Question

The three vertices of a parallelogram ABCD, taken in order are $A(1,-2)$, $B(3,6)$ and $C(5,10)$. Find the coordinates of the fourth vertex D.

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

Answer 1

The correct option is A

$D(3,2)$

Sol. Let $A(1,-2)$, $B(3,6)$ and $C(5,10)$ be the three vertices of a parallelogram ABCD and let its fourth vertex be $D(a,b)$.

Join AC and BD. Let AC and BD intersect at point O.

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

So, O is the midpoint of AC as well as that of BD.

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

$A(1,-2)\equiv (x_1,y_1)$, $C(5,10)\equiv (x_2,y_2)$

Midpoint of AC is $\left(\begin{array}{c}\frac{1+5}{2},\frac{-2+10}{2}\end{array}\right)$, i.e.,$(3,4)$

Midpoint of BD is $\left(\begin{array}{c}\frac{3+a}{2},\frac{6+b}{2}\end{array}\right)$

$\therefore$ $\frac{3+a}{2}=3$ and $\frac{6+b}{2}=4$

$3+a=6$ and $6+b=8$

$a=3$ and $b=2$

Hence, the fourth vertex is $D(3,2)$.