Question

What is the length of the shortest side of a triangle that has vertices at (-3,-2), (1,6), and (5,3)?

Answer 1

5 units

Explanation:

Answer 2

The length of shortest side is
`5\ units`

Explanation:

Let

`A(-3,-2),B(1,6),C(5,3)`

we know that

the formula to calculate the distance between two points is equal to

`d=\sqrt{(y2-y1)^(2)+(x2-x1)^(2)}`

step 1

Find the distance AB

`A(-3,-2),B(1,6)`

substitute in the formula

`AB=\sqrt{(6+2)^(2)+(1+3)^(2)}`

`AB=\sqrt{(8)^(2)+(4)^(2)}`

`AB=√(80)=8.94\ units`

step 2

Find the distance BC

`B(1,6),C(5,3)`

substitute in the formula

`BC=\sqrt{(3-6)^(2)+(5-1)^(2)}`

`BC=\sqrt{(-3)^(2)+(4)^(2)}`

`BC=√(25)=5\ units`

step 3

Find the distance AC

`A(-3,-2),C(5,3)`

substitute in the formula

`AC=\sqrt{(3+2)^(2)+(5+3)^(2)}`

`AC=\sqrt{(5)^(2)+(8)^(2)}`

`AC=√(89)=9.43\ units`

Compare the length sides

The length of shortest side is
`5\ units`