Question

The line $2x+3y-5=0$ cuts the line joining $A(8,-9)$ and $B(2,1)$ at point $P$. The coordinates of $P$ are:

• A. $(8,1)$
• B. $(\frac{8}{3},\frac{-1}{9})$
• C. $(\frac{1}{3},\frac{7}{5})$
• D. $(\frac{-6}{11},\frac{2}{5})$

Answer 1

The correct option is B $(\frac{8}{3},\frac{-1}{9})$

Let $P(x,y)$ be the point on the line $2x+3y-5=0$ which divides the line joining $A(8,-9)$ and $B(2,1)$ in the ratio $k:1$.
By section formula,
$P(x,y)=(\frac{[k{x}_2+x_1]}{[k+1]},\frac{[k{y}_2+y_1]}{[k+1]})$
Where $(x_1,y_1)=(8,-9)$ and $(x_2,y_2)=(2,1)$
$\Rightarrow P(x,y)=[\frac{(2k+8)}{(k+1)},\frac{(k-9)}{(k+1)}]$

Since $P(x,y)$ lies on the line $2x+3y-5=0$,we have:
$\frac{2(2k+8)}{(k+1)}+\frac{3(k-9)}{(k+1)}-5=0$
$\Rightarrow \frac{(4k+16+3k-27)}{(k+1)}=5$
$\Rightarrow 7k-11=5k+5$
$\Rightarrow k=8$

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