Question

Show that the points A(2, -2), B(14, 10), C(ll, 13) and D(-1, 1) are the vertices of a rectangle.​

Answer 1

Solution :-

We have to show that :-

The points A(2 , -2), B(14 , 10), C(11 , 13) and D(-1 , 1 ) are the vertices of a rectangle.

$\boxed{\bf Distance \ formula = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }$

$\bullet\sf \ AB = \sqrt{(14-2)^2+(10-(-2))^2}$

        $= \sf \sqrt{12^2+12^2} \\\\=\sqrt{144+144} \\\\= \sqrt{288} \ units$

$\bullet \ \sf BC = \sqrt{(11-14)^2 + (13-10)^2}$

        $= \sf \sqrt{(-3)^2+3^2} \\\\= \sqrt{9+9} \\\\= \sqrt{18} \ units$

$\bullet \sf \ CD = \sqrt{(-1-11)^2+(1-13)^2}$

        $= \sf \sqrt{(-12)^2+(-12)^2} \\\\= \sqrt{144+144} \\\\= \sqrt{288} \ units$

$\bullet \ \sf AD = \sqrt{(-1-2)^2+(1-(-2))^2}$

        $= \sf \sqrt{(-3)^2 +3^2} \\\\= \sqrt{9+9} \\\\= \sqrt{18} \ units$

AB = CD and BC = AD

That is,

Opposite sides are equal.

$\bullet \sf \ AC = \sqrt{(11-2)^2+(13-(-2))^2}$

        $= \sf \sqrt{9^2 +15^2} \\\\= \sqrt{81+225} \\\\= \sqrt{306} \ units$

$\bullet \sf \ BD = \sqrt{(-1-14)^2+(1-10)^2}$

        $= \sf \sqrt{(-15)^2+(-9)^2} \\\\= \sqrt{225+81} \\\\= \sqrt{306} \ units$

AC = BD

That is,

Diagonals are equal.

Since, opposite sides and diagonals are equal. The given points are the vertices of a rectangle.