Class 10: Distance and Section Formula

$\displaystyle \textbf{Note: }\text{If a point divides the line segment joining }(x_1,y_1)\text{ and }(x_2,y_2)\text{ in the ratio }$
$\displaystyle m_1:m_2, \ \text{then the coordinates of the point are}$
$\displaystyle x=\frac{m_1x_2+m_2x_1}{m_1+m_2},\qquad y=\frac{m_1y_2+m_2y_1}{m_1+m_2}$
$\\$

$\displaystyle \textbf{Question 1: }\text{Calculate the coordinates of the point }P\text{ which divides the line segment joining:}$
$\displaystyle \text{i) }A(1,3)\text{ and }B(5,9)\text{ in the ratio }1:2$
$\displaystyle \text{ii) }A(-4,6)\text{ and }B(3,-5)\text{ in the ratio }3:2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) Ratio }m_1:m_2=1:2$
$\displaystyle \text{Let the coordinates of }P\text{ be }(x,y).$
$\displaystyle x=\frac{1\times5+2\times1}{1+2}=\frac{7}{3}$
$\displaystyle y=\frac{1\times9+2\times3}{1+2}=\frac{15}{3}=5$
$\displaystyle \therefore P\left(\frac{7}{3},5\right)$
$\displaystyle \text{ii) Ratio }m_1:m_2=3:2$
$\displaystyle \text{Let the coordinates of }P\text{ be }(x,y).$
$\displaystyle x=\frac{2\times(-4)+3\times3}{3+2}=\frac{1}{5}$
$\displaystyle y=\frac{2\times6+3\times(-5)}{3+2}=-\frac{3}{5}$
$\displaystyle \therefore P\left(\frac{1}{5},-\frac{3}{5}\right)$
$\\$

$\displaystyle \textbf{Question 2: }\text{In what ratio is the line joining }(2,-3)\text{ and }(5,6)\text{ divided by the }x\text{-axis?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required ratio be }k:1\text{ and the point on the }x\text{-axis be }(x,0).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle y=\frac{ky_2+y_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{k\times6-3}{k+1}$
$\displaystyle \Rightarrow 6k-3=0$
$\displaystyle \Rightarrow k=\frac{1}{2}$
$\displaystyle \therefore m_1:m_2=1:2$
$\\$

$\displaystyle \textbf{Question 3: }\text{In what ratio is the line joining }(2,-4)\text{ and }(-3,6)\text{ divided by the }y\text{-axis?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required ratio be }k:1\text{ and the point on the }y\text{-axis be }(0,y).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle x=\frac{kx_2+x_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{k\times(-3)+2}{k+1}$
$\displaystyle \Rightarrow -3k+2=0$
$\displaystyle \Rightarrow 3k-2=0$
$\displaystyle \Rightarrow k=\frac{2}{3}$
$\displaystyle \therefore m_1:m_2=2:3$
$\\$

$\displaystyle \textbf{Question 4: }\text{In what ratio does the point }(1,a)\text{ divide the join of }(-1,4)\text{ and } \\ (4,-1)\text{? Also, find }a.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }(1,a)\text{ divide the join of }(-1,4)\text{ and }(4,-1)\text{ in the ratio }k:1.$
$\displaystyle \text{Using the section formula,}$
$\displaystyle x=\frac{kx_2+x_1}{k+1}$
$\displaystyle \Rightarrow 1=\frac{4k-1}{k+1}$
$\displaystyle \Rightarrow k+1=4k-1$
$\displaystyle \Rightarrow k=\frac{2}{3}$
$\displaystyle \therefore m_1:m_2=2:3$
$\displaystyle \text{Now,}$
$\displaystyle a=\frac{2\times(-1)+3\times4}{2+3}$
$\displaystyle =\frac{-2+12}{5}=2$
$\displaystyle \therefore a=2$
$\\$

$\displaystyle \textbf{Question 5: }\text{In what ratio does the point }(a,6)\text{ divide the join of }(-4,3)\text{ and } \\ (2,8)\text{? Also, find }a.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }(a,6)\text{ divide the join of }(-4,3)\text{ and }(2,8)\text{ in the ratio }k:1.$
$\displaystyle \text{Using the section formula,}$
$\displaystyle y=\frac{ky_2+y_1}{k+1}$
$\displaystyle \Rightarrow 6=\frac{8k+3}{k+1}$
$\displaystyle \Rightarrow 6k+6=8k+3$
$\displaystyle \Rightarrow k=\frac{3}{2}$
$\displaystyle \therefore m_1:m_2=3:2$
$\displaystyle \text{Now,}$
$\displaystyle a=\frac{3\times2+2\times(-4)}{3+2}$
$\displaystyle \Rightarrow 5a=6-8=-2$
$\displaystyle \Rightarrow a=-\frac{2}{5}$
$\\$

$\displaystyle \textbf{Question 6: }\text{In what ratio is the join of }(4,3)\text{ and }(2,-6)\text{ divided by} \\ \text{the } x\text{-axis? Also, find the coordinates of the point of intersection.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required ratio be }k:1\text{ and the point on the }x\text{-axis be }(x,0).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle y=\frac{ky_2+y_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{k\times(-6)+3}{k+1}$
$\displaystyle \Rightarrow -6k+3=0$
$\displaystyle \Rightarrow 6k-3=0$
$\displaystyle \Rightarrow k=\frac{1}{2}$
$\displaystyle \therefore m_1:m_2=1:2$
$\displaystyle \text{Now,}$
$\displaystyle x=\frac{1\times2+2\times4}{1+2}=\frac{10}{3}$
$\displaystyle y=\frac{1\times(-6)+2\times3}{1+2}=0$
$\displaystyle \therefore \text{Point of intersection }=\left(\frac{10}{3},0\right)$
$\\$

$\displaystyle \textbf{Question 7: }\text{Find the ratio in which the join of }(-4,7)\text{ and }(3,0)\text{ is divided} \\ \text{by the }y\text{-axis. Also, find the coordinates of the point of intersection.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required ratio be }k:1\text{ and the point on the }y\text{-axis be }(0,y).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle x=\frac{kx_2+x_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{3k-4}{k+1}$
$\displaystyle \Rightarrow 3k-4=0$
$\displaystyle \Rightarrow k=\frac{4}{3}$
$\displaystyle \therefore m_1:m_2=4:3$
$\displaystyle \text{Now,}$
$\displaystyle y=\frac{4\times0+3\times7}{4+3}=3$
$\displaystyle \therefore \text{Point of intersection }=(0,3)$
$\\$

$\displaystyle \textbf{Question 8: }\text{Points }A,B,C\text{ and }D\text{ divide the line segment joining }(5,-10) \\ \text{ and the origin into five equal parts. Find the coordinates of }B\text{ and }D.$
$\displaystyle \text{Answer:}$
$\displaystyle PB:BQ=2:3$
$\displaystyle PD:DQ=4:1$
$\displaystyle \text{For }B:$
$\displaystyle x=\frac{2\times5+3\times0}{2+3}=2$
$\displaystyle y=\frac{2\times(-10)+3\times0}{2+3}=-4$
$\displaystyle \therefore B=(2,-4)$
$\displaystyle \text{For }D:$
$\displaystyle x=\frac{4\times5+1\times0}{4+1}=4$
$\displaystyle y=\frac{4\times(-10)+1\times0}{4+1}=-8$
$\displaystyle \therefore D=(4,-8)$
$\\$

$\displaystyle \textbf{Question 9: }\text{The line joining }A(-3,-10)\text{ and }B(-2,6)\text{ is divided by } \\ P\text{ such that }\frac{PB}{AB}=\frac{1}{5}.\text{ Find the coordinates of }P.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{PB}{AB}=\frac{1}{5}$
$\displaystyle \Rightarrow AP:PB=4:1$
$\displaystyle x=\frac{4\times(-2)+1\times(-3)}{4+1}=-\frac{11}{5}$
$\displaystyle y=\frac{4\times6+1\times(-10)}{4+1}=\frac{14}{5}$
$\displaystyle \therefore P=\left(-\frac{11}{5},\frac{14}{5}\right)$
$\\$

$\displaystyle \textbf{Question 10: }P\text{ is a point on the line joining }A(4,3)\text{ and }B(-2,6)\text{ such that } \\ 5AP=2BP.\text{ Find the coordinates of }P.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }5AP=2BP$
$\displaystyle \Rightarrow \frac{AP}{BP}=\frac{2}{5}$
$\displaystyle \Rightarrow AP:PB=2:5$
$\displaystyle x=\frac{2\times(-2)+5\times4}{2+5}=\frac{16}{7}$
$\displaystyle y=\frac{2\times6+5\times3}{2+5}=\frac{27}{7}$
$\displaystyle \therefore P=\left(\frac{16}{7},\frac{27}{7}\right)$
$\\$

$\displaystyle \textbf{Question 11: }\text{Calculate the ratio in which the line joining }(-3,-1)\text{ and }(5,7) \\ \text{ is divided by the line }x=2.\text{ Also, find the coordinates of the point of intersection.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required ratio be }k:1\text{ and the point on the line }x=2\text{ be }(2,y).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle x=\frac{kx_2+x_1}{k+1}$
$\displaystyle \Rightarrow 2=\frac{5k-3}{k+1}$
$\displaystyle \Rightarrow 2k+2=5k-3$
$\displaystyle \Rightarrow k=\frac{5}{3}$
$\displaystyle \therefore m_1:m_2=5:3$
$\displaystyle \text{Now,}$
$\displaystyle y=\frac{5\times7+3\times(-1)}{5+3}=4$
$\displaystyle \therefore \text{Point of intersection }=(2,4)$
$\\$

$\displaystyle \textbf{Question 12: }\text{Calculate the ratio in which the line joining }A(6,5)\text{ and }B(4,-3) \\ \text{ is divided by the line }y=2.\hfill\text{[ICSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required ratio be }k:1\text{ and the point on the line }y=2\text{ be }(x,2).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle y=\frac{ky_2+y_1}{k+1}$
$\displaystyle \Rightarrow 2=\frac{k\times(-3)+5}{k+1}$
$\displaystyle \Rightarrow 2k+2=-3k+5$
$\displaystyle \Rightarrow k=\frac{3}{5}$
$\displaystyle \therefore m_1:m_2=3:5$
$\displaystyle \text{Now,}$
$\displaystyle x=\frac{3\times4+5\times6}{3+5}=\frac{42}{8}=\frac{21}{4}$
$\displaystyle \therefore \text{Point of intersection }=\left(\frac{21}{4},2\right)$
$\\$

$\displaystyle \textbf{Question 13: }\text{The point }P(5,-4)\text{ divides the line segment }AB\text{, as shown} \\ \text{in the figure, in the ratio }2:5.\text{ Find the coordinates of }A\text{ and }B.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Ratio }m_1:m_2=2:5$
$\displaystyle 5=\frac{2\times0+5x}{2+5}$
$\displaystyle \Rightarrow x=7$
$\displaystyle -4=\frac{2y+5\times0}{2+5}$
$\displaystyle \Rightarrow y=-14$
$\displaystyle \therefore A=(7,0)\text{ and }B=(0,-14)$
$\\$

$\displaystyle \textbf{Question 14: }\text{Find the coordinates of the points of trisection of the line joining } \\ (-3,0)\text{ and }(6,6).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the ratio }m_1:m_2=1:2$
$\displaystyle x=\frac{1\times6+2\times(-3)}{1+2}=0$
$\displaystyle y=\frac{1\times6+2\times0}{1+2}=2$
$\displaystyle \therefore \text{First trisection point }=(0,2)$
$\displaystyle \text{For the ratio }m_1:m_2=2:1$
$\displaystyle x=\frac{2\times6+1\times(-3)}{2+1}=3$
$\displaystyle y=\frac{2\times6+1\times0}{2+1}=4$
$\displaystyle \therefore \text{Second trisection point }=(3,4)$
$\\$

$\displaystyle \textbf{Question 15: }\text{Show that the line segment joining }(-5,8)\text{ and }(10,-4)\text{ is} \\ \text{trisected by the coordinate axes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the points on the }y\text{-axis and }x\text{-axis be }B(0,y)\text{ and }A(x,0)\text{ respectively.}$
$\displaystyle \text{For }B,\text{ the ratio }m_1:m_2=1:2$
$\displaystyle x=\frac{1\times10+2\times(-5)}{1+2}=0$
$\displaystyle y=\frac{1\times(-4)+2\times8}{1+2}=4$
$\displaystyle \therefore B=(0,4)$
$\displaystyle \text{For }A,\text{ the ratio }m_1:m_2=2:1$
$\displaystyle x=\frac{2\times10+1\times(-5)}{2+1}=5$
$\displaystyle y=\frac{2\times(-4)+1\times8}{2+1}=0$
$\displaystyle \therefore A=(5,0)$
$\displaystyle \therefore \text{The coordinate axes trisect the given line segment.}$
$\\$

$\displaystyle \textbf{Question 16: }\text{Show that }A(3,-2)\text{ is a point of trisection of the line segment} \\ \text{joining }(2,1)\text{ and }(5,-8).\text{ Also, find the coordinates of the other point of trisection.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(3,-2)\text{ divide the join of }(2,1)\text{ and }(5,-8)\text{ in the ratio }k:1.$
$\displaystyle \text{Using the section formula,}$
$\displaystyle x=\frac{kx_2+x_1}{k+1}$
$\displaystyle \Rightarrow 3=\frac{5k+2}{k+1}$
$\displaystyle \Rightarrow 3k+3=5k+2$
$\displaystyle \Rightarrow k=\frac{1}{2}$
$\displaystyle \therefore m_1:m_2=1:2$
$\displaystyle \text{Hence, }A(3,-2)\text{ is a point of trisection.}$
$\displaystyle \text{For the other point, }m_1:m_2=2:1$
$\displaystyle x=\frac{2\times5+1\times2}{2+1}=4$
$\displaystyle y=\frac{2\times(-8)+1\times1}{2+1}=-5$
$\displaystyle \therefore \text{The other point of trisection is }(4,-5)$
$\\$

$\displaystyle \textbf{Question 17: }\text{If }A=(-4,3)\text{ and }B=(8,-6):\hfill\text{[ICSE 2008]}$
$\displaystyle \text{(i) Find the length of }AB\qquad\text{(ii) In what ratio is }AB\text{ divided by the } \\ x\text{-axis?}$
$\displaystyle \text{Answer:}$
$\displaystyle AB=\sqrt{(8-(-4))^2+(-6-3)^2}$
$\displaystyle =\sqrt{12^2+(-9)^2}=\sqrt{144+81}=\sqrt{225}=15$
$\displaystyle \text{Let the required ratio be }k:1\text{ and the point on the }x\text{-axis be }(x,0).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle y=\frac{ky_2+y_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{k\times(-6)+3}{k+1}$
$\displaystyle \Rightarrow -6k+3=0$
$\displaystyle \Rightarrow 6k-3=0$
$\displaystyle \Rightarrow k=\frac{1}{2}$
$\displaystyle \therefore m_1:m_2=1:2$
$\\$

$\displaystyle \textbf{Question 18: }\text{The line segment joining }M(5,7)\text{ and }N(-3,2)\text{ is intersected} \\ \text{by the }y\text{-axis at }L.\text{ Write down the abscissa of }L.\text{ Hence, find the ratio in} \\ \text{which }L\text{ divides }MN.\text{ Also, find the coordinates of }L.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The abscissa of }L=0$
$\displaystyle \text{Let the required ratio be }k:1\text{ and }L=(0,y).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle x=\frac{kx_2+x_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{k\times(-3)+5}{k+1}$
$\displaystyle \Rightarrow 3k-5=0$
$\displaystyle \Rightarrow k=\frac{5}{3}$
$\displaystyle \therefore ML:LN=5:3$
$\displaystyle y=\frac{5\times2+3\times7}{5+3}=\frac{31}{8}$
$\displaystyle \therefore L=\left(0,\frac{31}{8}\right)$
$\\$

$\displaystyle \textbf{Question 19: }A(2,5),B(-1,2)\text{ and }C(5,8)\text{ are the vertices of }\triangle ABC.$
$\displaystyle \text{Points }P\text{ and }Q\text{ lie on }AB\text{ and }AC\text{ respectively such that } \\ AP:PB=AQ:QC=1:2.$
$\displaystyle \text{(i) Calculate the coordinates of }P\text{ and }Q.$
$\displaystyle \text{(ii) Show that }PQ=\frac{1}{3}BC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }P,\;m_1:m_2=1:2$
$\displaystyle x=\frac{1\times(-1)+2\times2}{1+2}=1$
$\displaystyle y=\frac{1\times2+2\times5}{1+2}=4$
$\displaystyle \therefore P=(1,4)$
$\displaystyle \text{For }Q,\;m_1:m_2=1:2$
$\displaystyle x=\frac{1\times5+2\times2}{1+2}=3$
$\displaystyle y=\frac{1\times8+2\times5}{1+2}=6$
$\displaystyle \therefore Q=(3,6)$
$\displaystyle PQ=\sqrt{(3-1)^2+(6-4)^2}=\sqrt{4+4}=2\sqrt{2}$
$\displaystyle BC=\sqrt{(5-(-1))^2+(8-2)^2}=\sqrt{36+36}=6\sqrt{2}$
$\displaystyle \therefore PQ=\frac{2\sqrt{2}}{6\sqrt{2}}\,BC=\frac{1}{3}BC$
$\displaystyle \therefore PQ=\frac{1}{3}BC$
$\\$

$\displaystyle \textbf{Question 20: }A(-3,4),B(3,-1)\text{ and }C(-2,4)\text{ are the vertices of }\triangle ABC. \\ \text{ Find the length of }AP,\text{ where }P\text{ lies on }BC\text{ such that }BP:PC=2:3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }P,\;m_1:m_2=2:3$
$\displaystyle x=\frac{2\times(-2)+3\times3}{2+3}=1$
$\displaystyle y=\frac{2\times4+3\times(-1)}{2+3}=1$
$\displaystyle \therefore P=(1,1)$
$\displaystyle AP=\sqrt{(1-(-3))^2+(1-4)^2}$
$\displaystyle =\sqrt{4^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5$
$\\$

$\displaystyle \textbf{Question 21: }\text{The line segment joining }A(2,3)\text{ and }B(6,-5)\text{ is intercepted} \\ \text{by the }x\text{-axis at }K.\text{ Write down the ordinate of }K.\text{ Hence, find the ratio} \\ \text{in which }K\text{ divides }AB.\text{ Also, find the coordinates of }K.\hfill\text{[ICSE 1990, 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The ordinate of }K=0$
$\displaystyle \text{Let the required ratio be }k:1\text{ and }K=(x,0).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle y=\frac{ky_2+y_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{k\times(-5)+3}{k+1}$
$\displaystyle \Rightarrow 5k-3=0$
$\displaystyle \Rightarrow k=\frac{3}{5}$
$\displaystyle \therefore AK:KB=3:5$
$\displaystyle x=\frac{3\times6+5\times2}{3+5}=\frac{28}{8}=\frac{7}{2}$
$\displaystyle \therefore K=\left(\frac{7}{2},0\right)$
$\\$

$\displaystyle \textbf{Question 22: }\text{The line segment joining }A(4,7)\text{ and }B(-6,-2)\text{ is intercepted} \\ \text{by the }y\text{-axis at }K.\text{ Write down the abscissa of }K.\text{ Hence, find the ratio in which } \\ K\text{ divides }AB.\text{ Also, find the coordinates of }K.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The abscissa of }K=0$
$\displaystyle \text{Let the required ratio be }k:1\text{ and }K=(0,y).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle x=\frac{kx_2+x_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{k\times(-6)+4}{k+1}$
$\displaystyle \Rightarrow 6k-4=0$
$\displaystyle \Rightarrow k=\frac{2}{3}$
$\displaystyle \therefore AK:KB=2:3$
$\displaystyle y=\frac{2\times(-2)+3\times7}{2+3}=\frac{17}{5}$
$\displaystyle \therefore K=\left(0,\frac{17}{5}\right)$
$\\$

$\displaystyle \textbf{Question 23: }\text{The line joining }P(-4,5)\text{ and }Q(3,2)\text{ intersects the }y\text{-axis at }R.$
$\displaystyle PM\text{ and }QN\text{ are perpendiculars from }P\text{ and }Q\text{ on the }x\text{-axis. Find:}$
$\displaystyle \text{(i) }PR:RQ\qquad\text{(ii) The coordinates of }R\qquad\text{(iii) The area of quadrilateral }PMNQ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required ratio be }k:1\text{ and }R=(0,y).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle x=\frac{kx_2+x_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{k\times3-4}{k+1}$
$\displaystyle \Rightarrow 3k-4=0$
$\displaystyle \Rightarrow k=\frac{4}{3}$
$\displaystyle \therefore PR:RQ=4:3$
$\displaystyle y=\frac{4\times2+3\times5}{4+3}=\frac{23}{7}$
$\displaystyle \therefore R=\left(0,\frac{23}{7}\right)$
$\displaystyle \text{Now, }PM=5,\;QN=2\text{ and }MN=3-(-4)=7$
$\displaystyle \text{Area of quadrilateral }PMNQ=\frac{1}{2}(PM+QN)\times MN$
$\displaystyle =\frac{1}{2}(5+2)\times7=\frac{49}{2}$
$\displaystyle \therefore \text{Area of quadrilateral }PMNQ=\frac{49}{2}\text{ sq. units}$
$\\$

$\displaystyle \textbf{Question 24: }\text{In the given figure, line }APB\text{ meets the }x\text{-axis at }A \\ \text{ and }y\text{-axis at }B.\text{ }P(-4,2)\text{ and }AP:PB=1:2.\text{ Find the coordinates} \\ \text{of }A\text{ and }B.\hfill\text{[ICSE 1999, 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }AP:PB=1:2$
$\displaystyle \text{Let }A=(x,0)\text{ and }B=(0,y).$
$\displaystyle -4=\frac{1\times0+2\times x}{1+2}$
$\displaystyle \Rightarrow x=-6$
$\displaystyle 2=\frac{1\times y+2\times0}{1+2}$
$\displaystyle \Rightarrow y=6$
$\displaystyle \therefore A=(-6,0)\text{ and }B=(0,6)$
$\\$

$\displaystyle \textbf{Question 25: }\text{Given a line segment }AB\text{ joining }A(-4,6)\text{ and }B(8,-3). \\ \text{ Find:}\hfill\text{[ICSE 2012]}$
$\displaystyle \text{(i) The ratio in which }AB\text{ is divided by the }y\text{-axis.}$
$\displaystyle \text{(ii) The coordinates of the point of intersection.}$
$\displaystyle \text{(iii) The length of }AB.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required ratio be }k:1\text{ and the point on the }y\text{-axis be }(0,y).$
$\displaystyle \text{Using the section formula,}$
$\displaystyle x=\frac{kx_2+x_1}{k+1}$
$\displaystyle \Rightarrow 0=\frac{k\times8-4}{k+1}$
$\displaystyle \Rightarrow 8k-4=0$
$\displaystyle \Rightarrow k=\frac{1}{2}$
$\displaystyle \therefore m_1:m_2=1:2$
$\displaystyle y=\frac{1\times(-3)+2\times6}{1+2}=3$
$\displaystyle \therefore \text{Point of intersection }=(0,3)$
$\displaystyle AB=\sqrt{(8-(-4))^2+(-3-6)^2}$
$\displaystyle =\sqrt{12^2+(-9)^2}=\sqrt{144+81}=\sqrt{225}=15\text{ units}$
$\\$