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

Step-by-step 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 = $\sqrt{36}$

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 = $\sqrt{64}$

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 = $\sqrt{100}$

D3 = 10

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