Question

if A(-5,7), B(a,0), C(4,b),(1,2) are the vertices of a parallelogram ABCD , find the values of a and b. Hence find the lengths of its sides.

Answer 1

if A(-5,7), B(a,0), C(4,b),(1,2) are the vertices of a parallelogram ABCD , find the values of a and b. Hence find the lengths of its sides

Answer 2

The lengths of its sides are $\sqrt{58}$ and $\sqrt{61}$.

Step-by-step explanation:

Given:

Parallelogram ABCD

A( -5, 7), B(a, 0), C(4, b), D(1, 2)

Now, Midpoint of AC = Midpoint of BD

(∵ diagonals of a parallelogram bisect each other)

Using Midpoint Formula, we get

The midpoint of the segment joining $(x_1,y_1)$  and $(x_2,y_2)$ is given by = $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Putting values,

The midpoint of AC = $(\frac{-5+4}{2},\frac{7+b}{2} )$

                                 =  $(\frac{-1}{2},\frac{7+b}{2} )$

The midpoint of BD = $(\frac{a+1}{2}, \frac{0+2}{2} )$                                         = $(\frac{a+1}{2}, 1)$

Now, $\frac{7+b}{2} = 1$

⇒ 7 + b =2

⇒ b = 2 - 7

⇒ b = - 5

And $\frac{a+1}{2}=\frac{-1}{2}$  

⇒ a+1 = -1

⇒ a = -1 - 1

⇒ a = -2

So,  A( -5, 7), B(-2, 0 ), C(4, -5), D(1, 2)

Using Distance Formula, we get

Distance between $(x_1, y_1)$ and $(x_2, y_2 )$ is given by,

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

Putting Values,

AB = $\sqrt{(-2-(-5))^2+(0-7)^2}$

     = $\sqrt{(-2+5)^2+(7)^2}$

     = $\sqrt{(3)^2+(7)^2}$

     = $\sqrt{9+49}$

     = $\sqrt{58}$

     = CD

AD = $\sqrt{(1-(-5))^2+(2-7)^2}$

     = $\sqrt{(1+5)^2+(-5)^2}$

     = $\sqrt{(6)^2+(-5)^2}$

     = $\sqrt{36+25}$

     = $\sqrt{61}$

     = BC

Therefore,

AB = CD = $\sqrt{58}$, AD = BC = $\sqrt{61}$