To classify triangle \( \triangle JOY \) formed by the points \( J(3, -2) \), \( O(2, -5) \), and \( Y(4, -5) \), we need to calculate the lengths of its sides.
We will use the distance formula, which is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Step 1: Calculate lengths of each side.
- Length of JO:
\[ d_{JO} = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
- Length of OY:
\[ d_{OY} = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \]
- Length of YJ:
\[ d_{YJ} = \sqrt{(4 - 3)^2 + (-5 - (-2))^2} = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
Step 2: Compare the lengths.
- \( JO = \sqrt{10} \)
- \( OY = 2 \)
- \( YJ = \sqrt{10} \)
Step 3: Classify the triangle.
- Since two sides are equal (\( JO = YJ \)), it is an isosceles triangle.
- For further classification, we check if it is a right triangle by using the Pythagorean theorem. We need the squares of the lengths:
\[ d_{JO}^2 + d_{OY}^2 \text{ should equal } d_{YJ}^2. \]
Calculating:
\[ (\sqrt{10})^2 + 2^2 = 10 + 4 = 14 \] \[ (\sqrt{10})^2 = 10 \]
Since \( 10 \neq 14 \), it is not a right triangle.
Conclusion:
The classification of triangle \( \triangle JOY \) is:
It is an isosceles triangle.
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