Question

A(-3,1),B(-6,-7),C(3,-9),D(6,-1) show that the following points taken in order form the vertices of a parallelogram​

Answer 1

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

• A (-3, 1)
• B (-6, 7)
• C (3, -9)
• D (6, -1)

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

• ABCD is a parallellogram

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Distance of AB

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

→ Substituting the datas we get,

 $AB=\sqrt{(-6+3)^{2}+(-7-1)^{2} }$

$AB=\sqrt{9+64}$

AB = √73

→ Distance of BC

$BC=\sqrt{(3+6)^{2}+(-9+7)^{2} }$

$BC=\sqrt{81+4}$

BC = √85

→ Distance of CD

$CD=\sqrt{(6-3)^{2}+(-1+9)^{2} }$

$CD=\sqrt{9+64}$

CD = √73

→ Distance of AD

$AD=\sqrt{(6+3)^{2}+(-1-1)^{2} }$

$AD=\sqrt{81+4}$

AD = √85

→ Here AB = CD and BC = AD.

→ Since opposite sides are equal, ABCD is a parallelogram.

Hence proved.

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The opposite sides of a parallelogram are equal and parallel to each other.

→ Opposite angles are equal in a parallelogram.

Answer 2

Solution :-

Given :-

• A = ( -3 , 1 )
• B = ( -6 , -7 )
• C = ( 3 , -9 )
• D = ( 6 , -1 )

We need to prove that the following points are taken in order of the vertices of a parallelogram.

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

$\bullet \sf \ AB = \sqrt{(-6+3^2)+(-7-1)^2} =\sqrt{9+64} =\sqrt{73}$

$\bullet\sf \ BC= \sqrt{(3-(-6)^2+(-9+7)^2} =\sqrt{81+4} =\sqrt{85}$

$\bullet\sf \ CD=\sqrt{(6-3)^2+(-1+9)^2}=\sqrt{9+64} =\sqrt{73}$

$\bullet\sf \ AD=\sqrt{(-1+3)^2+(6+3)^2} =\sqrt{4+81} =\sqrt{85}$

AB = CD and BC = AD

In a parallelogram opposite sides are equal .

So the following points are taken in the order of the vertices of a parallelogram .