Q.Find the equation of the straight line whose $x$-intercept is $4$ and $y$-intercept is $-3$.

The two axis-intercepts are given directly, so the intercept form $\dfrac{x}{a}+\dfrac{y}{b}=1$ applies immediately with $a=4$, $b=-3$.

Using $\dfrac{x}{4}+\dfrac{y}{-3}=1$, multiplying through by $12$: $3x-4y=12$, so $3x-4y-12=0$.

The intercept form of a line with $x$-intercept $a$ and $y$-intercept $b$ is:

Here $a=4$ and $b=-3$, so:

Multiplying every term by $12$ (the LCM of $4$ and $3$) to clear denominators:

Independent check. Setting $y=0$ in the derived equation: $3x-12=0 \Rightarrow x=4$, matching the given $x$-intercept of $4$ ✓. Setting $x=0$: $-4y-12=0 \Rightarrow y=-3$, matching the given $y$-intercept of $-3$ ✓. Both intercepts recovered exactly confirm the equation.

A common slip is writing $\dfrac{y}{-3}$ as $-\dfrac{y}{3}$ but then mishandling the sign when clearing denominators, producing $3x+4y=12$ instead of $3x-4y=12$ — it helps to substitute $y=0$ and $x=0$ back into the final equation immediately as a sign check, exactly as done above.