Question

Let $A(-6,-5)$ and $B(-6,4)$ be two points such that a point $P$ on the line $AB$ satisfies $AP=\frac{2}{9}AB$. Find the point $P$.

• A. $(-6,-3)$
• B. $(6,-3)$
• C. $(-6,3)$
• D. None of these

Answer 1

The correct option is A $(-6,-3)$

Given

$A(-6,-5)$ and $B(-6,4)$

$AP=\frac{2}{9}AB=>2AB=9AP$......$\left(1\right)$

Also$P$ lies on $AB$ so, $AP+PB=AB$.......$\left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right)$:

$\frac{AP}{PB}=\frac{2}{7}$. It means $P$ divides $AB$ in ratio $2:7$.

Take $P(x,y)$ and using section formula:

$x=\frac{m{x}_2+n{x}_1}{m+n}$ and $y=\frac{m{y}_2+n{y}_1}{m+n}$

$x=\frac{-42-12}{9}=-6$ and $y=\frac{-35+8}{9}=-3$

So, $P(x,y)=(-6,-3)$ is required point.