Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.

Distance between the points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

The length of PQ

$= \sqrt{(5 - 0)^2 + (10 - 5)^2}\\[1em] = \sqrt{5^2 + 5^2}\\[1em] = \sqrt{25 + 25}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em]$

The length of QR

$= \sqrt{(6 - 5)^2 + (3 - 10)^2}\\[1em] = \sqrt{1^2 + (-7)^2}\\[1em] = \sqrt{1 + 49}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em]$

The length of RP

$= \sqrt{(6 - 0)^2 + (3 - 5)^2}\\[1em] = \sqrt{6^2 + (-2)^2}\\[1em] = \sqrt{36 + 4}\\[1em] = \sqrt{40}\\[1em] = 2\sqrt{10}\\[1em]$

PQ = QR ⇒ the triangle is isosceles triangle

Hence, the triangle PQR is an isosceles triangle.