To determine the type of triangle and the lengths of its sides, we first need to find the distances between the vertices.
1. Finding the length of [tex]\(AB\)[/tex]:
The formula for the distance between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]
For points [tex]\(A (-2, 5)\)[/tex] and [tex]\(B (-4, -2)\)[/tex]:
[tex]\[
AB = \sqrt{((-4) - (-2))^2 + ((-2) - 5)^2} = \sqrt{(-2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53} \approx 7.2801
\][/tex]
2. Finding the length of [tex]\(AC\)[/tex]:
For points [tex]\(A (-2, 5)\)[/tex] and [tex]\(C (3, -4)\)[/tex]:
[tex]\[
AC = \sqrt{((3) - (-2))^2 + ((-4) - 5)^2} = \sqrt{(5)^2 + (-9)^2} = \sqrt{25 + 81} = \sqrt{106} \approx 10.2956
\][/tex]
3. Finding the length of [tex]\(BC\)[/tex]:
For points [tex]\(B (-4, -2)\)[/tex] and [tex]\(C (3, -4)\)[/tex]:
[tex]\[
BC = \sqrt{((3) - (-4))^2 + ((-4) - (-2))^2} = \sqrt{(7)^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \approx 7.2801
\][/tex]
Given these lengths:
- Length of [tex]\( AB \approx 7.2801 \)[/tex]
- Length of [tex]\( AC \approx 10.2956 \)[/tex]
- Length of [tex]\( BC \approx 7.2801 \)[/tex]
Next, we determine the type of triangle.
- If all three sides are equal, the triangle is equilateral.
- If exactly two sides are equal, the triangle is isosceles.
- If no sides are equal, the triangle is scalene.
Here, [tex]\( AB \approx 7.2801 \)[/tex] and [tex]\( BC \approx 7.2801 \)[/tex], while [tex]\( AC \approx 10.2956 \)[/tex]. Since two sides are approximately equal, the triangle is isosceles.
Therefore:
- The length of [tex]\(AB\)[/tex] is [tex]\( 7.2801 \)[/tex]
- The length of [tex]\(AC\)[/tex] is [tex]\(10.2956\)[/tex]
- The length of [tex]\(BC\)[/tex] is [tex]\( 7.2801 \)[/tex]
- The triangle is isosceles
