Question

Find the ratio in which the line $2x+3y-5=0$ divides the line segment joining the points $(8,-9)$ and $(2,1)$ . Also find the coordinates of the point of division .

Answer 1

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$, then

$(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Let the line given be equation $I$ divides AB at $P(x,y)$ in the ratio $k:1$ ,

Then , using the section formula, the coordinates of $P$ are given by

$\Rightarrow x=\frac{k\left(2\right)+1\left(8\right)}{(k+1)}$ and $y=\frac{k\left(1\right)+1(-9)}{k+1}$

$\Rightarrow x=\frac{2k+8}{k+1}$ and $y=\frac{k-9}{k+1}$

$\Rightarrow P(x,y)=(\frac{2k+8}{k+1},\frac{k-9}{k+1})$ lies on line $I$ so $P$ must satisfy equation $\left(I\right)$

So substitute $x=\frac{2k+8}{k+1}$ and $y=\frac{k-9}{k+1}$ in equation $I$

$\Rightarrow 2\left(\frac{2k+8}{k+1}\right)+3\left(\frac{k-9}{k+1}\right)-5=0$

On multiplying by $(k+1)$ in above equation both sides , we get

$2(2k+8)+3(k-9)-5(k+1)=0$

$\Rightarrow 4k+16+3k-27-5k-5=0$

$\Rightarrow 2k-16=0$

$\Rightarrow k=\frac{16}{2}=8$

$\therefore$ Point of intersection is given by $P(\frac{2k+8}{k+1},\frac{k-9}{k+1})$

$=P(\frac{2\times 8+8}{8+1},\frac{8-9}{8+1})$

$=P(\frac{16+8}{9},\frac{-1}{9})$

$=P(\frac{24}{9},\frac{-1}{9})$

$=P(\frac{8}{3},\frac{-1}{9})$

Hence , line of equation $\left(I\right)$ divides AB in the ratio $8:1$ at $P(\frac{8}{3},\frac{-1}{9})$. $x_3(y_1-y_2)]$

$\Rightarrow$ Area of $\Delta =\frac{1}{2}[-8(6-9)-6(9-4)-3(4-6\left)\right]$
$=\frac{1}{2}[-8(-3)-6(5)-3(-2\left)\right]$
$=\frac{1}{2}[24-30+6]=0$
Hence , the area of given triangle is zero.