Question

Show that the ▢PQRS formed by P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5) is a rectangle.

Answer 1

Distance Formula = `sqrt((x_2 - x_1)^2 +  (y_2 - y_1)^2)`

PQ = \[\sqrt{\left( - 1 - 2 \right)^2 + \left( 3 - 1 \right)^2} = \sqrt{9 + 4} = \sqrt{13}\]

QR = \[\sqrt{\left( - 5 + 1 \right)^2 + \left( - 3 - 3 \right)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\]

RS = \[\sqrt{\left( - 2 + 5 \right)^2 + \left( - 5 + 3 \right)^2} = \sqrt{9 + 4} = \sqrt{13}\]

PS = \[\sqrt{\left( - 5 - 1 \right)^2 + \left( - 2 - 2 \right)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}\]

PQ = RS and QR = PS

Opposite sides are equal.

PR \[= \sqrt{\left( - 3 - 1 \right)^2 + \left( - 5 - 2 \right)^2} = \sqrt{16 + 49} = \sqrt{65}\]

QS \[= \sqrt{\left( - 2 + 1 \right)^2 + \left( - 5 - 3 \right)^2} = \sqrt{1 + 64} = \sqrt{65}\]

Diagonals are equal.

Thus, the given vertices form a rectangle.