Question

Show that the quadrilateral ‘HOPE’ with vertices H(–2, 1), О(–1, 2), P(0, 1) and E(–1, 0) is a rhombus. Is it a square? Justify.

Answer 1

Given:

H(–2, 1), O(–1, 2), P(0, 1), E(–1, 0).

We use the distance formula to find side and diagonal lengths.

Step-wise calculation:

1. Side lengths distance formula `d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`

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

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

= `sqrt(2)`

`OP = sqrt((0 - (-1))^2 + (1 - 2)^2)`

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

= `sqrt(2)`

`PE = sqrt((-1 - 0)^2 + (0 - 1)^2)`

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

= `sqrt(2)`

`EH = sqrt((-2 - (-1))^2 + (1 - 0)^2)`

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

= `sqrt(2)`

All four sides are equal each = `sqrt(2)`.

2. Diagonals

HP: Between H(–2, 1) and P(0, 1):

`HP = sqrt((0 - (-2))^2 + (1 - 1)^2)`

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

= `sqrt(4)`

= 2

OE: Between O(–1, 2) and E(–1, 0):

`OE = sqrt((-1 - (-1))^2 + (0 - 2)^2)`

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

= `sqrt(4)`

= 2

The two diagonals are equal each = 2.

3. Right angle

Slope of HO = `(2 - 1)/(-1 - (-2))`

= `1/1`

= 1

Slope of OP = `(1 - 2)/(0 - (-1))`

= `(-1)/1`

= –1

Product of slopes = 1 × (–1) = –1

⇒ HO ⟂ OP, so angle at O is 90°.

Since all four sides are equal, HOPE is a rhombus.

Because adjacent sides are perpendicular there is a right angle equivalently, because the diagonals of this rhombus are equal HOPE is a square.