The area of the triangle, formed by the straight lines $y = 0, 12x - 5y = 0,$ and $3x + 4y = 7$ is

- Line 1: $y = 0$ (the x-axis) - Line 2: $12x - 5y = 0$ or $y = \dfrac{12x}{5}$ - Line 3: $3x + 4y = 7$ or $y = \dfrac{7-3x}{4}$ Find intersection points: Point A (lines 1 and 2): - When $y = 0$: $12x = 0 \implies x = 0$ So point A is $(0,0)$ Point B (lines 1 and 3): - When $y = 0$: $3x = 7 \implies x = \frac{7}{3}$ So point B is $(\frac{7}{3},0)$ - Point C (lines 2 and 3): $\frac{12x}{5} = \frac{7-3x}{4}$ $\frac{48x}{5} = 7-3x$ $48x + 15x = 35$ $63x = 35$ $x = \dfrac{35}{63} = \dfrac{5}{9}$ Substitute into line 2: $y = \dfrac{12 \cdot \frac{5}{9}}{5} = \dfrac{4}{3}$ So point C is $(\frac{5}{9}, \frac{4}{3})$ Calculate the area using the formula for triangle area given vertices: Area $=\dfrac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$ $=\dfrac{1}{2}|0 \cdot (0-\dfrac{4}{3}) + \dfrac{7}{3} \cdot \dfrac{4}{3} + \dfrac{5}{9} \cdot 0|$ $=\dfrac{1}{2} \cdot \dfrac{7 \cdot 4}{3 \cdot 3}$ $=\dfrac{1}{2} \cdot \dfrac{28}{9}$ $=\dfrac{14}{9}$ square units Alternatively, using $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$: Base along x-axis $=\frac{7}{3}$ Height to point C $=\frac{4}{3}$ Area $=\dfrac{1}{2} \cdot \dfrac{7}{3} \cdot \dfrac{4}{3} = \dfrac{14}{9}$ square units