Find the lengths of the median of a triangle whose vertices are $A (- 1, 1), B (5, - 3), C (3, 5)$

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Solution

$A (- 1, 1), B (5, - 3), C (3, 5)$

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Draw $A D ⊥ B C$

$B D + D C = B C$

$A C^{2} = A D^{2} + C D^{2}$

$A C^{2} = A D^{2} + {(\frac{B C}{2})}^{2}$ ...(1)

$A C^{2} = (3 + 1)^{2} + (5 - 1)^{2} = 32$

$B C^{2} = (3 - 5)^{2} + (5 + 3)^{2} = 68$

From (1)

$32 = A D^{2} + \frac{68}{4}$

$A D^{2} = 32 - 17 = 15$

$∴ A D = \sqrt{15}$

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