1. **State the problem:** We have a right triangle ABC with vertices A at the origin $(0,0)$, C on the x-axis, and B on the hypotenuse line $y = -0.5x + k$. The left leg is $y = 2x$, and the base is $y=0$ (x-axis). We want to find the coordinates of points B and C and the value of $k$ to solve the triangle. 2. **Identify points:** - Point A is at $(0,0)$. - Point C lies on the x-axis, so $y=0$. - Point B lies on the hypotenuse $y = -0.5x + k$ and also on the left leg $y=2x$ because B is on the hypotenuse between A and C. 3. **Find point C:** Since C is on the x-axis and on the hypotenuse line, set $y=0$ in $y = -0.5x + k$: $$0 = -0.5x + k \implies 0.5x = k \implies x = \frac{k}{0.5} = 2k$$ So, $C = (2k, 0)$. 4. **Find point B:** B lies on both $y=2x$ and $y = -0.5x + k$, so set them equal: $$2x = -0.5x + k$$ Add $0.5x$ to both sides: $$2x + 0.5x = k \implies 2.5x = k \implies x = \frac{k}{2.5} = 0.4k$$ Find $y$ using $y=2x$: $$y = 2 \times 0.4k = 0.8k$$ So, $B = (0.4k, 0.8k)$. 5. **Calculate the area of triangle ABC:** The base is segment AC along the x-axis from $(0,0)$ to $(2k,0)$, so base length is $2k$. The height is the vertical distance from B to the x-axis, which is $y$-coordinate of B, $0.8k$. Area formula: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2k \times 0.8k = 0.8k^2$$ 6. **Summary:** - $C = (2k, 0)$ - $B = (0.4k, 0.8k)$ - Area $= 0.8k^2$ Without additional information (like the area or length), $k$ remains a parameter. **Final answer:** $$B = \left(0.4k, 0.8k\right), \quad C = \left(2k, 0\right), \quad \text{Area} = 0.8k^2$$