Question

if A (-2, 1 ) , B (a, o) , C (4, b) and D (1, 2) are are the vertices of a parallelogram ABCD , finf the value a and b . hence find the lenght of of its sides?

do this !​

Answer 1

SOLUTION :-

Given that,

The points A ( -2 , 1),  ( a , 0 ), C ( 4 , b ) and D ( 1 , 2 ) are the vertices of a parallelogram.

We know that,

The diagonals of a parallelogram bisect each other.

That is,

Midpoint of AC = Midpoint of BD

$\boxed{\bf Midpoint \ formula = \left(\dfrac{x_1+x_2}{2} \ , \ \dfrac{y_1+y_2}{2} \right)}$

$\longrightarrow \sf \left( \dfrac{-2+4}{2} \ , \ \dfrac{1+b}{2} \right) = \left( \dfrac{1+a}{2} \ , \ \dfrac{2+0}{2} \right)$

$\hookrightarrow\sf \dfrac{-2+4}{2}=\dfrac{1+a}{2} \\\\ \hookrightarrow 2 = 1+a \\\\\hookrightarrow \bf a = 1$

$\hookrightarrow\sf \dfrac{1+b}{2} = \dfrac{2+0}{2} \\\\\hookrightarrow 1+b = 2 \\\\\hookrightarrow \bf b = 1$

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

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

      $= \sf \sqrt{(1+2)^2+(0-1)^2} \\\\= \sqrt{3^2+1^2} \\\\ = \sqrt{9+1} \\\\ = \sqrt{10} \ units$

$\bullet\sf \ BC = \sqrt{(4-1)^2+(1-0)^2}$

       $= \sf \sqrt{3^2+1^2} \\\\= \sqrt{9+1} \\\\= \sqrt{10} \ units$

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

        $= \sf \sqrt{(-3)^2+1^1} \\ \\= \sqrt{9+1} \\\\= \sqrt{10} \ units$

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

       $= \sf \sqrt{(1+2)^2+(2-1)^2} \\\\= \sqrt{3^2+1^2} \\\\= \sqrt{9+1} \\\\= \sqrt{10} \ units$