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(C) Let the intercept form of the line be $\frac{x}{a} + \frac{y}{b} = 1$. The intercepts are $(a, 0)$ and $(0, b)$.The point $(-5, 3)$ divides the li

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$9x - 20y + 105 = 0$

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(C) Let the intercept form of the line be $\frac{x}{a} + \frac{y}{b} = 1$. The intercepts are $(a, 0)$ and $(0, b)$.The point $(-5, 3)$ divides the line segment joining $(a, 0)$ and $(0, b)$ in the ratio $4:3$.Using the section formula, the coordinates of the point are $\left( \frac{4(0) + 3(a)}{4+3}, \frac{4(b) + 3(0)}{4+3} \right) = \left( \frac{3a}{7}, \frac{4b}{7} \right)$.Equating these to $(-5, 3)$:$\frac{3a}{7} = -5 \implies a = -\frac{35}{3}$$\frac{4b}{7} = 3 \implies b = \frac{21}{4}$Substituting $a$ and $b$ into the intercept form: $\frac{x}{-35/3} + \frac{y}{21/4} = 1$.$-\frac{3x}{35} + \frac{4y}{21} = 1$.Multiplying by $105$ (the $LCM$ of $35$ and $21$): $-9x + 20y = 105$.$9x - 20y + 105 = 0$.

The equation of the straight line passing through the point $(-5, 3)$ such that the portion of the line intercepted between the axes is divided by the point in the ratio $4:3$ (from the $X$-axis to the $Y$-axis) is...

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