Find the perimeter of triangle ABC with vertices A(6, 3), B(6, - 2) , and C(- 4, 3) .

26.2 units

Step-by-step explanation:

We are given the points/vertices

A(6, 3),

B(6, - 2) , and

C(- 4, 3)

Step two

Let us find the distances between the given points/vertices

A-B =A(6, 3) to B(6,-2)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((6-6)²+(-2-3)²)

d=√(-2-3)²)

d=√(-5)²)

d=5 units

B-C=B(6, - 2) to C(-4, 3)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((-4-6)²+(3+2)²)

d=√(-10)²+(5)²)

d=√100+25

d=√125

d=11.2 units

C-A=C(-4, 3) to A(6, 3)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((6+4)²+(3-3)²)

d=√(10)²

d=√100

d=10 units

Hence the perimeter is 5+11.2+10

P=26.2 units

so, converting firstly the mixed fractions to improper fractions and then "adding".