Q.Find the equation of the straight line passing through the point $(2,-3)$ with slope $4$.

One point and the slope are given directly, so the point-slope form $y-y_1=m(x-x_1)$ applies immediately without needing any other conversion.

Using $y-y_1=m(x-x_1)$ with $(x_1,y_1)=(2,-3)$, $m=4$: $y+3=4(x-2)$, so $y+3=4x-8$, giving $4x-y-11=0$.

The point-slope form of a line through $(x_1,y_1)$ with slope $m$ is:

Here $(x_1,y_1)=(2,-3)$ and $m=4$. Substituting:

Rearranging to general form: $4x - y - 8 - 3 = 0$, i.e.

Independent check. First, substitute the given point back in: $4(2)-(-3)-11 = 8+3-11=0$ ✓, confirming the line does pass through $(2,-3)$. Second, recover the slope from the general form using $m=-a/b$ with $a=4,b=-1$: $m = -4/(-1) = 4$ ✓, matching the given slope exactly.

A common sign slip is writing $y-(-3)$ as $y-3$ instead of $y+3$ — always rewrite $y-y_1$ carefully when $y_1$ is negative. Another is forgetting to distribute the $4$ across both terms inside $4(x-2)$.