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 $D$.

Answer 1

Hint: We will start this question by assuming the fourth vertex of a parallelogram be $D\left( x,y \right)$. We know that the diagonals of a parallelogram bisect each other, so we find the midpoint of both the diagonals by using the formula of midpoint. Then, by equating both equations we calculate the values of $x$ and $y$. We use the below formula to find the midpoint
$\left( X,Y \right)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$

Complete step by step solution:
Before starting to solve this question first we draw a diagram of a parallelogram $ABCD$.

Here, In above diagram $ABCD$ is a parallelogram with diagonals $AC$ and $BD$.
As we know that the diagonals of a parallelogram bisects each other. So, $O$ is the midpoint of $AC$ and $BD$.
To find the coordinates of $O$ we apply the formula of midpoint.
$\left( X,Y \right)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$
So, the midpoint of $AC$ where, $A\left( 1,-2 \right)$ $C(5,10)$ will be $\Rightarrow \left( \dfrac{1+5}{2},\dfrac{-2+10}{2} \right)=\left( \dfrac{6}{2},\dfrac{8}{2} \right)=\left( 3,4 \right)$
Also we have to find the midpoint of $BD$ where, $B\left( 3,6 \right)$ $D\left( x,y \right)$
$\Rightarrow \left( \dfrac{3+x}{2},\dfrac{6+y}{2} \right)$
Since both $AC$ and $BD$ has midpoint $O$, so we equate both coordinates we will get
$\left( \dfrac{3+x}{2} \right)=3$ and $\left( \dfrac{6+y}{2} \right)=4$
After cross multiplying, we get $3+x=6$ and $6+y=8$
Now, keeping variables on the LHS and transposing other terms, we get $x=6-3$ and $y=8-6$
Simplify further, we get
$x=3$ and $y=2$
So the coordinates of the fourth vertex $D$ $=\left( 3,2 \right)$

Note: Whenever such types of questions appear, first draw a diagram to understand the question. Here we use the concept that the diagonals of a parallelogram bisect each other. The possibility of mistake can be not using correct values in the formula. When substitute values in equation sign convention must be kept in mind.