Question

Show that the points (0, –1), (8, 3), (6, 7) and (–2, 3) are vertices of a rectangle.

Answer 1

Let the points be P(0, –1), Q(8, 3), R(6, 7), S(–2, 3)

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

∴ By distance formula,

d(P, Q) = `sqrt((8 - 0)^2 + [3 - (-1)]^2`

= `sqrt((8 - 0)^2 + (3 + 1)^2`

= `sqrt(8^2 + 4^2)`

= `sqrt(64 + 16)`

= `sqrt(80)`   ...(i)

d(Q, R) = `sqrt((6 - 8)^2 + (7 - 3)^2`

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

= `sqrt(4 + 16)`

= `sqrt(20)`   ...(ii)

d(R, S) = `sqrt([(-2) - 6]^2 + (3 - 7)^2`

= `sqrt((-8)^2 + (-4)^2`

= `sqrt(64 + 16)`

=`sqrt(80)`   ...(iii)

d(P, S) = `sqrt([(-2) - 0]^2 + [3 - (-1)^2]`

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

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

= `sqrt(4 + 16)`

= `sqrt(20)`   ...(iv)

In ▢PQRS,

∴ side PQ = side RS   ...[From (i) and (iii)]

side QR = side PS   ...[From (ii) and (iv)]

∴ ▢PQRS is a parallelogram   ...[A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent]

d(P, R) = `sqrt((6 - 0)^2 + [7 - (-1)]^2`

= `sqrt((6 - 0)^2 + (7 + 1)^2`

= `sqrt(6^2 + 8^2)`

= `sqrt(36 + 64)`

= `sqrt(100)`

= 10   ...(iv)

d(Q, S) = `sqrt([(-2) - 8]^2 + [3 - 3]^2`

= `sqrt((-10)^2 + (0)^2`

= `sqrt(100 + 0)`

= `sqrt(100)`

= 10   ...(vi)

In parallelogram PQRS,

PR = QS   ...[From (v) and (vi)]

∴ ▢PQRS is a rectangle.   ...[A parallelogram is a rectangle if its diagonals are equal]