Question

Show that the ▢PQRS formed by P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5) is a rectangle.

Answer 1

Given: P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5)

Distance Formula = `sqrt((x_2 - x_1)^2 +  (y_2 - y_1)^2)`

PQ = `sqrt((-1 - 2)^2 + (3 - 1)^2)`

= `sqrt((3-)^2 + (2)^2)`

= `sqrt(9 + 4)`

= `sqrt 13`    ...(i)

QR = `sqrt([-5 - (-1)]^2 + (-3 - 3)^2)`

= `sqrt((-4)^2 + (-6)^2)`

= `sqrt(16 + 36)`

= `sqrt 52`

= `sqrt(2 xx 2 xx 13)`

= 2`sqrt13`    ...(ii)

RS = `sqrt([-2 - (-5)]^2 + [-5 - (-3)]^2)`

= `sqrt((-2 + 5)^2 + (-5 + 3)^2)`

= `sqrt(3^2 + (-2)^2)`

= `sqrt(9 + 4)`

= `sqrt 13`    ...(iii)

PS = `sqrt((-2 - 2)^2 + (-5 - 1)^2)`

= `sqrt((-4)^2 + (-6)^2)`

= `sqrt(16 + 36)`

= `sqrt52`

= `sqrt(2 xx 2 xx 13)`

= 2`sqrt 13`    ...(iv)

In ▢PQRS,

PQ = RS    ...[From (i) and (iii)]

QR = PS    ...[From (ii) and (iv)]

∴ ▢PQRS is a parallelogram    ...(A quadrilateral is a parallelogram if its opposite sides are equal)

By distance formula,

PR = `sqrt((-5 - 2)^2 + (-3 - 1)^2)`

= `sqrt((-7)^2 + (-4)^2)`

= `sqrt(49 +16)`

= `sqrt 65`    ...(v)

QS = `sqrt([-2 - (-1)]^2 + (-5 - 3)^2)`

= `sqrt((-7)^2 + (-4)^2)`

= `sqrt(1 + 64)`

= `sqrt 65`    ...(vi)

In parallelogram PQRS,

PQ = QS    ...[From (v) and (vi)]

∴ ▢PQRS is a rectangle    ...(A parallelogram is a rectangle, if its diagonals are equal.)

P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5) are the vertices of a rectangle.