Question

show that the points A(-2,-1),B(-1,1),C(5,-2) and D(4,-4) form a rectangle​

Answer 1

SOLUTION :-

We have to show that,

The points A ( -2 , -1 ), B ( -1 , 1 ), C ( 5 , 2 ) and D ( 4 , -4 ) form a rectangle.

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

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

       $= \sf \sqrt{(-1+2)^2+(1+1)^2 } \\\\= \sqrt{1^2+2^2} \\\\= \sqrt{1+4} \\\\= \sqrt{5} \ units$

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

        $= \sf \sqrt{(5+1)^2+(-2-1)^2} \\\\= \sqrt{6^2+(-3)^2 } \\\\= \sqrt{36+9} \\\\= \sqrt{45} \\ \\= 3\sqrt{5} \ units$

$\bullet\sf \ CD = \sqrt{(4-5)^2+(-4-(-2))^2}$

        $= \sf \sqrt{(4-5)^2+(-4+2)^2} \\\\= \sqrt{(-1)^2+(-2)^2} \\\\= \sqrt{1+4} \\ \\= \sqrt{5} \ units$

$\bullet\sf \ AD = \sqrt{(4-(-2))^2+(-4-(-1))^2}$

        $= \sf \sqrt{(4-(-2))^2+(-4-(-1))^2} \\\\= \sqrt{(4+2)^2+(-4+1)^2} \\\\= \sqrt{6^2+(-3)^2} \\\\= \sqrt{36+9} \\\\= \sqrt{45} \\\\= 3\sqrt{5} \ units$

AB = CD and BC = AD

That is,

Opposite sides are equal.

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

        $= \sf \sqrt{(5+2)^2+(-2+1)^2} \\\\= \sqrt{7^2+(-1)^2} \\\\= \sqrt{49+1} \\\\= \sqrt{50} \ units$

$\bullet\sf \ BD = \sqrt{(4-(-1)^2+(-4-1)^2}$

       $= \sf \sqrt{(4+1)^2+(-4-1)^2} \\\\= \sqrt{5^2+(-5)^2} \\\\= \sqrt{25+25} \\\\= \sqrt{50} \ units$

AC = BD

That is,

Diagonals are equal.

Hence the given points form a rectangle.

Answer 2

Step-by-step explanation:

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