Question

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 $b.$

• A. $a=1,b=1$
• B. $a=2,b=2$
• C. $a=1,b=4$
• D. $a=1,b=2$

Answer 1

The correct option is A $a=1,b=1$

Midpoint formula $X=\left(\frac{x_1+x_2}{2}\right)andY=\left(\frac{y_1+y_2}{2}\right)$

Coordinates of $M$ are mid points of $AC$

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

As the diagonals of parallelogram bisect each other,

Coordinates of $M$ are mid points of $BD$ aswell

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

Equating $X$- coordinates:

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

$a=1$

Equating $Y$- coordinates:

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

$b=1$