Question

Show that the point a(5,6) b (1,5) c(2,1) d (6,2) are vertives of a square

Answer 1

Solution :-

We have to show that :-

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

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

$\bullet\sf \ AB = \sqrt{(1-5)^2+(5-6)^2}$

        $= \sf \sqrt{(-4)^2+(-1)^2} \\\\= \sqrt{16+1} \\\\= \sqrt{17} \ units$

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

        $= \sf \sqrt{1^2+(-4)^2} \\\\= \sqrt{1+16} \\\\=\sqrt{17} \ units$

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

        $= \sf \sqrt{4^2+1^2} \\\\= \sqrt{16+1} \\\\= \sqrt{17} \ units$

$\bullet \sf \ AD = \sqrt{(6-5)^2+(2-6)^2}$

        $= \sf \sqrt{1^2+(-4)^2} \\\\= \sqrt{1+16} \\\\= \sqrt{17} \ units$

AB = BC = CD = AD

That is,

Sides are equal.

$\bullet \sf \ AC = \sqrt{(2-5)^2+(1-6)^2}$

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

$\bullet \sf \ BD = \sqrt{(6-1)^2+(2-5)^2}$

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

AC = BD

That is,

Diagonals are equal.

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