Respuesta :
Using the formula for the distance between two points, it is found that:
- Since sides AB and BC have the same length, which is different of side AC, the triangle is an isosceles triangle.
- The area is of 38.63 units squared.
What is the distance between two points?
Suppose that we have two points, [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]. The distance between them is given by:
[tex]D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
In this problem, we use the formula to find the lengths for each side of the triangle.
For example, side AB has length given by the distance between (-6,3) and (3,5), hence:
[tex]AB = \sqrt{(-6 - 3)^2 + (3 - 5)^2} = 9.22[/tex]
Side AC has length given by:
[tex]AC = \sqrt{(-6 - 1)^2 + (3 - (-4))^2} = 9.9[/tex]
Side BC has length given by:
[tex]BC = \sqrt{(3 - 1)^2 + (5 - (-4))^2} = 9.22[/tex]
Since sides AB and BC have the same length, which is different of side AC, the triangle is an isosceles triangle.
What is the area of a triangle?
The area of a triangle is given by half the length of the base multiplied by the height.
The midpoint of side BC, of length of 9.22 units, is of (2,0.5). The height is the distance of this point from point A, hence:
[tex]d = \sqrt{(-6 - 2)^2 + (3 - 0.5)^2} = 8.38[/tex]
Hence the area is:
A = 0.5 x 9.22 x 8.38 = 38.63 units squared.
More can be learned about the distance between two points at https://brainly.com/question/18345417
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