Question

Triangle ABC
𝐴
𝐵
𝐶
has vertices point A(2, 1)
𝐴
(
2
,

1
)
, point B(−2, 2)
𝐵
(
−
2
,

2
)
, and point C(−3,−1)
𝐶

Answer 1

To analyze triangle ABC with vertices A(2, 1), B(−2, 2), and C(−3,−1), we can perform several calculations, such as determining the lengths of the sides, the perimeter, the area, and the type of triangle.

### Step 1: Calculate the lengths of the sides

We will use the distance formula, which is given by:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

#### Length of Side AB

\[
AB = \sqrt{(−2 - 2)^2 + (2 - 1)^2}
\]
\[
= \sqrt{(-4)^2 + (1)^2} = \sqrt{16 + 1} = \sqrt{17}
\]

#### Length of Side BC

\[
BC = \sqrt{(−3 - (−2))^2 + (−1 - 2)^2}
\]
\[
= \sqrt{(-1)^2 + (−3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]

#### Length of Side CA

\[
CA = \sqrt{(2 - (−3))^2 + (1 - (−1))^2}
\]
\[
= \sqrt{(5)^2 + (2)^2} = \sqrt{25 + 4} = \sqrt{29}
\]

### Step 2: Calculate the perimeter

\[
\text{Perimeter} = AB + BC + CA
\]
\[
= \sqrt{17} + \sqrt{10} + \sqrt{29}
\]

### Step 3: Calculate the area using the formula

The area of triangle ABC can be calculated using the determinant method:

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|
\]
where \( (x_1, y_1) = A(2, 1) \), \( (x_2, y_2) = B(-2, 2) \), \( (x_3, y_3) = C(-3, -1) \).

Substituting the coordinates:

\[
\text{Area} = \frac{1}{2} \left| 2(2 - (-1)) + (-2)(-1 - 1) + (-3)(1 - 2) \right|
\]
\[
= \frac{1}{2} \left| 2(3) + (-2)(-2) + (-3)(-1) \right|
\]
\[
= \frac{1}{2} \left| 6 + 4 + 3 \right| = \frac{1}{2} \left| 13 \right| = \frac{13}{2}
\]

### Summary

- Lengths of sides:
- \( AB = \sqrt{17} \)
- \( BC = \sqrt{10} \)
- \( CA = \sqrt{29} \)

- Perimeter: \( \sqrt{17} + \sqrt{10} + \sqrt{29} \)

- Area: \( \frac{13}{2} \)

If you need anything further regarding these calculations or more characteristics of triangle ABC, let me know!