using the point-slope form through }P(-1,4),$
$\displaystyle y-4=-1(x+1)$
$\displaystyle y-4=-x-1$
$\displaystyle x+y=3$
$\displaystyle \therefore \text{The equation of the line is }x+y=3.$
$\displaystyle \text{For }x\text{-intercept, put }y=0.$
$\displaystyle x+0=3$
$\displaystyle x=3$
$\displaystyle \therefore A=(3,0)$
$\displaystyle \text{For }y\text{-intercept, put }x=0.$
$\displaystyle 0+y=3$
$\displaystyle y=3$
$\displaystyle \therefore B=(0,3)$
$\displaystyle \text{Midpoint of }AB=M=\left(\frac{3+0}{2},\frac{0+3}{2}\right)=\left(\frac{3}{2},\frac{3}{2}\right)$
$\displaystyle \text{Hence, the equation of the line is }x+y=3,\ A=(3,0),\ B=(0,3)\text{ and }M=\left(\frac{3}{2},\frac{3}{2}\right).$
$\\$

$\displaystyle \textbf{Question 13. }\text{If the lines }y=3x+7\text{ and }2y+px=3\text{ are perpendicular to each} \\ \text{other, find the value of }p.\text{ [CBSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the line }y=3x+7,\ \text{slope }(m_1)=3.$
$\displaystyle \text{For the line }2y+px=3,$
$\displaystyle 2y=-px+3$
$\displaystyle y=-\frac{p}{2}x+\frac{3}{2}$
$\displaystyle \therefore \text{slope }(m_2)=-\frac{p}{2}$
$\displaystyle \text{Since the lines are perpendicular, }m_1\times m_2=-1$
$\displaystyle 3\times\left(-\frac{p}{2}\right)=-1$
$\displaystyle -\frac{3p}{2}=-1$
$\displaystyle 3p=2$
$\displaystyle p=\frac{2}{3}$
$\displaystyle \text{Hence, the required value of }p\text{ is }\frac{2}{3}.$
$\\$

$\displaystyle \textbf{Question 14. }\text{The line through }A(-2,3)\text{ and }B(4,b)\text{ is perpendicular to the line } \\ 2x-4y=5.\text{ Find the value of }b.\text{ [CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Slope of }AB=\frac{b-3}{4-(-2)}=\frac{b-3}{6}$
$\displaystyle \text{Given line is }2x-4y=5.$
$\displaystyle 4y=2x-5$
$\displaystyle y=\frac{1}{2}x-\frac{5}{4}$
$\displaystyle \therefore \text{Slope of the given line}=\frac{1}{2}$
$\displaystyle \text{Since the lines are perpendicular, }m_1\times m_2=-1$
$\displaystyle \frac{b-3}{6}\times\frac{1}{2}=-1$
$\displaystyle \frac{b-3}{12}=-1$
$\displaystyle b-3=-12$
$\displaystyle b=-9$
$\displaystyle \text{Hence, the required value of }b\text{ is }-9.$
$\\$

$\displaystyle \textbf{Question 15. }\text{(i) Find the equation of the line passing through }(5,-3)\text{ and} \\ \text{parallel to }x-3y=4.$
$\displaystyle \text{(ii) Find the equation of the line parallel to }3x+2y=8\text{ and passing through the} \\ \text{point }(0,1).\text{ [CBSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given point }(x_1,y_1)=(5,-3).$
$\displaystyle \text{Given line is }x-3y=4.$
$\displaystyle 3y=x-4$
$\displaystyle y=\frac{1}{3}x-\frac{4}{3}$
$\displaystyle \therefore \text{Slope }m=\frac{1}{3}$
$\displaystyle \text{Using the point-slope form,}$
$\displaystyle y-(-3)=\frac{1}{3}(x-5)$
$\displaystyle 3y+9=x-5$
$\displaystyle x-3y-14=0$
$\displaystyle \text{Hence, the required equation is }x-3y-14=0.$
$\displaystyle \text{(ii) Given point }(x_1,y_1)=(0,1).$
$\displaystyle \text{Given line is }3x+2y=8.$
$\displaystyle 2y=-3x+8$
$\displaystyle y=-\frac{3}{2}x+4$
$\displaystyle \therefore \text{Slope }m=-\frac{3}{2}$
$\displaystyle \text{Using the point-slope form,}$
$\displaystyle y-1=-\frac{3}{2}(x-0)$
$\displaystyle 2y-2=-3x$
$\displaystyle 3x+2y-2=0$
$\displaystyle \text{Hence, the required equation is }3x+2y-2=0.$
$\\$

$\displaystyle \textbf{Question 16. }\text{(i) Write down the equation of the line }AB\text{ through }(3,2)\text{ and} \\ \text{perpendicular to the line }2y=3x+5.$
$\displaystyle \text{(ii) }AB\text{ meets the }x\text{-axis at }A\text{ and the }y\text{-axis at }B.\text{ Write down} \\ \text{the coordinates of }A\text{ and }B.\text{ Calculate the area of triangle }OAB,\text{ where } \\ O\text{ is origin. [CBSE 1995]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given point is }(3,2).$
$\displaystyle \text{Given line is }2y=3x+5.$
$\displaystyle y=\frac{3}{2}x+\frac{5}{2}$
$\displaystyle \therefore \text{Slope of the given line}=\frac{3}{2}$
$\displaystyle \text{Slope of the line perpendicular to it}=-\frac{2}{3}$
$\displaystyle \text{Using the point-slope form,}$
$\displaystyle y-2=-\frac{2}{3}(x-3)$
$\displaystyle 3y-6=-2x+6$
$\displaystyle 2x+3y=12$
$\displaystyle \therefore \text{Equation of }AB\text{ is }2x+3y=12.$
$\displaystyle \text{(ii) For }x\text{-intercept, put }y=0.$
$\displaystyle 2x+3(0)=12$
$\displaystyle x=6$
$\displaystyle \therefore A=(6,0)$
$\displaystyle \text{For }y\text{-intercept, put }x=0.$
$\displaystyle 2(0)+3y=12$
$\displaystyle y=4$
$\displaystyle \therefore B=(0,4)$
$\displaystyle \text{Area of }\triangle OAB=\frac{1}{2}\times OA\times OB$
$\displaystyle =\frac{1}{2}\times 6\times 4$
$\displaystyle =12\text{ sq. units}$
$\displaystyle \text{Hence, }A=(6,0),\ B=(0,4)\text{ and area of }\triangle OAB=12\text{ sq. units.}$
$\\$

$\displaystyle \textbf{Question 17. }\text{Find the value of }a\text{ for which the points }A(a,3),\ B(2,1)\text{ and } \\ C(5,a)\text{ are collinear. Hence, find the equation of the line. [CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given points are }A(a,3),\ B(2,1)\text{ and }C(5,a).$
$\displaystyle \text{Slope of }AB=\frac{1-3}{2-a}=\frac{-2}{2-a}$
$\displaystyle \text{Slope of }BC=\frac{a-1}{5-2}=\frac{a-1}{3}$
$\displaystyle \text{Since }A,\ B\text{ and }C\text{ are collinear,}$
$\displaystyle \frac{-2}{2-a}=\frac{a-1}{3}$
$\displaystyle -6=(2-a)(a-1)$
$\displaystyle -6=2a-2-a^2+a$
$\displaystyle -6=3a-a^2-2$
$\displaystyle a^2-3a-4=0$
$\displaystyle (a-4)(a+1)=0$
$\displaystyle \therefore a=4\text{ or }a=-1$
$\displaystyle \text{For }a=4,\text{ the points are }(4,3)\text{ and }(2,1).$
$\displaystyle \text{Slope}=1$
$\displaystyle y-1=x-2$
$\displaystyle y=x-1$
$\displaystyle \text{For }a=-1,\text{ the points are }(-1,3)\text{ and }(2,1).$
$\displaystyle \text{Slope}=\frac{1-3}{2-(-1)}=-\frac{2}{3}$
$\displaystyle y-1=-\frac{2}{3}(x-2)$
$\displaystyle 3y-3=-2x+4$
$\displaystyle 2x+3y-7=0$
$\displaystyle \text{Hence, }a=4\text{ or }-1.\text{ The corresponding equations of the line are }y=x-1\text{ and } \\ 2x+3y-7=0\text{ respectively.}$
$\\$

$\displaystyle \textbf{Question 18. }\text{In }\triangle ABC,\ A=(3,5),\ B=(7,8)\text{ and }C=(1,-10).\text{ Find} \\ \text{the equation of the median through }A.\text{ [CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }D\text{ be the midpoint of }BC.$
$\displaystyle D=\left(\frac{7+1}{2},\frac{8+(-10)}{2}\right)=(4,-1)$
$\displaystyle \text{Slope of }AD=\frac{-1-5}{4-3}=\frac{-6}{1}=-6$
$\displaystyle \text{Using the point-slope form through }A(3,5),$
$\displaystyle y-5=-6(x-3)$
$\displaystyle y-5=-6x+18$
$\displaystyle y=-6x+23$
$\displaystyle 6x+y-23=0$
$\displaystyle \text{Hence, the equation of the median through }A\text{ is }6x+y-23=0.$
$\\$

$\displaystyle \textbf{Question 19. }\text{The line through }P(5,3)\text{ intersects the }y\text{-axis at }Q.$
$\displaystyle \text{(i) Write the slope of the line.}$
$\displaystyle \text{(ii) Write the equation of the line.}$
$\displaystyle \text{(iii) Find the coordinates of }Q.\text{ [CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given point is }P(5,3).$
$\displaystyle \text{From the figure, the inclination of the line is }45^\circ.$
$\displaystyle \therefore \text{Slope }m=\tan45^\circ=1$
$\displaystyle \text{Using the point-slope form through }P(5,3),$
$\displaystyle y-3=1(x-5)$
$\displaystyle y=x-2$
$\displaystyle \text{For the }y\text{-intercept, put }x=0.$
$\displaystyle y=-2$
$\displaystyle \therefore Q=(0,-2)$
$\displaystyle \text{Hence, the slope of the line is }1,\text{ its equation is }y=x-2,\text{ and }Q=(0,-2).$
$\\$

$\displaystyle \textbf{Question 20. }A(1,4),\ B(3,2)\text{ and }C(7,5)\text{ are the vertices of a triangle }ABC.\text{ Find:}$
$\displaystyle \text{(i) The coordinates of the centroid of triangle }ABC.$
$\displaystyle \text{(ii) The equation of a line through the centroid and parallel to }AB.\text{ [CBSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Let }G\text{ be the centroid of }\triangle ABC.$
$\displaystyle G=\left(\frac{1+3+7}{3},\frac{4+2+5}{3}\right)=\left(\frac{11}{3},\frac{11}{3}\right)$
$\displaystyle \therefore \text{The coordinates of the centroid are }\left(\frac{11}{3},\frac{11}{3}\right).$
$\displaystyle \text{(ii) Slope of }AB=\frac{2-4}{3-1}=\frac{-2}{2}=-1$
$\displaystyle \text{A line parallel to }AB\text{ has slope }-1\text{ and passes through }\left(\frac{11}{3},\frac{11}{3}\right).$
$\displaystyle y-\frac{11}{3}=-1\left(x-\frac{11}{3}\right)$
$\displaystyle 3y-11=-3x+11$
$\displaystyle 3x+3y-22=0$
$\displaystyle \text{Hence, the centroid is }\left(\frac{11}{3},\frac{11}{3}\right)\text{ and the required equation is }3x+3y-22=0.$
$\\$

$\displaystyle \textbf{Question 21. }ABC\text{ is a triangle whose vertices are }A(1,-1),\ B(0,4)\text{ and }C(-6,4).\text{ }D\text{ is the midpoint of }BC.\text{ Find: }$
$\displaystyle \text{(i) Coordinates of }D.\text{ (ii) Equation of the median }AD.\text{ [ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given vertices are }A(1,-1),\ B(0,4)\text{ and }C(-6,4).$
$\displaystyle \text{Midpoint of }BC=\left(\frac{0+(-6)}{2},\frac{4+4}{2}\right)=(-3,4)$
$\displaystyle \therefore D=(-3,4)$
$\displaystyle \text{Equation of median }AD\text{ is the equation of the line through }A(1,-1)\text{ and }D(-3,4).$
$\displaystyle \text{Using the two-point form,}$
$\displaystyle \frac{y-(-1)}{4-(-1)}=\frac{x-1}{-3-1}$
$\displaystyle \frac{y+1}{5}=\frac{x-1}{-4}$
$\displaystyle -4(y+1)=5(x-1)$
$\displaystyle -4y-4=5x-5$
$\displaystyle 5x+4y=1$
$\displaystyle \text{Hence, }D=(-3,4)\text{ and the equation of median }AD\text{ is }5x+4y=1.$
$\\$

$\displaystyle \textbf{Question 22. }A(4,2),\ B(6,8)\text{ and }C(8,4)\text{ are the vertices of a triangle }ABC.\\ \text{Write down the equation of the median of the triangle through }A.\text{ [ICSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }D\text{ be the midpoint of }BC.$
$\displaystyle D=\left(\frac{6+8}{2},\frac{8+4}{2}\right)=(7,6)$
$\displaystyle \text{The median through }A\text{ is }AD.$
$\displaystyle \text{Slope of }AD=\frac{6-2}{7-4}=\frac{4}{3}$
$\displaystyle \text{Using the point-slope form through }A(4,2),$
$\displaystyle y-2=\frac{4}{3}(x-4)$
$\displaystyle 3(y-2)=4(x-4)$
$\displaystyle 3y-6=4x-16$
$\displaystyle 4x-3y=10$
$\displaystyle \text{Hence, the equation of the median through }A\text{ is }4x-3y=10.$
$\\$

$\displaystyle \textbf{Question 23. }\text{In the diagram given below, equation of }AB\text{ is } \\ x-\sqrt{3}y+1=0\text{ and equation of }AC\text{ is }x-y-2=0.$
$\displaystyle \text{(i) Write down the angles that the lines }AC\text{ and }AB\text{ make with the positive direction of }x\text{-axis.}$
$\displaystyle \text{(ii) Find }\angle BAC.\text{ [ICSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For line }AB:x-\sqrt{3}y+1=0$
$\displaystyle \sqrt{3}y=x+1$
$\displaystyle y=\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}$
$\displaystyle \therefore \text{Slope of }AB=\frac{1}{\sqrt{3}}$
$\displaystyle \tan \theta_1=\frac{1}{\sqrt{3}}$
$\displaystyle \therefore \theta_1=30^\circ$
$\displaystyle \text{For line }AC:x-y-2=0$
$\displaystyle y=x-2$
$\displaystyle \therefore \text{Slope of }AC=1$
$\displaystyle \tan \theta_2=1$
$\displaystyle \therefore \theta_2=45^\circ$
$\displaystyle \therefore \angle BAC=45^\circ-30^\circ$
$\displaystyle =15^\circ$
$\displaystyle \text{Hence, }AB\text{ makes an angle }30^\circ\text{ and }AC\text{ makes an angle } \\ 45^\circ\text{ with the positive direction of }x\text{-axis, and }\angle BAC=15^\circ.$
$\\$

$\displaystyle \textbf{Question 24. }\text{The slope of a line joining }P(6,k)\text{ and }Q(1-3k,3)\text{ is }\frac{1}{2}.\text{ Find:}$
$\displaystyle \text{(i) The value of }k.$
$\displaystyle \text{(ii) The midpoint of }PQ,\text{ using the value of }k\text{ found in (i). [ICSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given points are }P(6,k)\text{ and }Q(1-3k,3).$
$\displaystyle \text{(i) Since the slope of the line joining these two points is }\frac{1}{2},$
$\displaystyle \frac{3-k}{(1-3k)-6}=\frac{1}{2}$
$\displaystyle \frac{3-k}{-3k-5}=\frac{1}{2}$
$\displaystyle 2(3-k)=-3k-5$
$\displaystyle 6-2k=-3k-5$
$\displaystyle k=-11$
$\displaystyle \text{(ii) When }k=-11,\text{ the coordinates of }P\text{ and }Q\text{ are }(6,-11)\text{ and }(34,3).$
$\displaystyle \text{Midpoint of }PQ=\left(\frac{6+34}{2},\frac{-11+3}{2}\right)$
$\displaystyle =\left(\frac{40}{2},\frac{-8}{2}\right)$
$\displaystyle =(20,-4)$
$\displaystyle \text{Hence, }k=-11\text{ and the midpoint of }PQ\text{ is }(20,-4).$
$\\$

$\displaystyle \textbf{Question 25. }\text{Find the equation of a line with }x\text{-intercept }=5\text{ and passing through the} \\ \text{point }(4,-7).\text{ [ICSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the }x\text{-intercept is }5,\text{ the line intersects the }x\text{-axis at }(5,0).$
$\displaystyle \text{Hence, the line passes through the points }(5,0)\text{ and }(4,-7).$
$\displaystyle \text{Slope of the line}=\frac{-7-0}{4-5}=\frac{-7}{-1}=7$
$\displaystyle \text{Using the point-slope form through }(5,0),$
$\displaystyle y-0=7(x-5)$
$\displaystyle y=7x-35$
$\displaystyle 7x-y-35=0$
$\displaystyle \text{Hence, the required equation of the line is }7x-y-35=0.$
$\\$

$\displaystyle \textbf{Question 26. }\text{Write down the equation of the line whose gradient is }\frac{3}{2}\text{ and which passes} \\ \text{through }P,\text{ where }P\text{ divides the line segment joining }A(-2,6)\text{ and }B(3,-4)\text{ in the} \\ \text{ratio }2:3.\text{ [ICSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the coordinates of }P\text{ be }(x,y).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle P=\left(\frac{2(3)+3(-2)}{2+3},\frac{2(-4)+3(6)}{2+3}\right)$
$\displaystyle =\left(\frac{6-6}{5},\frac{-8+18}{5}\right)=\left(0,\frac{10}{5}\right)=(0,2)$
$\displaystyle \text{The gradient of the required line is }\frac{3}{2}.$
$\displaystyle \text{Using the point-slope form through }P(0,2),$
$\displaystyle y-2=\frac{3}{2}(x-0)$
$\displaystyle 2(y-2)=3x$
$\displaystyle 2y-4=3x$
$\displaystyle 3x-2y+4=0$
$\displaystyle \text{Hence, the required equation of the line is }3x-2y+4=0.$
$\\$

$\displaystyle \textbf{Question 27. }A(2,5),\ B(-1,2)\text{ and }C(5,8)\text{ are the vertices of }\triangle ABC.\ M\text{ is a point on } \\ AB\text{ such that }AM:MB=1:2.\text{ Find the coordinates of }M.\text{ Hence, find the equation of the} \\ \text{line passing through }C\text{ and }M.\text{ [ICSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }A(2,5),\ B(-1,2)\text{ and }C(5,8).$
$\displaystyle \text{Since }AM:MB=1:2,\text{ using the section formula,}$
$\displaystyle M=\left(\frac{1(-1)+2(2)}{1+2},\frac{1(2)+2(5)}{1+2}\right)$
$\displaystyle =\left(\frac{-1+4}{3},\frac{2+10}{3}\right)=(1,4)$
$\displaystyle \text{Now, }M(1,4)\text{ and }C(5,8)\text{ are two points on the required line.}$
$\displaystyle \text{Slope}=\frac{8-4}{5-1}=\frac{4}{4}=1$
$\displaystyle \text{Using the point-slope form through }M(1,4),$
$\displaystyle y-4=x-1$
$\displaystyle y=x+3$
$\displaystyle x-y+3=0$
$\displaystyle \text{Hence, the coordinates of }M\text{ are }(1,4)\text{ and the required equation is }x-y+3=0.$
$\\$

$\displaystyle \textbf{Question 28:}~\text{Prove that }A(2,1),~B(0,3)\text{ and }C(-2,1)\text{ are the three vertices of} \\ \text{an isosceles right angled triangle.}~\text{Hence find the coordinates of a point }D,~\text{if } \\ ABCD\text{ is a square.}~\text{[ICSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, coordinates are }A(2,1),~B(0,3),~C(-2,1)$
$\displaystyle \text{We have by distance formula}$
$\displaystyle AB=\sqrt{(0-2)^2+(3-1)^2}=\sqrt{(-2)^2+2^2}=\sqrt{4+4}=2\sqrt{2}$
$\displaystyle BC=\sqrt{(-2-0)^2+(1-3)^2}=\sqrt{(-2)^2+(-2)^2}=\sqrt{4+4}=2\sqrt{2}$
$\displaystyle CA=\sqrt{(2+2)^2+(1-1)^2}=\sqrt{4^2+0^2}=4$
$\displaystyle \text{Here, }AB=BC\text{ and }AB^2+BC^2=CA^2$

$\displaystyle \therefore A(2,1),~B(0,3),~C(-2,1)\text{ are the three vertices of an isosceles right angled triangle}$
$\displaystyle \text{For coordinates of point }D\text{: Let the coordinates of }D\text{ be }(x,y)$
$\displaystyle \because ABCD\text{ is a square}$
$\displaystyle \text{The coordinates of }O\text{ by mid-point formula}$
$\displaystyle O=\left(\frac{-2+2}{2},\frac{1+1}{2}\right)=(0,1)$
$\displaystyle \text{Mid-point of }AC=\text{Mid-point of }BD$

$\displaystyle (0,1)=\left(\frac{0+x}{2},\frac{3+y}{2}\right)$
$\displaystyle \Rightarrow \frac{x}{2}=0\Rightarrow x=0$
$\displaystyle \Rightarrow \frac{3+y}{2}=1\Rightarrow 3+y=2\Rightarrow y=-1$
$\displaystyle \therefore \text{The coordinates of point }D\text{ are }(0,-1)$

$\displaystyle \textbf{Question 29:}~\text{In the given figure, }ABC\text{ is a triangle and }BC\text{ is parallel} \\ \text{to the Y-axis.}~AB\text{ and }AC\text{ intersect the Y-axis at }P\text{ and }Q\text{ respectively.}~\text{[ICSE 2015]}$

$\displaystyle \text{(i) Write the coordinates of }A.$
$\displaystyle \text{(ii) Find the length of }AB\text{ and }AC.$
$\displaystyle \text{(iii) Find the ratio in which }Q\text{ divides }AC.$
$\displaystyle \text{(iv) Find the equation of the line }AC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) From the given graph, the coordinates of }A\text{ are }(4,0)$
$\displaystyle \text{(ii) Clearly, the length of }AB=\sqrt{(4+2)^2+(0-3)^2}=\sqrt{6^2+(-3)^2}$
$\displaystyle =\sqrt{36+9}=\sqrt{45}=3\sqrt{5}\text{ units}$
$\displaystyle \text{and the length of }AC=\sqrt{(4+2)^2+(0+4)^2}=\sqrt{6^2+4^2}$
$\displaystyle =\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\text{ units}$
$\displaystyle \text{(iii) Let }Q\text{ divides }AC\text{ in the ratio }K:1$
$\displaystyle \text{Then, coordinates of }Q=\left(\frac{-2K+4}{K+1},\frac{-4K+0}{K+1}\right)$
$\displaystyle \text{But }Q\text{ lies on Y-axis, therefore x-coordinate of }Q\text{ will be }0$
$\displaystyle \Rightarrow \frac{-2K+4}{K+1}=0$
$\displaystyle \Rightarrow -2K+4=0\Rightarrow 2K=4\Rightarrow K=2$
$\displaystyle \therefore \text{Required ratio is }2:1$
$\displaystyle \text{(iv) Equation of line }AC\text{ is given by}$
$\displaystyle y-0=\frac{-4-0}{-2-4}(x-4)$
$\displaystyle \Rightarrow y=\frac{-4}{-6}(x-4)=\frac{2}{3}(x-4)$
$\displaystyle \Rightarrow 3y=2x-8$
$\displaystyle \Rightarrow 2x-3y-8=0$

$\displaystyle \textbf{Question 30:}~\text{The line through }P(5,3)\text{ intersects Y-axis at }Q.$

$\displaystyle \text{(i) Write the slope of the line.}$
$\displaystyle \text{(ii) Write the equation of the line.}$
$\displaystyle \text{(iii) Find the coordinates of }Q.~\text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) From the given figure, }\theta=45^\circ$
$\displaystyle \text{Slope of the line }PQ=m=\tan45^\circ=1$
$\displaystyle \text{(ii) The equation of line passing through }P(5,3)\text{ having slope }1$
$\displaystyle y-3=1(x-5)$
$\displaystyle \Rightarrow y-3=x-5$
$\displaystyle \Rightarrow y=x-2$
$\displaystyle \text{(iii) Since, }Q\text{ lies on Y-axis, therefore }x=0$
$\displaystyle \Rightarrow y=0-2=-2$
$\displaystyle \therefore \text{Coordinates of }Q\text{ are }(0,-2)$

$\displaystyle \textbf{Question 31:}~\text{Given equation of line }L_1\text{ is }y=4.$

$\displaystyle \text{(i) Write the slope of line }L_2,~\text{if }L_2\text{ is the bisector of }\angle O.$
$\displaystyle \text{(ii) Write the coordinates of point }P.$
$\displaystyle \text{(iii) Find the equation of }L_2.~\text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }L_2\text{ is a bisector of }\angle O$
$\displaystyle \text{Since, }\angle XOY=90^\circ$
$\displaystyle \therefore \angle POX=\angle POY=\frac{90^\circ}{2}=45^\circ=\theta\text{ (say)}$
$\displaystyle \text{Then, slope of }L_2=\tan45^\circ=1$
$\displaystyle \text{(ii) Let line }L_1\text{ intersects Y-axis at }M.\text{ Draw a line from }P,$
$\displaystyle \text{parallel to X-axis. Then, we get}$

$\displaystyle NP=OM=4\qquad\text{(i)}$
$\displaystyle \text{Also, in }\triangle ONP,$
$\displaystyle \angle PON+\angle ONP+\angle NPO=180^\circ$
$\displaystyle \Rightarrow 45^\circ+90^\circ+\angle NPO=180^\circ$
$\displaystyle \Rightarrow \angle NPO=180^\circ-90^\circ-45^\circ=45^\circ$
$\displaystyle \Rightarrow ON=NP\qquad[\text{sides opposite to equal angles are equal}]$
$\displaystyle \Rightarrow ON=4\qquad[\text{using Eq. (i)}]$
$\displaystyle \text{So, the coordinates of }N\text{ are }(4,0)\text{ and coordinates of }M\text{ are }(0,4)$
$\displaystyle \text{Hence, the coordinates of }P\text{ are }(4,4)$
$\displaystyle \text{(iii) Equation of line }L_2,\text{ which is passing through origin }(0,0)$
$\displaystyle \text{and having slope }1\text{ is}$
$\displaystyle y-0=1(x-0)$
$\displaystyle \Rightarrow y=x$
$\displaystyle \text{Hence, the equation of }L_2\text{ is }y=x$

$\displaystyle \textbf{Question 32:}~A\text{ and }B\text{ are two points on the X-axis and Y-axis respectively.} \\ \text{If }P(2,-3)\text{ is the mid-point of }AB,\text{ then find }$

$\displaystyle \text{(i) coordinates of }A\text{ and }B.$
$\displaystyle \text{(ii) slope of line }AB.$
$\displaystyle \text{(iii) equation of line }AB.~\text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Here, the line is intersecting the X-axis at }A$
$\displaystyle \text{and Y-axis at }B,\text{ so, let the coordinates of }A\text{ and }B\text{ be }(x_1,0)$
$\displaystyle \text{and }(0,y_1),\text{ respectively}$
$\displaystyle \text{So, the coordinates of mid-point of }A\text{ and }B\text{ are }\left(\frac{x_1+0}{2},\frac{0+y_1}{2}\right)$
$\displaystyle \left[\text{mid-point formula }=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right]$
$\displaystyle \text{But given coordinates of mid-point are }(2,-3)$
$\displaystyle \therefore (2,-3)=\left(\frac{x_1}{2},\frac{y_1}{2}\right)$
$\displaystyle \text{On equating the x and y-coordinates of both sides, we get}$
$\displaystyle 2=\frac{x_1}{2}\Rightarrow x_1=4\qquad\text{and}\qquad -3=\frac{y_1}{2}\Rightarrow y_1=-6$
$\displaystyle \text{Hence, the coordinates of }A\text{ are }(4,0)\text{ and }B\text{ are }(0,-6)$
$\displaystyle \text{(ii) Now, slope of line }AB,~m=\frac{y_2-y_1}{x_2-x_1}=\frac{-6-0}{0-4}=\frac{-6}{-4}=\frac{3}{2}$
$\displaystyle \text{(iii) Equation of line passing through the points }A(4,0)\text{ and }B(0,-6)\text{ is given by}$
$\displaystyle y-0=\frac{-6-0}{0-4}(x-4)$
$\displaystyle \Rightarrow y=\frac{-6}{-4}(x-4)$
$\displaystyle \Rightarrow y=\frac{3}{2}(x-4)$
$\displaystyle \Rightarrow 2y=3(x-4)\Rightarrow 2y=3x-12$
$\displaystyle \Rightarrow 3x-2y=12$
$\displaystyle \text{Hence, the equation of line }AB\text{ is }3x-2y=12$

$\displaystyle \textbf{Question 33:}~\text{Find the equation of a line passing through the point }(-2,3)\text{ and having the} \\ \text{x-intercept }4\text{ units.}~\text{[ICSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since x-intercept }=4,\text{ therefore the line meets the X-axis at }(4,0).$
$\displaystyle \text{Thus, the line passes through the points }(4,0)\text{ and }(-2,3).$
$\displaystyle \text{Using equation of a line passing through two points,}$
$\displaystyle y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$
$\displaystyle \Rightarrow y-0=\frac{3-0}{-2-4}(x-4)$
$\displaystyle \Rightarrow y=\frac{3}{-6}(x-4)$
$\displaystyle \Rightarrow y=-\frac{1}{2}(x-4)$
$\displaystyle \Rightarrow y=-\frac{1}{2}x+2$
$\displaystyle \Rightarrow x+2y-4=0$
$\displaystyle \text{Hence, the required equation of the line is }x+2y-4=0.$

$\displaystyle \textbf{Question 34:}~\text{A straight line passes through the points }P(-1,4)\text{ and }Q(5,-2).\text{ It} \\ \text{intersects the coordinate axes at points }A\text{ and }B.\text{ If }M\text{ is the mid-point of the} \\ \text{segment }AB,\text{ then find }$

$\displaystyle \text{(i) the equation of the line.}$
$\displaystyle \text{(ii) the coordinates of }A\text{ and }B.$
$\displaystyle \text{(iii) the coordinates of }M.~\text{[ICSE 2003]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) The equation of line passing through }P(-1,4)\text{ and }Q(5,-2)\text{ is}$
$\displaystyle y-4=\frac{-2-4}{5-(-1)}(x+1)$
$\displaystyle \Rightarrow y-4=\frac{-6}{6}(x+1)=\frac{-6}{6}(x+1)$
$\displaystyle \Rightarrow y-4=-1(x+1)$
$\displaystyle \Rightarrow y-4=-x-1$
$\displaystyle \Rightarrow y+x-3=0$
$\displaystyle \text{which is the required equation of line}$
$\displaystyle \text{(ii) Here, the line }x+y-3=0\text{ meets the X and Y-axes}$
$\displaystyle \text{Putting }y=0\text{ in Eq. (i), we get }x+0-3=0\Rightarrow x=3$
$\displaystyle \text{Thus, the line meets the X-axis at }A(3,0)$
$\displaystyle \text{Now, putting }x=0\text{ in Eq. (i), we get }0+y-3=0\Rightarrow y=3$
$\displaystyle \text{Thus, the line meets the Y-axis at }B(0,3)$
$\displaystyle \text{Hence, the coordinates of }A\text{ are }(3,0)\text{ and }B\text{ are }(0,3)$
$\displaystyle \text{(iii) The coordinates of mid point of }A(3,0)\text{ and }B(0,3)$
$\displaystyle =\left(\frac{3+0}{2},\frac{0+3}{2}\right)=\left(\frac{3}{2},\frac{3}{2}\right)$
$\displaystyle \text{Hence, the coordinates of }M\text{ are }\left(\frac{3}{2},\frac{3}{2}\right)$

$\displaystyle \textbf{Question 35:}~\text{If the lines }7y=ax+4\text{ and }2y=3-x\text{ are parallel to each other,} \\ \text{then the value of }a\text{ is}~\text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, lines are }7y=ax+4\text{ and }2y=3-x$
$\displaystyle \Rightarrow y=\frac{a}{7}x+\frac{4}{7}\text{ and }y=\frac{3}{2}-\frac{x}{2}$
$\displaystyle \text{On comparing with }y=m_1x+c_1\text{ and }y=m_2x+c_2\text{ respectively, we get}$
$\displaystyle m_1=\frac{a}{7}\text{ and }m_2=-\frac{1}{2}$
$\displaystyle \text{Since, the lines are parallel, }m_1=m_2$
$\displaystyle \Rightarrow \frac{a}{7}=-\frac{1}{2}$
$\displaystyle \Rightarrow a=-\frac{7}{2}$

$\displaystyle \textbf{Question 36:}~\text{If two lines are perpendicular to one another, then the relation between} \\ \text{their slopes }m_1\text{ and }m_2\text{ is}~\text{[ICSE Semester II 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) If two lines having slopes }m_1\text{ and }m_2\text{ are perpendicular,}$
$\displaystyle \text{then the angle between them is }90^\circ$
$\displaystyle \text{Also, }m_1\times m_2=-1$

$\displaystyle \textbf{Question 37:}~\text{Line }AB\text{ is perpendicular to }CD.\text{ Coordinates of }B,~C\text{ and }D\text{ are respectively } \\ (4,0),~(0,-1)\text{ and }(4,3).\text{ Find:}$

$\displaystyle \text{(i) slope of }CD.$
$\displaystyle \text{(ii) Equation of }AB.~\text{[ICSE Semester II 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Slope of }CD,~m=\frac{y_2-y_1}{x_2-x_1}$
$\displaystyle \Rightarrow m=\frac{3-(-1)}{4-0}=\frac{4}{4}=1$
$\displaystyle \text{(ii) Equation of }AB$
$\displaystyle y-y_1=-\frac{1}{m}(x-x_1)\qquad[\because AB\perp CD]$
$\displaystyle y-0=-\frac{1}{1}(x-4)$
$\displaystyle \Rightarrow y=-x+4$
$\displaystyle \Rightarrow x+y-4=0$

$\displaystyle \textbf{Question 38:}~\text{Find }a,\text{ if }A(2a+2,3),~B(7,4)\text{ and }C(2a+5,2)\text{ are collinear.} \\ \text{[ICSE Semester II 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, three points }A(2a+2,3),~B(7,4)\text{ and }C(2a+5,2)\text{ are collinear}$
$\displaystyle \therefore \text{Slope of }AB=\text{Slope of }BC$
$\displaystyle \Rightarrow \frac{4-3}{7-(2a+2)}=\frac{2-4}{(2a+5)-7}$
$\displaystyle \Rightarrow \frac{1}{5-2a}=\frac{-2}{2a-2}$
$\displaystyle \Rightarrow 2a-2=-2(5-2a)$
$\displaystyle \Rightarrow 2a-2=-10+4a$
$\displaystyle \Rightarrow 2a=8$
$\displaystyle \Rightarrow a=4$

$\displaystyle \textbf{Question 39:}~\text{(i) If the lines }kx-y+4=0\text{ and }2y=6x+7\text{ are perpendicular} \\ \text{to each other, then find the value of }k.$
$\displaystyle \text{(ii) Find the equation of a line parallel to }2y=6x+7\text{ and passing through }(-1,1). \\ \text{[ICSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }kx-y+4=0\qquad\text{and}\qquad 2y=6x+7$
$\displaystyle \text{From Eq. (i), }y=kx+4$
$\displaystyle \text{On comparing with }y=mx+c,\text{ slope }m_1=k$
$\displaystyle \text{From Eq. (ii), }y=3x+\frac{7}{2}$
$\displaystyle \text{On comparing with }y=mx+c,\text{ slope }m_2=3$
$\displaystyle \text{Since the lines are perpendicular, }m_1\times m_2=-1$
$\displaystyle \Rightarrow k\times 3=-1$
$\displaystyle \Rightarrow k=-\frac{1}{3}$
$\displaystyle \text{(ii) Let }6x-2y+\lambda=0\text{ be the equation of line parallel to }2y=6x+7$
$\displaystyle \text{and passing through }(-1,1)$
$\displaystyle \Rightarrow 6(-1)-2(1)+\lambda=0$
$\displaystyle \Rightarrow -6-2+\lambda=0$
$\displaystyle \Rightarrow \lambda=8$
$\displaystyle \text{Hence, the equation of line is }6x-2y+8=0$
$\displaystyle \text{or }2y=6x+8$

$\displaystyle \textbf{Question 40:}~\text{Find the equation of a line parallel to the line }2x+y-7=0\text{ and passing through the intersection of the lines }x+y-4=0\text{ and }2x-y=8.~\text{[ICSE Semester II 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given lines are }2x+y-7=0\qquad\text{(i)}$
$\displaystyle x+y-4=0\qquad\text{(ii)}$
$\displaystyle 2x-y-8=0\qquad\text{(iii)}$
$\displaystyle \text{From Eq. (i), }y=-2x+7$
$\displaystyle \Rightarrow \text{Slope, }m=-2$
$\displaystyle \text{Now, solving Eqs. (ii) and (iii), we get}$
$\displaystyle x+y-4=0\Rightarrow y=4-x$
$\displaystyle \text{Substituting in Eq. (iii): }2x-(4-x)-8=0$
$\displaystyle \Rightarrow 2x-4+x-8=0$
$\displaystyle \Rightarrow 3x-12=0\Rightarrow x=4$
$\displaystyle \Rightarrow y=4-4=0$
$\displaystyle \text{Thus, }x_1=4\text{ and }y_1=0$
$\displaystyle \text{Equation of required line }y-y_1=m(x-x_1)$
$\displaystyle y-0=-2(x-4)$
$\displaystyle \Rightarrow y=-2x+8$
$\displaystyle \Rightarrow 2x+y=8$

$\displaystyle \textbf{Question 41:}~\text{Find the value of }p\text{ if the lines }5x-3y+2=0\text{ and} \\ 6x-py+7=0 \text{ are perpendicular to each other. Hence, find the equation of a line} \\ \text{passing through }(-2,-1)\text{ and parallel to }6x-py+7=0.~\text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, equation of lines are }5x-3y+2=0\text{ and }6x-py+7=0$
$\displaystyle \Rightarrow -3y=-5x-2\Rightarrow y=\frac{5}{3}x+\frac{2}{3}$
$\displaystyle \Rightarrow -py=-6x-7\Rightarrow y=\frac{6}{p}x+\frac{7}{p}$
$\displaystyle \text{On comparing with }y=m_1x+c_1\text{ and }y=m_2x+c_2\text{ respectively}$
$\displaystyle m_1=\frac{5}{3}\text{ and }m_2=\frac{6}{p}$
$\displaystyle \text{Since, given lines are perpendicular}$
$\displaystyle m_1\times m_2=-1$
$\displaystyle \Rightarrow \frac{5}{3}\times\frac{6}{p}=-1$
$\displaystyle \Rightarrow \frac{10}{p}=-1$
$\displaystyle \Rightarrow p=-10$
$\displaystyle \text{When }p=-10,\text{ slope of line }6x-py+7=0$
$\displaystyle m=\frac{6}{-10}=-\frac{3}{5}$
$\displaystyle \text{Equation of a line passing through }(-2,-1)\text{ and parallel to }6x-py+7=0$
$\displaystyle y+1=-\frac{3}{5}(x+2)$
$\displaystyle \Rightarrow 5y+5=-3x-6$
$\displaystyle \Rightarrow 3x+5y+11=0$

$\displaystyle \textbf{Question 42:}~\text{From the above figure, find:}$

$\displaystyle \text{(i) coordinates of points }P,~Q,~R.$
$\displaystyle \text{(ii) equation of the line through }P\text{ and parallel to }QR.~\text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) From the graph, }P(4,-3),~Q(-2,-2)\text{ and }R(2,0)$
$\displaystyle \text{(ii) Slope of line joining }QR=\frac{0-(-2)}{2-(-2)}=\frac{2}{4}=\frac{1}{2}$
$\displaystyle \text{Since, required line is parallel to }QR,\text{ slope }=\frac{1}{2}$
$\displaystyle \text{Equation of line passing through }P(4,-3)\text{ and having slope }\frac{1}{2}$
$\displaystyle y-(-3)=\frac{1}{2}(x-4)$
$\displaystyle \Rightarrow y+3=\frac{1}{2}(x-4)$
$\displaystyle \Rightarrow 2y+6=x-4$
$\displaystyle \Rightarrow x-2y-10=0$

$\displaystyle \textbf{Question 43:}~\text{Find the equation of the perpendicular bisector of line segment} \\ \text{joining }A(4,2)\text{ and }B(-3,-5).~\text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }A(4,2)\text{ and }B(-3,-5)$
$\displaystyle \text{Mid-point of line segment joining }A\text{ and }B$
$\displaystyle =\left(\frac{4+(-3)}{2},\frac{2+(-5)}{2}\right)=\left(\frac{1}{2},-\frac{3}{2}\right)$
$\displaystyle \left[\text{mid-point formula }=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right]$
$\displaystyle \text{Slope of line joining }A(4,2)\text{ and }B(-3,-5)$
$\displaystyle =\frac{-5-2}{-3-4}=\frac{-7}{-7}=1$
$\displaystyle \left[\text{slope }=\frac{y_2-y_1}{x_2-x_1}\right]$
$\displaystyle \text{Since, required line is perpendicular to }AB$
$\displaystyle \text{Slope of required line }=-\frac{1}{1}=-1$
$\displaystyle \text{Equation of line passing through }(x_0,y_0)\text{ and slope }m\text{ is}$
$\displaystyle y-y_0=m(x-x_0)$
$\displaystyle \text{Required line passes through }\left(\frac{1}{2},-\frac{3}{2}\right)\text{ and slope }-1$
$\displaystyle \Rightarrow y+\frac{3}{2}=-1\left(x-\frac{1}{2}\right)$
$\displaystyle \Rightarrow 2y+3=-2x+1$
$\displaystyle \Rightarrow 2x+2y+2=0\Rightarrow x+y+1=0$

$\displaystyle \textbf{Question 44:}~\text{The vertices of }\triangle ABC\text{ are }A(3,8),~B(-1,2)\text{ and }C(6,-6).$
$\displaystyle \text{(i) slope of }BC.$
$\displaystyle \text{(ii) equation of a line perpendicular to }BC\text{ and passing through }A.~\text{[ICSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given vertices of }\triangle ABC\text{ are }A(3,8),~B(-1,2)\text{ and }C(6,-6)$
$\displaystyle \text{(i) Now, slope of }BC,~m=\frac{-6-2}{6-(-1)}=\frac{-8}{7}$
$\displaystyle \text{(ii) Equation of line passing through }A(3,8)\text{ and perpendicular to }BC$
$\displaystyle \text{Slope of perpendicular line }=\frac{7}{8}$
$\displaystyle y-8=\frac{7}{8}(x-3)$
$\displaystyle \Rightarrow 8y-64=7x-21$
$\displaystyle \Rightarrow 7x-8y+43=0$

$\displaystyle \textbf{Question 45:}~\text{If the straight lines }3x-5y=7\text{ and }4x+ay+9=0\text{ are perpendicular} \\ \text{to one another, then find the value of }a.~\text{[ICSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, equation of straight lines are }3x-5y=7\text{ and }4x+ay+9=0$
$\displaystyle \Rightarrow 5y=3x-7\Rightarrow y=\frac{3}{5}x-\frac{7}{5}\qquad\text{(i)}$
$\displaystyle \Rightarrow ay=-4x-9\Rightarrow y=-\frac{4}{a}x-\frac{9}{a}\qquad\text{(ii)}$
$\displaystyle \text{On comparing Eq. (i) by }y=m_1x+c_1\text{ and Eq. (ii) by }y=m_2x+c_2$
$\displaystyle m_1=\frac{3}{5}\text{ and }m_2=-\frac{4}{a}$
$\displaystyle \text{Since, given straight lines are perpendicular}$
$\displaystyle m_1\times m_2=-1$
$\displaystyle \Rightarrow \frac{3}{5}\times\left(-\frac{4}{a}\right)=-1$
$\displaystyle \Rightarrow -\frac{12}{5a}=-1$
$\displaystyle \Rightarrow 5a=12\Rightarrow a=\frac{12}{5}$

$\displaystyle \textbf{Question 46:}~\text{A line }AB\text{ meets X-axis at }A\text{ and Y-axis at }B\text{ and } \\ P(4,-1)\text{ divides }AB\text{ in the ratio }1:2.\text{ Find:}$

$\displaystyle \text{(i) the coordinates of }A\text{ and }B.$
$\displaystyle \text{(ii) the equation of a line through }P\text{ and perpendicular to }AB.~\text{[ICSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Let coordinates of }A\text{ on X-axis be }(x,0)\text{ and }B\text{ on Y-axis be }(0,y)$

$\displaystyle \text{Given }P(4,-1)\text{ divides }AB\text{ in ratio }1:2$
$\displaystyle \text{By section formula, }x_P=\frac{1\times0+2\times x}{1+2}$
$\displaystyle 4=\frac{2x}{3}\Rightarrow x=6$
$\displaystyle \text{Similarly, }y_P=\frac{1\times y+2\times0}{1+2}$
$\displaystyle -1=\frac{y}{3}\Rightarrow y=-3$
$\displaystyle \text{Hence, coordinates of }A\text{ and }B\text{ are }(6,0)\text{ and }(0,-3)$
$\displaystyle \text{(ii) Clearly, slope of line }AB=\frac{-3-0}{0-6}=\frac{-3}{-6}=\frac{1}{2}$
$\displaystyle \text{Let }m\text{ be slope of desired line perpendicular to }AB$
$\displaystyle \frac{1}{2}\times m=-1\Rightarrow m=-2$
$\displaystyle \text{Equation of line passing through }(4,-1)\text{ and slope }-2$
$\displaystyle y+1=-2(x-4)$
$\displaystyle \Rightarrow y+1=-2x+8$
$\displaystyle \Rightarrow 2x+y-7=0$

$\displaystyle \textbf{Question 47:}~\text{Find the value of }a\text{ for which the following points }A(a,3),~B(2,1)\text{ and } \\ C(5,a)\text{ are collinear. Hence, find the equation of the line.}~\text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{If three points }A,~B\text{ and }C\text{ are collinear, then use the relation}$
$\displaystyle \text{slope of }AB=\text{slope of }BC$
$\displaystyle \text{where slope of a line is given by }\frac{y_2-y_1}{x_2-x_1}$
$\displaystyle \text{Further, use equation }(y-y_1)=m(x-x_1)\text{ to get the required}$
$\displaystyle \text{equation of line}$
$\displaystyle \text{Given, points }A(a,3),~B(2,1)\text{ and }C(5,a)\text{ are collinear}$
$\displaystyle \text{So, slope of }AB=\text{slope of }BC$
$\displaystyle \therefore \frac{1-3}{2-a}=\frac{a-1}{5-2}$
$\displaystyle \Rightarrow \frac{-2}{2-a}=\frac{a-1}{3}$
$\displaystyle \Rightarrow -2\times3=(a-1)(2-a)$
$\displaystyle \Rightarrow -6=2a-a^2-2+a$
$\displaystyle \Rightarrow -6=-a^2+3a-2$
$\displaystyle \Rightarrow a^2-3a-4=0$
$\displaystyle \Rightarrow a^2-4a+a-4=0\qquad[\text{splitting the middle term}]$
$\displaystyle \Rightarrow a(a-4)+1(a-4)=0$
$\displaystyle \Rightarrow (a+1)(a-4)=0$
$\displaystyle \Rightarrow a=-1,~4$
$\displaystyle \text{So, the given points become either }A(-1,3),~B(2,1),~C(5,-1)$
$\displaystyle \text{or }A(4,3),~B(2,1),~C(5,4)$
$\displaystyle \text{Now, equations of line is:}$
$\displaystyle \text{Case I: When collinear points are }A(-1,3),~B(2,1)\text{ and }C(5,-1)$
$\displaystyle \text{Then, equation of line passing through }A(-1,3)\text{ and }B(2,1)\text{ is}$
$\displaystyle y-3=\frac{1-3}{2-(-1)}(x+1)$
$\displaystyle \Rightarrow y-3=\frac{-2}{3}(x+1)$
$\displaystyle \Rightarrow 3(y-3)=-2(x+1)$
$\displaystyle \Rightarrow 3y-9=-2x-2$
$\displaystyle \Rightarrow 3y+2x=7$
$\displaystyle \text{Case II: When collinear points are }A(4,3),~B(2,1)\text{ and }C(5,4)$
$\displaystyle \text{Then, equation of line passing through }A(4,3)\text{ and }B(2,1)\text{ is}$
$\displaystyle y-3=\frac{1-3}{2-4}(x-4)$
$\displaystyle \Rightarrow y-3=\frac{-2}{-2}(x-4)$
$\displaystyle \Rightarrow y-3=x-4$
$\displaystyle \Rightarrow y=x-1$
$\displaystyle \text{Hence, equations of the line is }3y+2x=7\text{ or }y=x-1$

$\displaystyle \textbf{Question 48:}~\text{The line through }A(-2,3)\text{ and }B(4,b)\text{ is perpendicular to the line } \\ 2x-4y=5.\text{ Find the value of }b.~\text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, points are }A(-2,3)\text{ and }B(4,b)$
$\displaystyle \text{Now, slope of line joining the points }A\text{ and }B\text{ is}$
$\displaystyle m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{b-3}{4-(-2)}=\frac{b-3}{6}\qquad\text{(i)}$
$\displaystyle \text{Also, given equation of line is }2x-4y=5$
$\displaystyle \Rightarrow 4y=2x-5$
$\displaystyle \Rightarrow y=\frac{2x-5}{4}\qquad[\text{dividing both sides by }4]$
$\displaystyle \Rightarrow y=\frac{1}{2}x-\frac{5}{4}$
$\displaystyle \text{On comparing with }y=m_2x+c\text{ we get }m_2=\frac{1}{2}\qquad\text{(ii)}$
$\displaystyle \text{According to the question, the lines are perpendicular, therefore}$
$\displaystyle m_1\times m_2=-1$
$\displaystyle \Rightarrow \frac{b-3}{6}\times\frac{1}{2}=-1$
$\displaystyle \Rightarrow b-3=-12$
$\displaystyle \Rightarrow b=-9$
$\displaystyle \text{Hence, the value of }b\text{ is }-9$

$\displaystyle \textbf{Question 49:}~\text{The equation of a line is }3x+4y-7=0.\text{ Find:}$
$\displaystyle \text{(i) the slope of the line.}$
$\displaystyle \text{(ii) the equation of a line perpendicular to the given line and passing through the} \\ \text{intersection of the lines }x-y+2=0\text{ and }3x+y-10=0.~\text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given equation of line is }3x+4y-7=0$
$\displaystyle \Rightarrow 4y=-3x+7$
$\displaystyle \Rightarrow y=-\frac{3}{4}x+\frac{7}{4}\qquad[\text{dividing both sides by }4]$
$\displaystyle \text{On comparing with }y=mx+c,\text{ slope }(m)=-\frac{3}{4}$
$\displaystyle \text{Hence, the slope of line is }-\frac{3}{4}$
$\displaystyle \text{(ii) Now, slope of required line perpendicular to given line is}$
$\displaystyle m_1=-\frac{1}{m}=-\frac{1}{-\frac{3}{4}}=\frac{4}{3}\qquad[\because m_1m_2=-1]$
$\displaystyle \text{Also, given lines are }x-y+2=0\qquad\text{(i)}$
$\displaystyle \text{and }3x+y-10=0\qquad\text{(ii)}$
$\displaystyle \text{On adding Eqs. (i) and (ii), we get}$
$\displaystyle 4x-8=0\Rightarrow x=2$
$\displaystyle \text{On putting }x=2\text{ in Eq. (i), we get }2-y+2=0$
$\displaystyle \Rightarrow y=4$
$\displaystyle \text{So, point of intersection of lines (i) and (ii) is }(2,4)$
$\displaystyle \text{Equation of line passing through }(2,4)\text{ and slope }\frac{4}{3}\text{ is}$
$\displaystyle y-4=\frac{4}{3}(x-2)$
$\displaystyle \Rightarrow 3(y-4)=4x-8$
$\displaystyle \Rightarrow 3y-12=4x-8$
$\displaystyle \Rightarrow 3y=4x+4$
$\displaystyle \Rightarrow 4x-3y+4=0$
$\displaystyle \text{Hence, equation of required line is }3y=4x+4\text{ or }4x-3y+4=0$

$\displaystyle \textbf{Question 50:}~\text{Find the value of }p\text{ for which the lines }2x+3y-7=0\text{ and } \\ 4y-px-12=0\text{ are perpendicular to each other.}~\text{[ICSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{If lines having slopes }m_1\text{ and }m_2\text{ are perpendicular, then }m_1m_2=-1$
$\displaystyle \text{Given lines are }2x+3y-7=0\text{ and }4y-px-12=0$
$\displaystyle \Rightarrow 3y=-2x+7\Rightarrow y=-\frac{2}{3}x+\frac{7}{3}$
$\displaystyle \Rightarrow 4y=px+12\Rightarrow y=\frac{p}{4}x+3$
$\displaystyle \text{On comparing with }y=m_1x+c_1\text{ and }y=m_2x+c_2$
$\displaystyle m_1=-\frac{2}{3}\text{ and }m_2=\frac{p}{4}$
$\displaystyle \text{Since, lines are perpendicular}$
$\displaystyle m_1m_2=-1$
$\displaystyle \Rightarrow -\frac{2}{3}\times\frac{p}{4}=-1$
$\displaystyle \Rightarrow -\frac{2p}{12}=-1$
$\displaystyle \Rightarrow -\frac{p}{6}=-1$
$\displaystyle \Rightarrow p=6$
$\displaystyle \text{Hence, value of }p\text{ is }6$

$\displaystyle \textbf{Question 51:}~\text{Find the equation of the line parallel to }3x+2y=8\text{ and passing} \\ \text{through the point }(0,1).~\text{[ICSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation of line is }3x+2y=8$
$\displaystyle \Rightarrow 2y=-3x+8\Rightarrow y=-\frac{3}{2}x+4$
$\displaystyle \text{On comparing with }y=mx+c,\text{ slope }m=-\frac{3}{2}$
$\displaystyle \text{Since, required line is parallel, slope remains }-\frac{3}{2}$
$\displaystyle \text{Equation passing through }(0,1)\text{ is}$
$\displaystyle y-1=-\frac{3}{2}(x-0)$
$\displaystyle \Rightarrow 2(y-1)=-3x$
$\displaystyle \Rightarrow 2y-2=-3x$
$\displaystyle \Rightarrow 3x+2y-2=0$

$\displaystyle \textbf{Question 52:}~\text{If the lines }y=3x+7\text{ and }2y+px=3\text{ are perpendicular to each} \\ \text{other, then find the value of }p.~\text{[ICSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given lines are }y=3x+7\text{ and }2y+px=3$
$\displaystyle \text{First line is already in the form }y=mx+c$
$\displaystyle \therefore m_1=3$
$\displaystyle \text{From second equation, }2y= -px+3$
$\displaystyle \Rightarrow y=-\frac{p}{2}x+\frac{3}{2}$
$\displaystyle \therefore m_2=-\frac{p}{2}$
$\displaystyle \text{Since the lines are perpendicular, }m_1m_2=-1$
$\displaystyle \Rightarrow 3\left(-\frac{p}{2}\right)=-1$
$\displaystyle \Rightarrow -\frac{3p}{2}=-1$
$\displaystyle \Rightarrow -3p=-2$
$\displaystyle \Rightarrow p=\frac{2}{3}$
$\displaystyle \text{Hence, the value of }p\text{ is }\frac{2}{3}.$

$\displaystyle \textbf{Question 53:}~\text{In the given figure, }ABC\text{ is a triangle, where }B(4,-4)\text{ and }C(-4,-2). \\ D\text{ is a point on }AC.$

$\displaystyle \text{(i) Write down the coordinates of }A\text{ and }D.$
$\displaystyle \text{(ii) Find the coordinates of the centroid of }\triangle ABC.$
$\displaystyle \text{(iii) If }D\text{ divides }AC\text{ in the ratio }k:1,\text{ then find the value of }k.$
$\displaystyle \text{(iv) Find the equation of the line }BD.~\text{[ICSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Coordinates of }A\text{ are }(0,6)\text{ and }D\text{ are }(-3,0)$
$\displaystyle \text{(ii) Coordinates of vertices of }\triangle ABC\text{ are }A(0,6),~B(4,-4),~C(-4,-2)$
$\displaystyle \text{Centroid of }\triangle ABC=\left(\frac{0+4-4}{3},\frac{6-4-2}{3}\right)$
$\displaystyle =\left(\frac{0}{3},\frac{0}{3}\right)=(0,0)$
$\displaystyle \text{Thus, origin is the centroid}$
$\displaystyle \text{(iii) Given }D\text{ divides }AC\text{ in ratio }k:1$
$\displaystyle (-3,0)=\left(\frac{k(-4)+1(0)}{k+1},\frac{k(-2)+1(6)}{k+1}\right)$
$\displaystyle \text{On comparing y-coordinates, }0=\frac{-2k+6}{k+1}$
$\displaystyle \Rightarrow -2k+6=0$
$\displaystyle \Rightarrow 2k=6$
$\displaystyle \Rightarrow k=3$
$\displaystyle \text{(iv) The line }BD\text{ passes through the points }(4,-4)\text{ and }(-3,0)$
$\displaystyle \text{Slope of }BD=\frac{0-(-4)}{-3-4}=\frac{4}{-7}=-\frac{4}{7}$
$\displaystyle \text{Equation of line }BD\text{ is}$
$\displaystyle y+4=-\frac{4}{7}(x-4)$
$\displaystyle \Rightarrow 7(y+4)=-4(x-4)$
$\displaystyle \Rightarrow 7y+28=-4x+16$
$\displaystyle \Rightarrow 4x+7y+12=0$
$\displaystyle \text{which is the required equation of line }BD$

$\displaystyle \textbf{Question 54:}~A\text{ and }B\text{ are two points on the X-axis and Y-axis, respectively.}$

$\displaystyle \text{(i) Write down the coordinates of }A\text{ and }B.$
$\displaystyle \text{(ii) }P\text{ is a point on }AB\text{ such that }AP:PB=3:1.\text{ Using section formula, find} \\ \text{the coordinates of point }P.$
$\displaystyle \text{(iii) Find the equation of a line passing through }P\text{ and perpendicular to }AB.~\text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Coordinates of }A=(4,0)\text{ and coordinates of }B=(0,4)$
$\displaystyle \text{(ii) Given that, }P\text{ divides }AB\text{ internally in ratio }3:1$
$\displaystyle \text{By section formula, }P=\left(\frac{3\times0+1\times4}{3+1},\frac{3\times4+1\times0}{3+1}\right)$
$\displaystyle =\left(\frac{4}{4},\frac{12}{4}\right)=(1,3)$
$\displaystyle \text{(iii) Equation of line }AB$
$\displaystyle y-0=\frac{4-0}{0-4}(x-4)$
$\displaystyle \Rightarrow y=\frac{4}{-4}(x-4)$
$\displaystyle \Rightarrow y=-(x-4)$
$\displaystyle \Rightarrow y=4-x$
$\displaystyle \text{On comparing with }y=mx+c,\text{ slope }m=-1$
$\displaystyle \text{Slope of line perpendicular to }AB\text{ is }1$
$\displaystyle \text{Equation passing through }P(1,3)$
$\displaystyle y-3=1(x-1)$
$\displaystyle \Rightarrow y=x+2$

$\displaystyle \textbf{Question 55:}~ABCD\text{ is a square, where }B(1,3)\text{ and }D(3,2)\text{ are the end points} \\ \text{of the diagonal }BD.\text{ Find:}$
$\displaystyle \text{(i) the coordinates of point of intersection of the diagonals }AC\text{ and }BD.$
$\displaystyle \text{(ii) the equation of the diagonal }AC.~\text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Diagonal }AC\text{ is perpendicular bisector of }BD$

$\displaystyle \text{Coordinates of }O=\left(\frac{1+3}{2},\frac{3+2}{2}\right)=\left(2,\frac{5}{2}\right)$
$\displaystyle \text{Hence, coordinates of intersection of diagonals }AC\text{ and }BD\text{ are }\left(2,\frac{5}{2}\right)$
$\displaystyle \text{(ii) Slope of }BD=\frac{2-3}{3-1}=-\frac{1}{2}$
$\displaystyle \text{Since, }AC\perp BD,\text{ slope of }AC=2$
$\displaystyle y-\frac{5}{2}=2(x-2)$
$\displaystyle \Rightarrow y-\frac{5}{2}=2x-4$
$\displaystyle \Rightarrow 2y-5=4x-8$
$\displaystyle \Rightarrow 4x-2y-3=0$

$\displaystyle \textbf{Question 56:}~\text{From the given figure:}$

$\displaystyle \text{(i) Write down the coordinates of }A\text{ and }B.$
$\displaystyle \text{(ii) If }P\text{ divides }AB\text{ in the ratio }2:3,\text{ then find the coordinates of point }P.$
$\displaystyle \text{(iii) Find the equation of a line parallel to line }AB\text{ and passing through origin.}~\text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) From figure, line }AB\text{ intersects X-axis at }(5,0)\text{ and Y-axis at }(0,3)$
$\displaystyle \text{Hence, coordinates of }A(5,0)\text{ and }B(0,3)$
$\displaystyle \textbf{(ii) }P\text{ divides }AB\text{ in the ratio }2:3$
$\displaystyle \text{Let }m_1=2\text{ and }m_2=3$
$\displaystyle \text{Coordinates of }P=\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)$
$\displaystyle \text{Here, }(x_1,y_1)=(5,0)\text{ and }(x_2,y_2)=(0,3)$
$\displaystyle \Rightarrow P=\left(\frac{2\times0+3\times5}{2+3},\frac{2\times3+3\times0}{2+3}\right)$
$\displaystyle =\left(\frac{15}{5},\frac{6}{5}\right)=\left(3,\frac{6}{5}\right)$
$\displaystyle \text{Hence, coordinates of }P\text{ are }\left(3,\frac{6}{5}\right)$
$\displaystyle \textbf{(iii) Now, slope of line }AB=\frac{3-0}{0-5}=-\frac{3}{5}$
$\displaystyle \text{Let equation of line parallel to }AB\text{ be }y=mx+c$
$\displaystyle \text{Since line passes through origin, }0=m\cdot0+c$
$\displaystyle \Rightarrow c=0\Rightarrow y=mx$
$\displaystyle \text{Slope }m=-\frac{3}{5}$
$\displaystyle \therefore \text{Equation of line is }y=-\frac{3}{5}x$