Question

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

• A. $\frac{x}{2}-\frac{y}{4}=1$ and $\frac{x}{-2}+\frac{y}{3}=1$
• B. $\frac{x}{2}-\frac{y}{5}=1$ and $\frac{x}{-2}+\frac{y}{4}=1$
• C. $\frac{x}{2}+\frac{y}{3}=1$ and $\frac{x}{-1}+\frac{y}{1}=1$
• D. $\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$

Answer 1

The correct option is D $\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$

Let ${}^{\prime }{a}^{\prime }$ be the $x$ intercept of the line.
Hence y intercept will be $-1-a$.
Therefore the equation of the line in slope intercept form will be
$\frac{x}{a}+\frac{y}{-1-a}=1$
Since it passes through $(4,3)$
$\frac{4}{a}+\frac{3}{-1-a}=1$
$4(-1-a)+3\left(a\right)=(-1-a)a$
$-4-4a+3a=-a-a^2$
$-4-a=-a-a^2$
$a^2=4$
$a=\pm 2$
Hence the equations of the possible lines are
$\frac{x}{-2}+\frac{y}{1}=1$ and
$\frac{x}{2}+\frac{y}{-3}=1$