Question

A(1, 1) and B(2, -3) are two points and D is a point on AB produced such that AD = 3 AB Then the co-ordinates of D is

• A. (4, 11)
• B. (4, -11)
• C. (-2, 5)
• D. (-4, -11)

Answer 1

The correct option is B (4, -11)

Since, $AD=3AB=>\frac{AD}{BD}=\frac{3}{2}$

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
$C=(\frac{3\left(2\right)-2\left(1\right)}{3-2},\frac{3(-3)-2\left(1\right)}{3-2})=(4,-11)$