Question

Show that A(6,4), B(5,-2) and C(7,-2) are the vertices of an isosceles triangle. Also, find the length of the median through A.

Answer 1

$AB=\sqrt{(6-5{)}^2+(4+2{)}^2}=\sqrt{37}$

$AC=\sqrt{(6-7{)}^2+(4+2{)}^2}=\sqrt{37}$

$AB=AC$

So, $△ABC$ is isosceles.

Let D be the mid-point of BC. Then, coordinates of D are $(\frac{5+7}{2},\frac{-2-2}{2})i.e.,(6,-2)$.

Therefore,

$AD=\sqrt{(6-6{)}^2+(4+2{)}^2}=\sqrt{36}=6$