Question

Show that the quadrilateral ‘PLOT’ with vertices P(1, 1), L(–5, 1), O(–5, 4) and T(1, 4) is a rectangle. Is it a square? Justify.

Answer 1

To P.T. PLOT is a rectangle.

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

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

= `sqrt((1 + 5)^2 + 0)`

= `sqrt(6^2)`

= `sqrt(36)`

= 6

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

= `sqrt((-5 - (5))^2 + (4 - 1)^2`

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

= `sqrt(3^2)`

= `sqrt(9)`

= 3

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

= `sqrt((1 - (-5))^2 + (4 - 4)^2`

= `sqrt((1 + 5)^2 + 0^2)`

= `sqrt(6^2)`

= `sqrt(36)`

= 6

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

= `sqrt((1 - 1)^2 + (1 - 4)^2`

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

= `sqrt(9)`

= 3

∴ PL = OT = 6

OL = TP = 3

∴ Opposite sides are equal   ...(i)

Now, Diagonal PO = `sqrt((1 - (-5))^2 + (1 - 4)^2`

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

= `sqrt(6^2 + 9)`

= `sqrt(36 + 9)`

= `sqrt(45)`

Diagonal LT = `sqrt((-5 - 1)^2 + (1 - 4)^2`

= `sqrt((-6)^2 + (-3)^2`

= `sqrt(36 + 9)`

= `sqrt(45)`

∴ PO = LT = `sqrt(45)`

∴ Diagonals are equal.    ...(ii)

∴ From equations (i) and (ii),

PLOT is a rectangle.

PLOT is not a square as adjacent sides.

PL ≠ OL.

∴ It is not square.