Question

Three consecutive vertices of a parallelogram are $(1,-2)$, $(3,6)$ and $(5,10)$. The coordinates of the fourth vertex are:

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

Answer 1

Answer: (C)

$(3,2)$

Let $ABCD$ be the parallelogram, having three consecutive vertices as $A(1,-2),B(3,6)$ and $C(5,10)$.

Also, let the fourth vertex be $D(x,y)$

We know that, the diagonals of a parallelogram bisect each other.

So, the midpoint of $AC$ is same as the midpoint of $BD$.

We also know that, the coordinates of the midpoint of the line segment joining $(x_1,y_1)and(x_2,y_2)$ are:

$P(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

So, midpoint of $AC=$ Midpoint of $BD$
$\Rightarrow (\frac{1+5}{2},\frac{-2+10}{2})=(\frac{3+x}{2},\frac{6+y}{2})$
$\Rightarrow (\frac{6}{2},\frac{8}{2})=(\frac{3+x}{2},\frac{6+y}{2})$
$\Rightarrow 3+x=6;6+y=8$
$\Rightarrow x=3;y=2$
$=(3,2)$

Hence, the coordinates of the fourth vertex are $(3,2)$.