If $A = (3, - 4)$ and the midpoints of $A B, A C$ are $(2, - 1), (4, - 5)$ respectively then the mid.point of $B C$ is

(1, 2)

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(3, - 2)

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(- 1, 2)

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(0, - 3)

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Solution

The correct option is C $(3, - 2)$
$A (3, - 4)$
Mid point of $A B = (2, - 1)$
$B (x_{2}, y_{2})$
$(\frac{3 + x_{2}}{2}, \frac{y_{2} - 4}{2}) = (2, - 1)$
$x_{2} + 3 = 4$ $y_{2} - 4 = - 2$
$x_{2} = 1$ $y_{2} = 2$
$∴ B (1, 2)$
Mid point of AC $= (4, - 5)$
$C (x_{3}, y_{3})$
$(\frac{3 + x_{3}}{2}, \frac{y_{3} - 4}{2}) = (4, - 5)$
$x_{3} = 8 - 3$ , $y_{3} = - 10 + 4$
$C (5, - 6)$
Mid point of BC $= (\frac{1 + 5}{2}, \frac{2 - 6}{2})$
$= (\frac{6}{2}, \frac{- 4}{2})$
$= (3, - 2)$ .

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