To determine the correct answers, we need to calculate the lengths of the sides of triangle \( \triangle ABC \) using the distance formula. The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Let's calculate the distances:
1. Length of \( AB \)
The coordinates of points \( A \) and \( B \) are \((-2, 5)\) and \((-4, -2)\), respectively.
[tex]\[
AB = \sqrt{((-4) - (-2))^2 + ((-2) - 5)^2} = \sqrt{(-2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53} \approx 7.2801
\][/tex]
2. Length of \( AC \)
The coordinates of points \( A \) and \( C \) are \((-2, 5)\) and \((3, -4)\), respectively.
[tex]\[
AC = \sqrt{((3) - (-2))^2 + ((-4) - 5)^2} = \sqrt{(5)^2 + (-9)^2} = \sqrt{25 + 81} = \sqrt{106} \approx 10.2956
\][/tex]
3. Length of \( BC \)
The coordinates of points \( B \) and \( C \) are \((-4, -2)\) and \((3, -4)\), respectively.
[tex]\[
BC = \sqrt{((3) - (-4))^2 + ((-4) - (-2))^2} = \sqrt{(7)^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \approx 7.2801
\][/tex]
Having calculated the side lengths:
- \( AB \approx 7.2801 \)
- \( AC \approx 10.2956 \)
- \( BC \approx 7.2801 \)
To classify the triangle, we compare the side lengths:
- Since \( AB \approx 7.2801 \) and \( BC \approx 7.2801 \) and \( AC \approx 10.2956 \), we notice that only two sides are equal.
- This makes \( \triangle ABC \) an isosceles triangle.
So, the final answer is:
- The length of \( AB \) is \( \approx 7.2801 \)
- The length of \( AC \) is \( \approx 10.2956 \)
- The length of \( BC \) is \( \approx 7.2801 \)
- Therefore, the triangle is isosceles
