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

We have
$AB=\sqrt{{(6-5)}^2+{(4+2)}^2}=\sqrt{37}$$AC=\sqrt{{(6-7)}^2+{(4+2)}^2}=\sqrt{37}$
$\therefore$ AB=ACSo, $△$ ABC is isosceles.Let D be the mid-point of BC. Then, coordinates of D are $(\frac{5+7}{2},\frac{-2-2}{2})i.e(6,-2)$
$\therefore$ $AD=\sqrt{{(6-6)}^2+{(4+2)}^2}=\sqrt{36}=6$