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(C) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$. Given that the sum of intercepts is $a + b = -1$, so $b = -1 -

Vedclass · Accepted Answer

$\frac{x}{-2} + \frac{y}{1} = 1$ and $\frac{x}{2} + \frac{y}{-3} = 1$

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(C) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$. Given that the sum of intercepts is $a + b = -1$, so $b = -1 - a$. The line passes through $(4, 3)$, so $\frac{4}{a} + \frac{3}{b} = 1$. Substituting $b = -(1 + a)$, we get $\frac{4}{a} - \frac{3}{1 + a} = 1$. $4(1 + a) - 3a = a(1 + a) \Rightarrow 4 + 4a - 3a = a + a^2$. $a^2 = 4 \Rightarrow a = 2$ or $a = -2$. If $a = 2$, then $b = -1 - 2 = -3$. The equation is $\frac{x}{2} + \frac{y}{-3} = 1$. If $a = -2$, then $b = -1 - (-2) = 1$. The equation is $\frac{x}{-2} + \frac{y}{1} = 1$.

The equation of the straight line passing through the point $(4, 3)$ and making intercepts on the coordinate axes whose sum is $-1$ is

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