The equation of the straight line passing thr | vedclass

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Question

(A) Let the intercepts on the $x$-axis and $y$-axis be $a$ and $b$ respectively. Given that $a + b = -1$,so $b = -1 - a$. The equation of the line in

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the intercepts on the $x$-axis and $y$-axis be $a$ and $b$ respectively. Given that $a + b = -1$,so $b = -1 - a$. The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$. Substituting $b$,we get $\frac{x}{a} + \frac{y}{-(1+a)} = 1$,which is $\frac{x}{a} - \frac{y}{1+a} = 1$. Since the line passes through $(4, 3)$,we have $\frac{4}{a} - \frac{3}{1+a} = 1$. Multiplying by $a(1+a)$,we get $4(1+a) - 3a = a(1+a)$. $4 + 4a - 3a = a + a^2$. $4 + a = a + a^2$,which simplifies to $a^2 = 4$,so $a = 2$ or $a = -2$. If $a = 2$,then $b = -1 - 2 = -3$. The equation is $\frac{x}{2} + \frac{y}{-3} = 1$,or $\frac{x}{2} - \frac{y}{3} = 1$. 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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