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Mathematics

If the points A(-2, -1), B(1, 0), C(a, 3) and D(1, b) form a parallelogram, find the values of a and b.

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Answer

We know that,

Diagonals of a parallelogram bisect each other.

Since, ABCD is a // gm.

Thus, Mid-point of AC = Mid-point of BD.

Given,

A(-2, -1), B(1, 0), C(a, 3) and D(1, b)

If the points A(-2, -1), B(1, 0), C(a, 3) and D(1, b) form a parallelogram, find the values of a and b. Reflection, RSA Mathematics Solutions ICSE Class 10.

By using mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Mid-point of AC :

MAC=(2+a2,1+32)MAC=(2+a2,22)MAC=(2+a2,1).\Rightarrow M{AC} = \Big(\dfrac{-2 + a}{2}, \dfrac{-1 + 3}{2}\Big) \\[1em] \Rightarrow M{AC} = \Big(\dfrac{-2 + a}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow M_{AC} = \Big(\dfrac{-2 + a}{2}, 1\Big).

Midpoint of Diagonal BD :

MBD=(1+12,0+b2)MBD=(22,b2)MBD=(1,b2).\Rightarrow M{BD} = \Big(\dfrac{1 + 1}{2}, \dfrac{0 + b}{2}\Big) \\[1em] \Rightarrow M{BD} = \Big(\dfrac{2}{2}, \dfrac{b}{2}\Big) \\[1em] \Rightarrow M_{BD} = \Big(1, \dfrac{b}{2}\Big) .

Equating the x-coordinates of mid-points of AC and BD, we get :

2+a2\dfrac{-2 + a}{2} = 1

⇒ -2 + a = 2

⇒ a = 2 + 2

⇒ a = 4.

Equating the y-coordinates:

⇒ 1 = b2\dfrac{b}{2}

⇒ b = 2.

Hence, a = 4 and b = 2.

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