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, diagonals bisect each other. This means the midpoint of diagonal \(AC\) is the same as the midpoint of diagonal \(BD\). Given \(A = (6, 1)\) and \(C = (9, 4)\), the midpoint of \(AC\) is: \[ \left(\frac{6 + 9}{2}, \frac{1 + 4}{2}\right) = \left(\frac{15}{2}, \frac{5}{2}\right) \] Let \(B = (p, 2)\) and \(D = (7, q)\). The midpoint of \(BD\) is: \[ \left(\frac{p + 7}{2}, \frac{2 + q}{2}\right) \] Equating the midpoints gives: \[ \frac{p + 7}{2} = \frac{15}{2} \Rightarrow p = 8 \] \[ \frac{2 + q}{2} = \frac{5}{2} \Rightarrow q = 3 \] Therefore, \(p = 8\), \(q = 3\). To check if the parallelogram is a rectangle, verify if adjacent sides are perpendicular (dot product equals 0). Vectors: \[ \vec{AB} = B - A = (8 - 6, 2 - 1) = (2, 1) \] \[ \vec{BC} = C - B = (9 - 8, 4 - 2) = (1, 2) \] Dot product: \[ \vec{AB} \cdot \vec{BC} = 2 \cdot 1 + 1 \cdot 2 = 2 + 2 = 4 eq 0 \] Hence, ABCD is not a rectangle.