Question

8. The points A (-4,-2), B (4, 6),
C (-4,6) and D (4,-2) are the
vertices of a​

Answer 1

SOLUTION :-

Given that,

• A = ( -4 , -2 )

• B = ( 4 , 6 )

• C = ( -4 , 6 )

• D = ( 4 , -2 )

$\boxed{\bf Distance \ formula = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} }$

$\bullet \sf \ AB = \sqrt{(4-(-4))^2+(6-(-2))^{2}}$

        $= \sf \sqrt{(4+4)^2+(6+2)^2} \\\\= \sqrt{8^{2}+8^{2}} \\\\= \sqrt{64+64} \\\\= \sqrt{128} \\\\= 8\sqrt{2} \ units$

$\bullet\sf \ BC = \sqrt{(-4-4)^{2}-(6-6)^{2}}$

       $= \sf \sqrt{8^2+0} \\\\= \sqrt{64} \\\\= 8 \ units$

$\bullet\sf \ CD = \sqrt{(4-(-4))^{2}+(6-(-2))^2}$

       $= \sf \sqrt{(4+4)^{2}+(6+2)^{2} }\\\\= \sqrt{8^{2}+8^{2}} \\\\= \sqrt{64+64} \\\\=\sqrt{128} \\\\= 8 \sqrt{2} \ units$

$\bullet \sf \ AD = \sqrt{(4-(-4))^{2}+(-2-(-2))^{2}}$

       $= \sf \sqrt{(4+4)^{2}(-2+2)^{2}} \\\\= \sqrt{8^{2}+0} \\\\= \sqrt{64} \\\\= 8 \ units$

The points A, B, C and D are the points of a rectangle.

Since AB = CD and BC = AD .