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

Solution

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} = 2sqrt{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} = 2sqrt{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)