Question

Show that the points A(3, 5), B(6, 0), C(1, -3) and D(-2, 2) are the vertices of a square ABCD​

Answer 1

Solution :-

We have to show that :-

The points A ( 3 , 5 ), B ( 6 , 0 ), C ( 1 , -3 ) and D ( -2 , 2 ) are the vertices of a square ABCD.

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

$\bullet \ \sf A B = \sqrt{(6-3)^2+(0-5)^2}$

        $= \sf \sqrt{3^2+(-5)^2} \\\\= \sqrt{9+25} \\\\= \sqrt{34} \ units$

$\bullet \sf \ BC = \sqrt{(1-6)^2+(-3-0)^3}$

        $= \sf \sqrt{(-5)^2+(-3)^2} \\\\= \sqrt{25+9} \\\\= \sqrt{34} \ units$

$\bullet \sf \ CD = \sqrt{(-2-1)^2+(2-(-3))^2}$

        $= \sf \sqrt{(-3)^2+5^2} \\\\= \sqrt{9+25} \\\\= \sqrt{34} \ units$

$\bullet \sf \ AD = \sqrt{(-2-3)^2+(2-5)^2}$

        $= \sf \sqrt{(-5)^2+(-3)^2} \\\\=\sqrt{25+9 } \\\\= \sqrt{34} \ units$

∴ AB = BC = CD = AD

That is,

Sides are of equal length.

$\bullet \sf \ Diagonal \ AC = \sqrt{(1-3)^2+(-3-5)^2}$

                      $= \sf \sqrt{(-2)^2+(-8)^2} \\\\= \sqrt{4+64} \\\\= \sqrt{68} \ units$

$\bullet \sf \ Diagonal \ BD = \sqrt{(-2-6)^2+(2-0)^2}$

                      $= \sf \sqrt{(-8)^2+2^2} \\\\=\sqrt{64+4} \\\\= \sqrt{68} \ units$

∴ AC = BD

That is,

Diagonals are equal.

Since sides and diagonals are equal, the given points are the vertices of the square ABCD.