To determine the lengths of the sides of triangle [tex]\( \triangle ABC \)[/tex] with vertices [tex]\( A(-2, 5) \)[/tex], [tex]\( B(-4, -2) \)[/tex], and [tex]\( C(3, -4) \)[/tex], and to classify the type of triangle, proceed with the following steps:
### 1. Calculate the Length of [tex]\( AB \)[/tex]
The distance formula 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]
Plugging in the coordinates for [tex]\( A(-2, 5) \)[/tex] and [tex]\( B(-4, -2) \)[/tex]:
[tex]\[ AB = \sqrt{((-4) - (-2))^2 + ((-2) - 5)^2} \][/tex]
[tex]\[ AB = \sqrt{(-2)^2 + (-7)^2} \][/tex]
[tex]\[ AB = \sqrt{4 + 49} \][/tex]
[tex]\[ AB = \sqrt{53} \][/tex]
[tex]\[ AB \approx 7.280 \][/tex]
### 2. Calculate the Length of [tex]\( AC \)[/tex]
Using the same distance formula, plug in the coordinates for [tex]\( A(-2, 5) \)[/tex] and [tex]\( C(3, -4) \)[/tex]:
[tex]\[ AC = \sqrt{((3) - (-2))^2 + ((-4) - 5)^2} \][/tex]
[tex]\[ AC = \sqrt{(5)^2 + (-9)^2} \][/tex]
[tex]\[ AC = \sqrt{25 + 81} \][/tex]
[tex]\[ AC = \sqrt{106} \][/tex]
[tex]\[ AC \approx 10.296 \][/tex]
### 3. Calculate the Length of [tex]\( BC \)[/tex]
Again, using the distance formula with the coordinates for [tex]\( B(-4, -2) \)[/tex] and [tex]\( C(3, -4) \)[/tex]:
[tex]\[ BC = \sqrt{((3) - (-4))^2 + ((-4) - (-2))^2} \][/tex]
[tex]\[ BC = \sqrt{(7)^2 + (-2)^2} \][/tex]
[tex]\[ BC = \sqrt{49 + 4} \][/tex]
[tex]\[ BC = \sqrt{53} \][/tex]
[tex]\[ BC \approx 7.280 \][/tex]
### 4. Classify the Triangle
Now, check the lengths of [tex]\( AB \)[/tex], [tex]\( AC \)[/tex], and [tex]\( BC \)[/tex]:
- [tex]\( AB \approx 7.280 \)[/tex]
- [tex]\( AC \approx 10.296 \)[/tex]
- [tex]\( BC \approx 7.280 \)[/tex]
Since two sides [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] are equal in length and different from the third side [tex]\( AC \)[/tex], the triangle is classified as an Isosceles Triangle.
### Final Answer:
Using the drop-down menus:
The length of [tex]\( AB \)[/tex] is [tex]\( \approx 7.280 \)[/tex]
The length of [tex]\( AC \)[/tex] is [tex]\( \approx 10.296 \)[/tex]
The length of [tex]\( BC \)[/tex] is [tex]\( \approx 7.280 \)[/tex]
Therefore, the triangle is Isosceles.
