Question

If A $(1,0)$, B$(5,3)$, C $(2,7)$ and D $(x,y$) are vertices of a parallelogram ABCD, the co-ordinates of D are

• A. $(-2,-3)$
• B. $(-2,4)$
• C. $(2,-3)$
• D. $(3,5)$

Answer 1

Answer: (B)

$(-2,4)$

We know that the diagonals of a parallelogram bisect each other. So, the
the midpoint of AC is the same as the midpoint of BD.

Mid point of two points ${(x}_1,y_1)$ and ${(x}_2,y_2)$ is calculated by the formula $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

So, midpoint of $AC=$ Mid point of $BD$

$=>(\frac{1+2}{2},\frac{0+7}{2})=(\frac{5+x}{2},\frac{3+y}{2})$

$=>(\frac{3}{2},\frac{7}{2})=(\frac{5+x}{2},\frac{3+y}{2})$

$=>5+x=3;3+y=7$

$=>x=-2;y=4$

Hnce, $D=(-2,4)$