Quadrilaterals - Verify that the shape with vertices at P(-5,2), Q(-1,3), R(-2,-1) and S(-6,-2) is a rhombus.

Question

Verify that the shape with vertices at P(-5,2), Q(-1,3), R(-2,-1) and S(-6,-2) is a rhombus.

Answer 1

A rhombus has four equal sides. So we can prove this is a rhombus by finding the lengths of the sides. We will use the distance formula to solve this.

d = √(x₂-x₁)² + (y₂-y₁)²

Step 1: Find the distance between P & Q. Consider (x₁, y₁) to be P( -5, 2) and (x₂, y₂) to be Q(-1, 3).

d = √(x₂-x₁)² + (y₂-y₁)²

d = √(-1-(-5))² + (3-2)²

d = √(4)² + (1)²

d = √16 + 1

d = √17

Step 2: Find the distance between Q & R. Consider (x₁, y₁) to be Q(-1, 3) and (x₂, y₂) to be R(-2, -1).

d = √(x₂-x₁)² + (y₂-y₁)²

d = √(-2-(-1))² + (-1-(3))²

d = √(-1)² + (-4)²

d = √1 + 16

d = √17

Step 3: Find the distance between R & S. Consider (x₁, y₁) to be R(-2, -1) and (x₂, y₂) to be S(-6, -2).

d = √(x₂-x₁)² + (y₂-y₁)²

d = √(-6-(-2))² + (-2-(-1))²

d = √(-4)² + (-1)²

d = √16 + 1

d = √17

Step 4: Find the distance between S & P. Consider (x₁, y₁) to be S(-6, -2) and (x₂, y₂) to be P( -5, 2).

d = √(x₂-x₁)² + (y₂-y₁)²

d = √(-5-(-6))² + (2-(-2))²

d = √(1)² + (4)²

d = √1 + 16

d = √17

Since all of the side lengths are √17, this is a rhombus.