The points `(8,2)`, `(5,-3)` and `(0,0)` are the vertices of a/an

Options

A. Isosceles Triangle
B. Equilateral Triangle
C. Right Angled Triangle
D. None

Answer

To determine if the points (8, 2), (5, -3), and (0, 0) are the vertices of an isosceles triangle, we need to check two conditions:

1. The points must be non-collinear.
2. At least two sides of the triangle formed by these points must be of equal length.

### Step-by-Step Solution:

**Step 1: Assign the points**
Let:
- A = (8, 2)
- B = (5, -3)
- C = (0, 0)

**Step 2: Check for collinearity using slopes**
To check if the points are collinear, we will calculate the slopes of the lines AB, BC, and AC.

- **Slope of AB (m_AB)**:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 2}{5 - 8} = \frac{-5}{-3} = \frac{5}{3} \]

- **Slope of BC (m_BC)**:
\[ m_{BC} = \frac{y_3 - y_2}{x_3 - x_2} = \frac{0 - (-3)}{0 - 5} = \frac{3}{-5} = -\frac{3}{5} \]

- **Slope of AC (m_AC)**:
\[ m_{AC} = \frac{y_3 - y_1}{x_3 - x_1} = \frac{0 - 2}{0 - 8} = \frac{-2}{-8} = \frac{1}{4} \]

Since \( m_{AB} \neq m_{BC} \neq m_{AC} \), the points are non-collinear.

**Step 3: Calculate the lengths of the sides**
Now, we will calculate the lengths of the sides AB, BC, and AC using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

- **Length of AB (d_AB)**:
\[ d_{AB} = \sqrt{(5 - 8)^2 + (-3 - 2)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \]

- **Length of BC (d_BC)**:
\[ d_{BC} = \sqrt{(0 - 5)^2 + (0 - (-3))^2} = \sqrt{(-5)^2 + (3)^2} = \sqrt{25 + 9} = \sqrt{34} \]

- **Length of AC (d_AC)**:
\[ d_{AC} = \sqrt{(0 - 8)^2 + (0 - 2)^2} = \sqrt{(-8)^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68} \]

**Step 4: Check the lengths for equality**
We found:
- \( d_{AB} = \sqrt{34} \)
- \( d_{BC} = \sqrt{34} \)
- \( d_{AC} = \sqrt{68} \)

Since \( d_{AB} = d_{BC} \), we have two sides of equal length.

### Conclusion:
The points (8, 2), (5, -3), and (0, 0) are the vertices of an isosceles triangle.

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