Right Triangle Slopes Be7144

1. **State the problem:** We need to show that $\triangle ABC$ with vertices $A(0,0)$, $B(3,2)$, and $C(-2,3)$ is a right triangle by finding the slopes of sides $AB$ and $AC$ and verifying that their product is $-1$. 2. **Formula for slope:** The slope $m$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ 3. **Calculate slope $m_1$ of $AB$:** $$m_1 = \frac{2 - 0}{3 - 0} = \frac{2}{3}$$ 4. **Calculate slope $m_2$ of $AC$:** $$m_2 = \frac{3 - 0}{-2 - 0} = \frac{3}{-2} = -\frac{3}{2}$$ 5. **Verify the product $m_1 m_2$:** $$m_1 m_2 = \frac{2}{3} \times \left(-\frac{3}{2}\right) = \frac{2 \times (-3)}{3 \times 2} = \frac{-6}{6} = -1$$ 6. **Interpretation:** Since the product of the slopes of $AB$ and $AC$ is $-1$, the lines $AB$ and $AC$ are perpendicular. 7. **Conclusion:** Because two sides of $\triangle ABC$ are perpendicular, $\triangle ABC$ is a right triangle with the right angle at vertex $A$. **Final numeric answers:** $m_1 = \frac{2}{3}$, $m_2 = -\frac{3}{2}$