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 $y = \dfrac{7-3x}{4}$

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Find intersection points:

Point A (lines 1 and 2):

• When $y = 0$:

$12x = 0 \implies x = 0$

So point A is $(0,0)$

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Point B (lines 1 and 3):

• When $y = 0$:

$3x = 7 \implies x = \frac{7}{3}$

So point B is $(\frac{7}{3},0)$

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• 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 $(\frac{5}{9}, \frac{4}{3})$

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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

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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