Question

Triangle ABC, with vertices A(2,-8), B(5,-9), and C(4,-6), is drawn on the coordinate grid below. y 2 1 X -2 -1 1 N 4 5.6 7 8 9 10 -1 -2 -3 -4

Answer 1

First, we have to calculate the measures of the sides of the triangle.

Apply the distance formula:

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

Replace with the coordinate points given:

A= (2,-8)

B= (5,-9)

C= (4,-6)

Side AC

`d=\sqrt[]{(4-2)^2+(-6-(-8))^2}=\sqrt[]{2^2+2^2}=\sqrt[]{8}`

Side CB

`d=\sqrt[]{(5-4)^2+(-9-(-6))^2}=\sqrt[]{1+9}=\sqrt[]{10}`

Side AB

`d=\sqrt[]{(5-2)^2+(-9-(-8))}=\sqrt[]{9-1}=\sqrt[]{10}`

Apply Heron's formula:

`A\text{ =}\sqrt[]{s(s-a)(s-b)(s-c)}`

`s=(a+b+c)/(2)`

a, b, c are the distances ,CB , AC and AB

`s=\frac{\sqrt[]{10}+\sqrt[]{10}+\sqrt[]{8}}{2}=\frac{2\sqrt[]{10}+\sqrt[]{8}}{2}=4.58`