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If $A (- 2, 1), B (a, 0), C (4, b)$ and $D (1, 2)$ are the vertices of a parallelogram ABCD taken in order, find the values of $a$ and $b .$

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a = 1, b = 1

a = 2, b = 2

a = 1, b = 4

a = 1, b = 2

answer is A.

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Detailed Solution

Given, ABCD is a parallelogram,

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

and

D (1, 2)

.
Drawing the required figure,
Question Image

Midpoint of two point

{(x}_{1}, y_{1})

{(x}_{2}, y_{2})

is ,

(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})

Coordinates of

M

are mid points of

AC

M = (\frac{4 - 2}{2}, \frac{b + 1}{2})

…….(1)
As the diagonals of parallelogram bisect each other,
Coordinates of

M

are mid points of

BD

as well,

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

……..(2)
Equating x- coordinates of equation (1) and (2),

\Rightarrow \frac{4 - 2}{2} = \frac{a + 1}{2}

\Rightarrow a + 1 = 2

\Rightarrow a = 1

Equating y- coordinates of equation (1) and (2),

\Rightarrow \frac{b + 1}{2} = \frac{0 + 2}{2}

\Rightarrow b + 1 = 2

\Rightarrow b = 1

Thus, the value of a is 1 and b is 1.
Therefore, option (1) is correct.

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