Question

If 𝑨(−𝟐, 𝟏), 𝑩(𝒂, 𝟎), 𝑪(𝟒, 𝒃) and 𝑫(𝟏, 𝟐) are the vertices of a parallelogram 𝑨𝑩𝑪𝑫, find the values of 𝒂 and 𝒃. Also, find the lengths of its sides.

Answer 1

We know that the diagonals of a parallelogram bisect each other. Therefore, the coordinates of the mid-point of AC are same as the coordinates of the mid-point of BD.

The coordinates of the mid-point of a line formed by joining two points $(x_1,y_1)and(x_2,y_2)$ are $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

Midpoint of AC = $(\frac{-2+4}{2},\frac{1+b}{2})$

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

⇒ $(\frac{-2+4}{2},\frac{1+b}{2})$ = $(\frac{a+1}{2},\frac{0+2}{2})$
⇒ $(1,\frac{b+1}{2})=(\frac{a+1}{2},1)$
⇒ $\frac{a+1}{2}=1$ and $\frac{b+1}{2}=1$
⇒ $a+1=2$ and $b+1=2$
⇒ $a=1$ and $b=1$

So, the coordinates of the vertices of the parallelogram ABCD are A(-2, 1), B(1, 0), C(4, 1) and D(1, 2).

Now, length of side $AB=DC=\sqrt{(1+2{)}^2+(0-1{)}^2}=\sqrt{9+1}=\sqrt{10}$ units
Length of side $AD=BC=\sqrt{(1+2{)}^2+(2-1{)}^2}=\sqrt{9+1}=\sqrt{10}$ units