Show that the c PQRS formed by P(2,1), Q(-1,3), R(-5,-3) and S(-2,-5) is a rectangle
12. Show that the c PQRS formed by P(2,1), Q(-1,3), R(-5,-3) and S(-2,-5) is a rectangle
The given points are P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5).
PQ = \[\sqrt{\left( - 1 - 2 \right)^2 + \left( 3 - 1 \right)^2} = \sqrt{9 + 4} = \sqrt{13}\]
QR = \[\sqrt{\left( - 5 + 1 \right)^2 + \left( - 3 - 3 \right)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\]
RS = \[\sqrt{\left( - 2 + 5 \right)^2 + \left( - 5 + 3 \right)^2} = \sqrt{9 + 4} = \sqrt{13}\]
PS = \[\sqrt{\left( - 5 - 1 \right)^2 + \left( - 2 - 2 \right)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}\]
PQ = RS and QR = PS
Opposite sides are equal.
\[PR = \sqrt{\left( - 3 - 1 \right)^2 + \left( - 5 - 2 \right)^2} = \sqrt{16 + 49} = \sqrt{65}\]
\[QS = \sqrt{\left( - 2 + 1 \right)^2 + \left( - 5 - 3 \right)^2} = \sqrt{1 + 64} = \sqrt{65}\]
Diagonals are equal.
Thus, the given vertices form a rectangle.
Explanation:-
The given points are P(2, 1), Q(-1, 3), R(-5, -3), and S(-2, -5).
To find the length of the sides, we use the distance formula.
PQ = √[(-1-2)² + (3-1)²] = √[9+4] = √13
QR = √[(-5+1)² + (-3-3)²] = √[16+36] = √52 = 2√13
RS = √[(-2+5)² + (-5+3)²] = √[9+4] = √13
PS = √[(-5-1)² + (-2-2)²] = √[36+16] = √52 = 2√13
We observe that PQ = RS and QR = PS. Opposite sides are equal.
To check if the given vertices form a rectangle, we need to check if the diagonals are equal.
PR = √[(-3-1)² + (-5-2)²] = √[16+49] = √65
QS = √[(-2+1)² + (-5-3)²] = √[1+64] = √65
Since PR = QS, the diagonals are equal. Thus, the given vertices form a rectangle.
Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)