Question

What is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)?

Answer 1

Answer: 6,8 and 10

Explanation:

To find the length , all we need to find is the distance between each point ,

the formula for calculating distance between two points is given by :

D =
`\sqrt{(x_(2)-x_(1)) ^(2)+(y_(2)-y_(1)) ^(2)}`

Let the points be :

A ( -5,-1)

B(-5,5)

C(3,-1)

Calculating the length AB , we have

D1 =
`\sqrt{(x_(2)-x_(1)) ^(2)+(y_(2)-y_(1)) ^(2)}`

D1 =
`\sqrt{(-5+5)^(2)+(5+1)^(2)}`

D1 =
`√(36)`

D1= 6

Calculating the length AC , we have

D2 =
`\sqrt{(x_(2)-x_(1)) ^(2)+(y_(2)-y_(1)) ^(2)}`

D2 =
`\sqrt{(3+5)^(2)+(-1+1)^(2)}`

D2 =
`√(64)`

D2 = 8

Calculating the length BC , we have

D3 =
`\sqrt{(x_(2)-x_(1)) ^(2)+(y_(2)-y_(1)) ^(2)}`

D3 =
`\sqrt{(3+5)^(2)+(-1-5)^(2)}`

D3 =
`√(100)`

D3 = 10

Therefore ,the length of the sides of the triangle are 6,8 and 10