Question

If $A(3,3)$ and $B(6,2)$ are in $I{R}^2$ ,find $P$ on $\overleftrightarrow{AB}$ such that $AP=4AB$

Answer 1

Let $P(h,k)$ be the point such that,

$AP=4AB$

So, $AP=AB+BP$

$\Rightarrow 4AB-AB=BP$

$\Rightarrow BP=3AB$

So, $AP:BP=4:3$

Now the point $P$ divides the line externally so we have,

$AB:BP=1:3$

So, $6=\frac{3\times 3+h\times 1}{3+1}$

$\Rightarrow 6\times 4-3\times 3=h$

$\Rightarrow h=24-9=15$

and $2=\frac{3\times 3+k+1}{3+1}$

$\Rightarrow 8-3\times 3=k$

$\Rightarrow k=8-9=-1$

So, we have $P(15,-1)$