To determine the distance between the vertices and the type of the triangle based on its side lengths, let's follow these steps:
1. Calculate the length of [tex]\( AB \)[/tex]:
The coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are [tex]\( A (-2, 5) \)[/tex] and [tex]\( B (-4, -2) \)[/tex].
The formula to calculate the distance between two points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is:
[tex]\[
\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]
Plugging in the coordinates of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[
\text{Length of } AB = \sqrt{(-4 - (-2))^2 + (-2 - 5)^2} = \sqrt{(-2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53} \approx 7.28
\][/tex]
2. Calculate the length of [tex]\( AC \)[/tex]:
The coordinates of points [tex]\( A \)[/tex] and [tex]\( C \)[/tex] are [tex]\( A (-2, 5) \)[/tex] and [tex]\( C (3, -4) \)[/tex].
Using the distance formula again:
[tex]\[
\text{Length of } AC = \sqrt{(3 - (-2))^2 + (-4 - 5)^2} = \sqrt{(5)^2 + (-9)^2} = \sqrt{25 + 81} = \sqrt{106} \approx 10.30
\][/tex]
3. Calculate the length of [tex]\( BC \)[/tex]:
The coordinates of points [tex]\( B \)[/tex] and [tex]\( C \)[/tex] are [tex]\( B (-4, -2) \)[/tex] and [tex]\( C (3, -4) \)[/tex].
Using the distance formula again:
[tex]\[
\text{Length of } BC = \sqrt{(3 - (-4))^2 + (-4 - (-2))^2} = \sqrt{(7)^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \approx 7.28
\][/tex]
4. Determine the type of the triangle:
Now we compare the lengths of [tex]\( AB \)[/tex], [tex]\( AC \)[/tex], and [tex]\( BC \)[/tex]:
- Length of [tex]\( AB \approx 7.28 \)[/tex]
- Length of [tex]\( AC \approx 10.30 \)[/tex]
- Length of [tex]\( BC \approx 7.28 \)[/tex]
Since we have two sides of the same length ([tex]\( AB \)[/tex] and [tex]\( BC \)[/tex]) and one side of a different length ([tex]\( AC \)[/tex]), the triangle is an isosceles triangle.
Therefore, the correct answers are:
- The length of [tex]\( AB \)[/tex] is [tex]\( 7.28 \)[/tex]
- The length of [tex]\( AC \)[/tex] is [tex]\( 10.30 \)[/tex]
- The length of [tex]\( BC \)[/tex] is [tex]\( 7.28 \)[/tex]
- Therefore, the triangle is isosceles
