37
Prove that $A(4, 3)$, $B(6, 4)$, $C(5, 6)$, $D(3, 5)$ are the vertices of a square ABCD.
Show SolutionHide Solution↓
$AB = \sqrt{(6-4)^2 + (4 - 3)^2} = \sqrt{5}$ units
$BC = \sqrt{(5-6)^2 + (6 - 4)^2} = \sqrt{5}$ units
$CD = \sqrt{(3-5)^2 + (5 - 6)^2} = \sqrt{5}$ units
$DA = \sqrt{(4-3)^2 + (3 - 5)^2} = \sqrt{5}$ units
$AC = \sqrt{(5-4)^2 + (6 - 3)^2} = \sqrt{10}$ units
$BD = \sqrt{(3 - 6)^2 + (5 - 4)^2} = \sqrt{10}$ units
As $AB = BC = CD = DA$ and $AC = BD$, so ABCD is a square.
$BC = \sqrt{(5-6)^2 + (6 - 4)^2} = \sqrt{5}$ units
$CD = \sqrt{(3-5)^2 + (5 - 6)^2} = \sqrt{5}$ units
$DA = \sqrt{(4-3)^2 + (3 - 5)^2} = \sqrt{5}$ units
$AC = \sqrt{(5-4)^2 + (6 - 3)^2} = \sqrt{10}$ units
$BD = \sqrt{(3 - 6)^2 + (5 - 4)^2} = \sqrt{10}$ units
As $AB = BC = CD = DA$ and $AC = BD$, so ABCD is a square.