NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.3

Question 3

Step 1: Use the standard coordinate-geometry tests (as in the method shown).

• A quadrilateral is a parallelogram if the midpoints of diagonals \(AC\) and \(BD\) are the same.
• A parallelogram is a rectangle if its diagonals are equal, i.e., \(AC = BD\) (or \(AC^2 = BD^2\)).
• If in that rectangle adjacent sides are equal (\(AB = BC\)), then it becomes a square.

Step 2: Check parallelogram condition by midpoints of diagonals.

Midpoint of \(AC\):

\(A(2,-2),\ C(11,-1)\)

\(M_{AC} = \left(\dfrac{2+11}{2},\dfrac{-2+(-1)}{2}\right) = \left(\dfrac{13}{2},\dfrac{-3}{2}\right)\)

Midpoint of \(BD\):

\(B(7,3),\ D(6,-6)\)

\(M_{BD} = \left(\dfrac{7+6}{2},\dfrac{3+(-6)}{2}\right) = \left(\dfrac{13}{2},\dfrac{-3}{2}\right)\)

Since \(M_{AC} = M_{BD}\), the diagonals bisect each other. Hence, \(ABCD\) is a parallelogram.

Step 3: Check whether the parallelogram is a rectangle by comparing diagonals.

Compute squared lengths (to avoid roots):

\(AC^2 = (11-2)^2 + \big((-1)-(-2)\big)^2 = 9^2 + 1^2 = 81 + 1 = 82\)

\(BD^2 = (6-7)^2 + \big((-6)-3\big)^2 = (-1)^2 + (-9)^2 = 1 + 81 = 82\)

So \(AC^2 = BD^2\) → \(AC = BD\). Therefore, the parallelogram is a rectangle (it could be a square only if adjacent sides are equal).

Step 4: Check if it is a square by testing adjacent sides \(AB\) and \(BC\).

\(AB^2 = (7-2)^2 + (3-(-2))^2 = 5^2 + 5^2 = 25 + 25 = 50\)

\(BC^2 = (11-7)^2 + \big((-1)-3\big)^2 = 4^2 + (-4)^2 = 16 + 16 = 32\)

Since \(AB^2 \ne BC^2\), we have \(AB \ne BC\). So it is not a square.

Final Conclusion: \(ABCD\) is a parallelogram with equal diagonals, but adjacent sides are not equal. Hence, \(ABCD\) is a rectangle.