Question

If anyone knows this can you please help i have about an hour left to submit this (:

Find the area of a triangle with the given vertices.

Part I: Graph the following points on the coordinate grid below.
(1, -3), (3, -1), (5, -3)

Part II: Find the area of the triangle. Show your work.

Answer 1

Part 1 : Figure show the graph of triangle

Part 2 : The area of triangle is 4 sqaure units

Explanation:

Given points A(1, -3), B(3, -1) and C(5, -3) make triangle.

Part 1:

Figure show the graph of triangle with A(1, -3), B(3, -1) and C(5, -3) as vertices.

Part 2: Find the area of the triangle.

The area of triangle is given by A=
`((Base)(height))/(2)`

From figure, Take base as length of AC

Length of line is given by L=
`\sqrt{(X1-X2)^(2)+(Y1-Y2)^(2) }`

Now, Base = length of AC

Base =
`\sqrt{(X1-X2)^(2)+(Y1-Y2)^(2) }`

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

=
`\sqrt{(-4)^(2)+(0)^(2)}`

=
`√(16)`

=4units

and Height as difference of y-component of point A and point B

Height = (y of component of point B)- (y of component of point A)

= (-1)- (-3)

= 2units

Therefore, The area of triangle is given by A=
`((Base)(height))/(2)`

A=
`((4)(2))/(2)`

A=4 sqaure units