If the points $\mathbf{P}=\hat{i}+2 \hat{j}, \mathbf{Q}=4 \hat{i}+6 \hat{j}$, $\mathbf{R}=5 \hat{i}+7 \hat{j}, \mathbf{S}=a \hat{i}+b \hat{j}$ are the consecutive vertices of a parallelogram $P Q R S$, then

Question

Aicharya · Accepted Answer

Given, $\begin{aligned} & \mathbf{P}=\hat{i}+2 \hat{j} \\ & \mathbf{Q}=4 \hat{i}+6 \hat{j} \\ & \mathbf{R}=5 \hat{i}+7 \hat{j} \\ & \mathbf{S}=a \hat{i}+b \hat{j}\end{aligned}$

Let $A$ be the point of intersection of diagonals $P R$ and $Q S$. $\Rightarrow A$ is the mid-point of $P R$ and $Q S$ both. Now, $A=\left(\frac{1+5}{2}, \frac{2+7}{2}\right) \equiv\left(3, \frac{9}{2}\right)$ $[\because A$ is the mid-point of $P R]$ Also, $A \equiv\left(\frac{4+a}{2}, \frac{6+b}{2}\right) \equiv\left(3, \frac{9}{2}\right)$ $\Rightarrow \quad \frac{4+a}{2}=3 \Rightarrow a=2$ and $\frac{6+b}{2}=\frac{9}{2} \Rightarrow b=3$ Thus, $(a, b) \equiv(2,3)$

If the points $\mathbf{P}=\hat{i}+2 \hat{j}, \mathbf{Q}=4 \hat{i}+6 \hat{j}$, $\mathbf{R}=5 \hat{i}+7…

Solution