Question

If A and B are the points (-3, 4) and (2, 1) then the co ordinates of the point C on AB produced such that AC = 2BC are

• A. (2, 4)
• B. (3, 7)
• C. (7, -2)
• D. $(\frac{1}{2},\frac{5}{2})$

Answer 1

Answer: (C)

(7, -2)

Since $AC=2BC=>\frac{AC}{BC}=2:1$
This means, C divides AB externally in the ratio $2:1$
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 $(x_1,y_1)=(-3,4)$ and $(x_2,y_2)=(2,1)$ and $m=2,n=1$ in the section formula, we get
$C=(\frac{2\left(2\right)-1(-3)}{2-1},\frac{2\left(1\right)-1\left(4\right)}{2-1})=(7,-2)$