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:

In a parallelogram, diagonals bisect each other. Use this to find unknown coordinates. For rectangles, adjacent sides must be perpendicular.

Answer 1

In a parallelogram, the diagonals bisect each other. So, the midpoint of diagonal \(AC\) must equal the midpoint of diagonal \(BD\). Coordinates of \(A = (6, 1)\), \(C = (9, 4)\) Midpoint of \(AC\) is: \[ \left(\frac{6 + 9}{2}, \frac{1 + 4}{2}\right) = \left(\frac{15}{2}, \frac{5}{2}\right) \] Let \(B = (p, 2)\), \(D = (7, q)\) Midpoint of \(BD\) is: \[ \left(\frac{p + 7}{2}, \frac{2 + q}{2}\right) \] Equating midpoints: \[ \frac{p + 7}{2} = \frac{15}{2} \Rightarrow p = 8 \\ \frac{2 + q}{2} = \frac{5}{2} \Rightarrow q = 3 \] Therefore, \(p = 8\), \(q = 3\) Now to check if it's a rectangle, check if adjacent sides are perpendicular (dot product = 0). Vectors: \[ \vec{AB} = B - A = (8 - 6, 2 - 1) = (2, 1) \\ \vec{BC} = C - B = (9 - 8, 4 - 2) = (1, 2) \] Dot product: \[ \vec{AB} \cdot \vec{BC} = 2 \cdot 1 + 1 \cdot 2 = 2 + 2 = 4 \neq 0 \] Hence, ABCD is not a rectangle.