Class 10: Equation of a Line

$\displaystyle \textbf{Question 1: }\text{Find the equation of a line whose y-intercept is }2\text{ and slope is }3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given y-intercept }=2,\text{ the line passes through }(0,2).$
$\displaystyle \text{Also, slope }m=3.$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-2=3(x-0)$
$\displaystyle y=3x+2$
$\displaystyle \therefore \text{The required equation is }y=3x+2.$
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$\displaystyle \textbf{Question 2: }\text{Find the equation of a line whose y-intercept is }-1\text{ and inclination is }45^\circ.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Slope }m=\tan45^\circ=1.$
$\displaystyle \text{Since the y-intercept is }-1,\text{ the line passes through }(0,-1).$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-(-1)=1(x-0)$
$\displaystyle y+1=x$
$\displaystyle y=x-1$
$\displaystyle \therefore \text{The required equation is }y=x-1.$
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$\displaystyle \textbf{Question 3: }\text{Find the equation of the line whose slope is }-\frac{4}{3}\text{ and which} \\ \text{passes through }(-3,4).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, slope }m=-\frac{4}{3}\text{ and the point }(-3,4).$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-4=-\frac{4}{3}(x+3)$
$\displaystyle 3y-12=-4x-12$
$\displaystyle 4x+3y=0$
$\displaystyle \therefore \text{The required equation is }4x+3y=0.$
$\\$

$\displaystyle \textbf{Question 4: }\text{Find the equation of the line passing through }(5,4)\text{ and making an} \\ \text{angle of }60^\circ\text{ with the positive }x\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Slope }m=\tan60^\circ=\sqrt{3}.$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-4=\sqrt{3}(x-5)$
$\displaystyle \sqrt{3}x-y=5\sqrt{3}-4$
$\displaystyle y=\sqrt{3}x+4-5\sqrt{3}$
$\displaystyle \therefore \text{The required equation is }y=\sqrt{3}x+4-5\sqrt{3}.$
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$\displaystyle \textbf{Question 5: }\text{Find the equation of the line passing through:}$
$\displaystyle \text{i) }(0,1)\text{ and }(1,2)\qquad \text{ii) }(-1,-4)\text{ and }(3,0)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Slope }m=\frac{2-1}{1-0}=1$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-1=x-0$
$\displaystyle y=x+1$
$\displaystyle \therefore \text{The required equation is }y=x+1.$
$\displaystyle \text{ii) Slope }m=\frac{0-(-4)}{3-(-1)}=\frac{4}{4}=1$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-0=1(x-3)$
$\displaystyle y=x-3$
$\displaystyle \therefore \text{The required equation is }y=x-3.$
$\\$

$\displaystyle \textbf{Question 6: }\text{The co-ordinates of }P\text{ and }Q\text{ are }(2,6)\text{ and }(-3,5)\text{ respectively. Find:}$
$\displaystyle \text{i) The gradient of }PQ\qquad\text{ii) The equation of }PQ\qquad\text{iii) The point where }PQ\text{ intersects the }x\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Slope of }PQ=m=\frac{5-6}{-3-2}=\frac{-1}{-5}=\frac{1}{5}$
$\displaystyle \text{ii) Using }y-y_1=m(x-x_1),$
$\displaystyle y-5=\frac{1}{5}(x+3)$
$\displaystyle 5y-25=x+3$
$\displaystyle x-5y+28=0$
$\displaystyle \therefore \text{The equation of }PQ\text{ is }x-5y+28=0.$
$\displaystyle \text{iii) On the }x\text{-axis, }y=0.$
$\displaystyle x-5(0)+28=0$
$\displaystyle x=-28$
$\displaystyle \therefore \text{The line intersects the }x\text{-axis at }(-28,0).$
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$\displaystyle \textbf{Question 7: }\text{The co-ordinates of }A\text{ and }B\text{ are }(-3,4)\text{ and }(2,-1).\text{ Find:}$
$\displaystyle \text{i) The equation of }AB\qquad\text{ii) The point where }AB\text{ intersects the }y\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Slope of }AB=m=\frac{-1-4}{2-(-3)}=\frac{-5}{5}=-1$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y+1=-1(x-2)$
$\displaystyle y+1=-x+2$
$\displaystyle x+y=1$
$\displaystyle \therefore \text{The equation of }AB\text{ is }x+y=1.$
$\displaystyle \text{On the }y\text{-axis, }x=0.$
$\displaystyle 0+y=1$
$\displaystyle y=1$
$\displaystyle \therefore \text{The line intersects the }y\text{-axis at }(0,1).$
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$\displaystyle \textbf{Question 8: }\text{The figure shows two straight lines }AB\text{ and }CD \\ \text{ intersecting at }P(3,4).\text{ Find the equations of }AB\text{ and }CD.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The point of intersection is }P(3,4).$
$\displaystyle \text{Slope of }AB=m_1=\tan45^\circ=1$
$\displaystyle \text{Slope of }CD=m_2=\tan60^\circ=\sqrt{3}$
$\displaystyle \text{Equation of }AB:$
$\displaystyle y-4=1(x-3)$
$\displaystyle y=x+1$
$\displaystyle \text{Equation of }CD:$
$\displaystyle y-4=\sqrt{3}(x-3)$
$\displaystyle y=\sqrt{3}x+4-3\sqrt{3}$
$\displaystyle \therefore \text{The required equations are }y=x+1\text{ and }y=\sqrt{3}x+4-3\sqrt{3}.$
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$\displaystyle \textbf{Question 9: }\text{In }A=(3,5),\;B=(7,8)\text{ and }C=(1,-10),\text{ find the equation} \\ \text{of the median through }A.\hfill\text{[ICSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }D\text{ be the midpoint of }BC.$
$\displaystyle D=\left(\frac{7+1}{2},\frac{8+(-10)}{2}\right)=(4,-1)$
$\displaystyle \text{Slope of }AD=\frac{-1-5}{4-3}=\frac{-6}{1}=-6$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-5=-6(x-3)$
$\displaystyle y-5=-6x+18$
$\displaystyle 6x+y=23$
$\displaystyle \therefore \text{The required equation of the median is }6x+y=23.$
$\\$

$\displaystyle \textbf{Question 10: }\text{The figure shows a parallelogram }ABCD\text{ with } \\ AB\parallel x\text{-axis and }C=(7,5).\text{ Find the equations of }BC\text{ and }CD.$
$\displaystyle \text{Answer:}$
$\displaystyle AB\parallel x\text{-axis}$
$\displaystyle \text{Slope of }BC=\tan60^\circ=\sqrt{3}$
$\displaystyle \text{Equation of }BC:$
$\displaystyle y-5=\sqrt{3}(x-7)$
$\displaystyle y=\sqrt{3}x+5-7\sqrt{3}$
$\displaystyle \text{Slope of }CD=\tan0^\circ=0$
$\displaystyle \text{Equation of }CD:$
$\displaystyle y-5=0(x-7)$
$\displaystyle y=5$
$\displaystyle \therefore \text{The required equations are }y=\sqrt{3}x+5-7\sqrt{3}\text{ and }y=5.$
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$\displaystyle \textbf{Question 11: }\text{Find the equation of the straight line passing through the origin and} \\ \text{the point of intersection of }x+2y=7\text{ and }x-y=4.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Solving }x+2y=7\text{ and }x-y=4,\text{ we get}$
$\displaystyle x=5,\quad y=1$
$\displaystyle \therefore \text{The point of intersection is }(5,1).$
$\displaystyle \text{The line passes through }(0,0)\text{ and }(5,1).$
$\displaystyle \text{Slope }m=\frac{1-0}{5-0}=\frac{1}{5}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-0=\frac{1}{5}(x-0)$
$\displaystyle 5y=x$
$\displaystyle x-5y=0$
$\displaystyle \therefore \text{The required equation is }x-5y=0.$
$\\$

$\displaystyle \textbf{Question 12: }\text{In }\triangle ABC,\text{ the vertices }A,\;B\text{ and }C\text{ are } \\ (4,7),\;(-2,3)\text{ and }(0,1)\text{ respectively. Find the equation of the median through }A. \\ \text{Also, find the equation of the line through }B\text{ and parallel to }AC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }D\text{ be the midpoint of }BC.$
$\displaystyle D=\left(\frac{-2+0}{2},\frac{3+1}{2}\right)=(-1,2)$
$\displaystyle \text{Slope of }AD=\frac{2-7}{-1-4}=\frac{-5}{-5}=1$
$\displaystyle \text{Equation of median }AD:$
$\displaystyle y-7=1(x-4)$
$\displaystyle y-x=3$
$\displaystyle \therefore \text{The equation of the median through }A\text{ is }y-x=3.$
$\displaystyle \text{Slope of }AC=\frac{1-7}{0-4}=\frac{-6}{-4}=\frac{3}{2}$
$\displaystyle \text{Since the required line is parallel to }AC,\text{ its slope is }\frac{3}{2}.$
$\displaystyle \text{The required line passes through }B(-2,3).$
$\displaystyle y-3=\frac{3}{2}(x+2)$
$\displaystyle 2y-6=3x+6$
$\displaystyle 2y-3x=12$
$\displaystyle \therefore \text{The required equation is }2y-3x=12.$
$\\$

$\displaystyle \textbf{Question 13: }\text{The points }A,\;B\text{ and }C\text{ have coordinates }(0,3), \\ \;(4,4)\text{ and }(8,0)\text{ respectively. Find the equation of the line through }A\text{ and} \\ \text{perpendicular to }BC.$
$\displaystyle \text{Answer:}$
$\displaystyle A=(0,3),\;B=(4,4)\text{ and }C=(8,0)$
$\displaystyle \text{Slope of }BC=\frac{0-4}{8-4}=\frac{-4}{4}=-1$
$\displaystyle \text{Slope of the line perpendicular to }BC=-\frac{1}{-1}=1$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-3=1(x-0)$
$\displaystyle y=x+3$
$\displaystyle \therefore \text{The required equation is }y=x+3.$
$\\$

$\displaystyle \textbf{Question 14: }\text{Find the equation of the perpendicular dropped from the point } \\ (-1,2)\text{ onto the line joining }(1,4)\text{ and }(2,3).$
$\displaystyle \text{Answer:}$
$\displaystyle A=(1,4),\;B=(2,3)\text{ and }C=(-1,2)$
$\displaystyle \text{Slope of }AB=\frac{3-4}{2-1}=\frac{-1}{1}=-1$
$\displaystyle \text{Slope of the line perpendicular to }AB=-\frac{1}{-1}=1$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-2=1(x+1)$
$\displaystyle y=x+3$
$\displaystyle \therefore \text{The required equation is }y=x+3.$
$\\$

$\displaystyle \textbf{Question 15: }\text{Find the equation of the line whose:}$
$\displaystyle \text{i) x-intercept }=5\text{ and y-intercept }=3\qquad \text{ii) x-intercept }=-4\text{ and} \\ \text{y-intercept }=6\qquad \text{iii) x-intercept }=-8\text{ and y-intercept }=-4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) The line passes through }(5,0)\text{ and }(0,3).$
$\displaystyle \text{Slope }m=\frac{3-0}{0-5}=-\frac{3}{5}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-3=-\frac{3}{5}(x-0)$
$\displaystyle 5y-15=-3x$
$\displaystyle 3x+5y=15$
$\displaystyle \therefore \text{The required equation is }3x+5y=15.$
$\displaystyle \text{ii) The line passes through }(-4,0)\text{ and }(0,6).$
$\displaystyle \text{Slope }m=\frac{6-0}{0-(-4)}=\frac{6}{4}=\frac{3}{2}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-6=\frac{3}{2}(x-0)$
$\displaystyle 2y-12=3x$
$\displaystyle 2y=3x+12$
$\displaystyle \therefore \text{The required equation is }2y=3x+12.$
$\displaystyle \text{iii) The line passes through }(-8,0)\text{ and }(0,-4).$
$\displaystyle \text{Slope }m=\frac{-4-0}{0-(-8)}=\frac{-4}{8}=-\frac{1}{2}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y+4=-\frac{1}{2}(x-0)$
$\displaystyle 2y+8=-x$
$\displaystyle x+2y+8=0$
$\displaystyle \therefore \text{The required equation is }x+2y+8=0.$
$\\$

$\displaystyle \textbf{Question 16: }\text{Find the equation of the line whose slope is }6\text{ and} \\ \text{x-intercept is }6.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given slope }m=6\text{ and x-intercept }(6,0).$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-0=6(x-6)$
$\displaystyle y=6x-36$
$\displaystyle 6x-y=36$
$\displaystyle \therefore \text{The required equation is }y=6x-36.$
$\\$

$\displaystyle \textbf{Question 17: }\text{Find the equation of the line with x-intercept }5\text{ and passing} \\ \text{through }(-3,2).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The line passes through }(5,0)\text{ and }(-3,2).$
$\displaystyle \text{Slope }m=\frac{2-0}{-3-5}=\frac{2}{-8}=-\frac{1}{4}$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-2=-\frac{1}{4}(x+3)$
$\displaystyle 4y-8=-x-3$
$\displaystyle x+4y=5$
$\displaystyle \therefore \text{The required equation is }x+4y=5.$
$\\$

$\displaystyle \textbf{Question 18: }\text{Find the equation of the line through }(1,3)\text{ and making an} \\ \text{intercept of }5\text{ on the }y\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The line passes through }(1,3)\text{ and }(0,5).$
$\displaystyle \text{Slope }m=\frac{5-3}{0-1}=\frac{2}{-1}=-2$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-5=-2(x-0)$
$\displaystyle y+2x=5$
$\displaystyle \therefore \text{The required equation is }y+2x=5.$
$\\$

$\displaystyle \textbf{Question 19: }\text{Find the equations of the lines passing through }(-2,0)\text{ and equally} \\ \text{inclined to the coordinate axes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{A line equally inclined to the coordinate axes has slope }1\text{ or }-1.$
$\displaystyle \text{For }m=1:$
$\displaystyle y-0=1(x+2)$
$\displaystyle y=x+2$
$\displaystyle \text{For }m=-1:$
$\displaystyle y-0=-1(x+2)$
$\displaystyle x+y+2=0$
$\displaystyle \therefore \text{The required equations are }y=x+2\text{ and }x+y+2=0.$
$\\$

$\displaystyle \textbf{Question 20: }\text{The line through }P(5,3)\text{ intersects the }y\text{-axis at }Q.\text{ Find:}$
$\displaystyle \text{i) The slope of the line}\qquad \text{ii) The equation of the line}\qquad \\ \text{iii) The coordinates of }Q.\hfill\text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given point }P=(5,3).$
$\displaystyle \text{i) Slope }m=\tan45^\circ=1$
$\displaystyle \text{ii) Using }y-y_1=m(x-x_1),$
$\displaystyle y-3=1(x-5)$
$\displaystyle y=x-2$
$\displaystyle \therefore \text{The required equation is }y=x-2.$
$\displaystyle \text{iii) On the }y\text{-axis, }x=0.$
$\displaystyle y=0-2=-2$
$\displaystyle \therefore Q=(0,-2).$
$\\$

$\displaystyle \textbf{Question 21: }\text{Write down the equation of the line whose gradient is }-\frac{2}{5} \\ \text{ and which passes through }P,\text{ where }P\text{ divides }A(4,-8)\text{ and }B(12,0)\text{ in the ratio }3:1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given gradient }m=-\frac{2}{5}.$
$\displaystyle \text{Let }P=(x,y)\text{ divide }A(4,-8)\text{ and }B(12,0)\text{ in the ratio }3:1.$
$\displaystyle x=\frac{3(12)+1(4)}{3+1}=\frac{40}{4}=10$
$\displaystyle y=\frac{3(0)+1(-8)}{3+1}=\frac{-8}{4}=-2$
$\displaystyle \therefore P=(10,-2)$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-(-2)=-\frac{2}{5}(x-10)$
$\displaystyle y+2=-\frac{2}{5}(x-10)$
$\displaystyle 5y+10=-2x+20$
$\displaystyle 2x+5y=10$
$\displaystyle \therefore \text{The required equation is }2x+5y=10.$
$\\$

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

$\displaystyle \textbf{Question 23: }A(7,-1),\;B(4,1)\text{ and }C(-3,4)\text{ are the vertices of }\triangle ABC. \\ \text{ Find the equation of the line through }B\text{ and the point }P\text{ on }AC,\text{ such that }AP:PC=2:3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The points are }A(7,-1)\text{ and }C(-3,4).$
$\displaystyle \text{Since }AP:PC=2:3,\text{ the coordinates of }P\text{ are}$
$\displaystyle x=\frac{2(-3)+3(7)}{2+3}=3$
$\displaystyle y=\frac{2(4)+3(-1)}{2+3}=1$
$\displaystyle \therefore P=(3,1)$
$\displaystyle \text{Slope of }BP=\frac{1-1}{3-4}=0$
$\displaystyle \text{Using }y-y_1=m(x-x_1),$
$\displaystyle y-1=0(x-3)$
$\displaystyle y=1$
$\displaystyle \therefore \text{The required equation is }y=1.$
$\\$