Find the co-ordinates of a point C on AB produced so that 3AB = AC, where A = (3 ,2) and B = (-2 , 4)

(- 13, 2)

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(8, 12)

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(12, 8)

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(- 8, 12)

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Solution

The correct option is A $(- 13, 2)$
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 $(x, y) = (\frac{m x_{2} - n x_{1}}{m - n}, \frac{m y_{2} - n y_{1}}{m - n})$
Since, $3 A B = A C => \frac{A B}{A C} = \frac{1}{3}$
$=> \frac{B C}{A C} = \frac{3 - 1}{3} = \frac{2}{3}$
Substituting $(x_{1}, y_{1}) = (3, 2)$ and $(x_{2}, y_{2}) = (- 2, 4)$ and $m = 2, n = 3$ in the section formula, we get
$C = (\frac{2 (- 2) - 3 (3)}{2 - 3}, \frac{2 (4) - 3 (2)}{2 - 3}) = (- 13, - 2)$

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