Question

Find the distance, c, between (3, –4) and (–7, 1) on the coordinate plane. Round to the nearest tenth if necessary.

Answer 1

`\sqrt{(-7-3)^(2) + (1-(-4))^(2) } = \sqrt{(-10)^(2) + (5)^(2) } = √(100 + 25) \\`

`= √(125) = 5√(5) =11.2`

Answer 2

c ≈ 11.2

Explanation:

Using the distance formula, we have:

c = √[(x2 - x1)² + (y2 - y1)²]

c = √[(-7 - 3)² + (1 - (-4))²]

c = √[(-10)² + 5²]

c = √(100 + 25)

c = √125

c ≈ 11.2

Therefore, the distance between the two points is approximately 11.2 units.