Question

4. Find the lengths of the medians of a A ABC whose vertices are A (7, -3), B (5,3) and C (3,-1).

​

Answer 1

Answer:

AB mid point

$d = ( \frac{7 + 5}{2} . \frac{ - 3 + 3}{2} ) \\ = (6 \: . \: 0)$

So length of median CD -

$= \sqrt{ {(3 - 6)}^{2} + {( - 1 - 0)}^{2} }$

$= \sqrt{ 9 + 1 } = \sqrt{10}$

Next,

AC mid point

$e = ( \frac{7 + 3}{2}. \frac{ - 3 - 1}{2}) \\ = (5 \: \: . \: \: - 2)$

So length of median BE -

$= \sqrt{ {(5 - 5)}^{2} + {(3 + 2})^{2} }$

$= \sqrt{25} = 5$

Finally,

Mid point of BC

$f = ( \frac{5 + 3}{2}. \frac{3 - 1}{2}) \\ = (4 \: \: . \: \: 1)$

So length of median AF -

$= \sqrt{ {(7 - 4)}^{2} + {( - 3 - 1)}^{2} }$

$\sqrt{9 + 16} = \sqrt{25} = 5$