Question

If two adjacent vertices of a parallelogram are (3, 2) and (–1, 0) and the diagonals intersect at (2, –5) then find the coordinates of the other two vertices. [CBSE 2017]

Answer 1

Let ABCD be the parallelogram with two adjacent vertices A(3, 2) and B(−1, 0). Suppose O(2, −5) be the point of intersection of the diagonals AC and BD.
Let C(x₁, y₁) and D(x₂, y₂) be the coordinates of the other vertices of the parallelogram.

We know that the diagonals of the parallelogram bisect each other. Therefore, O is the mid-point of AC and BD.
Using the mid-point formula, we have
$$\begin{gathered} \left(\frac{x_1+3}{2},\frac{y_1+2}{2}\right)=\left(2,-5\right) \\ \Rightarrow \frac{x_1+3}{2}=2\mathrm{and}\frac{y_1+2}{2}=-5 \\ \Rightarrow x_1+3=4\mathrm{and}{y}_1+2=-10 \\ \Rightarrow x_1=4-3=1\mathrm{and}{y}_1=-10-2=-12 \end{gathered}$$
So, the coordinates of C are (1, −12).
Also,
$$\begin{gathered} \left(\frac{x_2+\left(-1\right)}{2},\frac{y_2+0}{2}\right)=\left(2,-5\right) \\ \Rightarrow \frac{x_2-1}{2}=2\mathrm{and}\frac{y_2}{2}=-5 \\ \Rightarrow x_2-1=4\mathrm{and}{y}_2=-10 \\ \Rightarrow x_2=4+1=5\mathrm{and}{y}_2=-10 \end{gathered}$$
So, the coordinates of D are (5, −10).