Question

If A (–2, –1), B (a, 0), C (4, b) and D (1, 2) are the vertices of a parallelogram, find the values of a and b.

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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 $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$\left[1\right]$


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

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

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

⇒ $\frac{a+1}{2}$ = 1 and $\frac{b-1}{2}$ = 1

⇒ a + 1 = 2 and b - 1 = 2

⇒ a = 1 and b = 3
$\left[1\right]$

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

⇒ $(\frac{-2+4}{2},\frac{-1+b}{2})$ = $(\frac{a+1}{2},\frac{0+2}{2})$
$\left[1\right]$

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

⇒ $\frac{a+1}{2}$ = 1 and $\frac{b-1}{2}$ = 1

⇒ a + 1 = 2 and b - 1 = 2

⇒ a = 1 and b = 3
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