Question

If the points $A(6, 1)$, $B(p, 2)$, $C(9, 4)$ and $D(7, q)$ are the vertices of a parallelogram $ABCD$, then find the values of $p$ and $q$. Hence, check whether $ABCD$ is a rectangle or not.

Hint:

Use midpoint formula to solve parallelogram diagonal problems; use dot product to test right angles.

Answer 1

In parallelogram, diagonals bisect each other. Midpoint of $AC =$ midpoint of $BD$
Midpoint of $AC = \left( \dfrac{6 + 9}{2}, \dfrac{1 + 4}{2} \right) = (7.5, 2.5)$
Midpoint of $BD = \left( \dfrac{p + 7}{2}, \dfrac{2 + q}{2} \right)$
Equating: \[ \dfrac{p + 7}{2} = 7.5 \Rightarrow p + 7 = 15 \Rightarrow p = 8\\ \dfrac{2 + q}{2} = 2.5 \Rightarrow 2 + q = 5 \Rightarrow q = 3 \] Check if $ABCD$ is rectangle: \[ AB = \sqrt{(8 - 6)^2 + (2 - 1)^2} = \sqrt{4 + 1} = \sqrt{5}\\ BC = \sqrt{(9 - 8)^2 + (4 - 2)^2} = \sqrt{1 + 4} = \sqrt{5}\\ Dot product of $\vec{AB}$ and $\vec{BC}$: (2,1) • (1,2) = 2 + 2 = 4 ≠ 0 ⇒ Not perpendicular \] Answer: $p = 8$, $q = 3$; $ABCD$ is a parallelogram but not a rectangle.