Question

$A(1,1)$ and $B(2,-3)$ are two points and $D$ is a point on AB produced such that $AD=3AB$. If the coordinates of $D$ are given by $(p,q)$ then $|p+q|=?$

Answer 1

Given: $AD=3AB$

$\Rightarrow AB+BD=3AB$

Therefore, $BD=2AB$

Thus, D divides AB externally in the ratio $AD:BD=3:2$
Using the section formula, if a point $(x,y)$ divides the line joining the points
$(x_1,y_1)$ and $(x_2,y_2)$externally in the ratio $m:n$, then $(x,y)=(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n})$
Substituting $(x_1,y_1)=(1,1)$ and $(x_2,y_2)=(2,-3)$ and $m=3,n=2$ in the section formula, we get
$D=(\frac{3\left(2\right)-2\left(1\right)}{3-2},\frac{3(-3)-2\left(1\right)}{3-2})=(4,-11)$

So, $|p+q|=|4-11|=|-7|=7$