Question

Use analytical geometry to prove that the mid-point of the hypotenuse of a right-angled triangle is equidistant from its vertices.

Answer 1

Let AOB be a right-angled triangle with base OA taken along x-axis and the perpendicular OB taken along y-axis. Let OA=a and OB=b.Let D be the mid-point of the hypotenuse AB. Then, the coordinates of A,B and D are respectively (a, O), (O, b) and (a/2,b/2).

Now,

$DO=\sqrt{{(\frac{a}{2}-0)}^2+{(\frac{b}{2}-0)}^2}=\frac{1}{2}\sqrt{a^2+b^2},$
$DA=\sqrt{{(\frac{a}{2}-a)}^2+{(\frac{b}{2}-0)}^2}=\frac{1}{2}\sqrt{a^2+b^2}$

$and,DB=\sqrt{{(\frac{a}{2}-0)}^2+{(\frac{b}{2}-b)}^2}=\frac{1}{2}\sqrt{a^2+b^2}$

Hence, $DA=DB+DC$ i.e., D is equidistant from the vertices of triangle ABC.