1. **State the problem:** We have line L with equation $3x + 2y = 17$, point A at $(0,2)$, and line M perpendicular to L passing through A. We need to find the area of triangle ABC where B is the y-intercept of L and C is the intersection of L and M. 2. **Find point B (y-intercept of L):** Set $x=0$ in $3x + 2y = 17$. $$3(0) + 2y = 17 \implies 2y = 17 \implies y = \frac{17}{2} = 8.5$$ So, $B = (0, 8.5)$. 3. **Find slope of L:** Rewrite $3x + 2y = 17$ as $y = mx + c$. $$2y = 17 - 3x \implies y = \frac{17}{2} - \frac{3}{2}x$$ Slope of L is $m_L = -\frac{3}{2}$. 4. **Find slope of M (perpendicular to L):** Slope of M is negative reciprocal of $m_L$. $$m_M = -\frac{1}{m_L} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3}$$ 5. **Equation of M passing through A(0,2):** $$y - 2 = \frac{2}{3}(x - 0) \implies y = \frac{2}{3}x + 2$$ 6. **Find point C (intersection of L and M):** Solve system: $$\begin{cases} y = \frac{17}{2} - \frac{3}{2}x \\ y = \frac{2}{3}x + 2 \end{cases}$$ Set equal: $$\frac{17}{2} - \frac{3}{2}x = \frac{2}{3}x + 2$$ Multiply both sides by 6 to clear denominators: $$6 \times \left(\frac{17}{2} - \frac{3}{2}x\right) = 6 \times \left(\frac{2}{3}x + 2\right)$$ $$3 \times 17 - 3 \times 3x = 4x + 12$$ $$51 - 9x = 4x + 12$$ Add $9x$ to both sides: $$51 = 13x + 12$$ Subtract 12: $$51 - 12 = 13x \implies 39 = 13x$$ Divide both sides by 13: $$x = \frac{\cancel{39}^{3}}{\cancel{13}^{1}} = 3$$ Find $y$: $$y = \frac{2}{3} \times 3 + 2 = 2 + 2 = 4$$ So, $C = (3,4)$. 7. **Calculate area of triangle ABC:** Points are $A(0,2)$, $B(0,8.5)$, $C(3,4)$. Use formula for area with coordinates: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute: $$= \frac{1}{2} |0(8.5 - 4) + 0(4 - 2) + 3(2 - 8.5)|$$ $$= \frac{1}{2} |0 + 0 + 3(-6.5)| = \frac{1}{2} |-19.5| = \frac{19.5}{2} = 9.75$$ **Final answer:** The area of triangle ABC is $9.75$ square units.