Question

Show that points P(2, –2), Q(7, 3), R(11, –1) and S (6, –6) are vertices of a parallelogram.

Answer 1

The given points are P(2, –2), Q(7, 3), R(11, –1) and S(6, –6). 

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

By distance formula,

PQ = `sqrt((7 - 2)^2 + [3 - ( - 2)]^2)`

∴ PQ = `sqrt((7 - 2)^2 + (3 + 2)^2)`

∴ PQ = `sqrt((5)^2 + (5)^2)`

∴ PQ = `sqrt(25 + 25)`

∴ PQ = `sqrt(50)`

∴ PQ = `sqrt(25 × 2)`

∴ PQ = `5sqrt(2)`        ...(1)

QR = `sqrt((11 - 7)^2 + (-1 - 3)^2)`

∴ QR = `sqrt((4)^2 + (-4)^2)`

∴ QR = `sqrt(16 + 16)`

∴ QR = `sqrt(32)`

∴ QR = `sqrt(16 × 2)`

∴ QR = `4sqrt(2)`      ...(2)

RS = `sqrt((6 - 11)^2 + [-6 - (- 1)]^2)`

∴ RS = `sqrt((- 5)^2 + (-6 + 1)^2)`

∴ RS = `sqrt((- 5)^2 + (-5)^2)`

∴ RS = `sqrt(25 + 25)`

∴ RS = `sqrt(50)`

∴ RS = `sqrt(25 × 2)`

∴ RS = `5sqrt(2)`        ...(3)

PS = `sqrt((6 - 2)^2 + [-6 - (- 2)]^2)`

∴ PS = `sqrt((6 - 2)^2 + [-6 + 2]^2)`

∴ PS = `sqrt((4)^2 + (-4)^2)`

∴ PS = `sqrt(16 + 16)`

∴ PS = `sqrt(32)`

∴ PS = `sqrt(16 × 2)`

∴ PS = `4sqrt(2)`      ...(4)

In □ PQRS,
PQ = RS       ...[From (1) and (3)]
QR = PS       ...[From (2) and (4)]

A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.

Checking for slopes,

Slope of a line between two points (x₁, y₁) and (x₂, y₂) is

m = `(y_2 - y_1)/(x_2 - x_1)`

Slope PQ = `(7 - 2)/[3 - (- 2)] = 1`

Slope QR = `(11 - 7)/[- 1 - 3] = - 1`

Slope RS = `(6 - 11)/[- 6 - (- 1)] = 1`

Slope SP = `(6 - 2)/[- 6 - (- 2)] = - 1`

As PQ = RS and their slope = 1 And QR = SP and their slope = -1.

∴ □ PQRS is parallelogram.

∴ P, Q, R, and S are vertices of a parallelogram.