Question

Show that points P(1, –2), Q(5, 2), R(3, –1), S(–1, –5) are the vertices of a parallelogram.

Answer 1

Let, P(1, −2) = (x₁, y₁)

Q(5, 2) = (x₂, y₂)

R(3, −1) = (x₃, y₃)

S(−1, −5) = (x₄, y₄)

By the distance formula,

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

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

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

= `sqrt(16 + 16)`

= `sqrt32`

= `sqrt(2 xx 2 xx 2 xx 2 xx 2)`

= `4sqrt2`    ...(1)

QR = `sqrt((x_3 - x_2)^2 + (y_3 - y_2)^2)`

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

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

= `sqrt(4 + 9)`

= `sqrt13`    ...(2)

RS = `sqrt((x_4 - x_3)^2 + (y_4 - y_3)^2)`

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

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

= `sqrt(16 + 16)`

= `sqrt32`

= `sqrt(2 xx 2 xx 2 xx 2 xx 2)`

= `4sqrt2`    ...(3)

PS = `sqrt((x_4 - x_1)^2 + (y_4 - y_1)^2)`

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

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

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

= `sqrt(4 + 9)`

= `sqrt13`    ...(4)

In `square`PQRS, 

PQ = RS    ...[From (1) and (3)]

QR = PS    ...[From (2) and (4)]

`square`PQRS is a parallelogram.    ...(A quadrilateral is a parallelogram if both pairs of its opposite sides are congruent.)

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

Hence proved.