Once you know the slope of line $m$, think about how the slope of a perpendicular line is related to it (how do you get the negative reciprocal?).

Find which choices have the correct perpendicular slope, then use the fact that line $n$ passes through $(3,-1)$ to test those equations.

For any candidate equation, substitute $x=3$ and $y=-1$. If the equation is true, then the line passes through $(3,-1)$; if not, it doesn’t.

Type 4x-3y=15 into Desmos. This will draw line $m$.

Add the point (3,-1) as a separate expression. Make sure it lies on line $m$ so you can visually confirm the setup.

Enter each option as a separate line: 4x-3y=17, 4x+3y=7, 3x-4y=5, and 3x+4y=5. Look for the line that both passes through the point (3,-1) and forms a right angle (perpendicular) with line $m$.

We are told:

• Line $m$ goes through $(3,-1)$ and has equation $4x-3y=15$.
• Line $n$ is perpendicular to line $m$ and also passes through $(3,-1)$.

We need to find which choice could be the equation of line $n$.

Rewrite $4x-3y=15$ in slope-intercept form $y=mx+b$:

Subtract $4x$ from both sides:

Divide by $-3$:

So the slope of line $m$ is $\dfrac{4}{3}$.

For two lines to be perpendicular, their slopes must be negative reciprocals of each other. That means:

• If one slope is $\dfrac{4}{3}$,
• The perpendicular slope must be $-\dfrac{3}{4}$.

So line $n$ must have slope $-\dfrac{3}{4}$.

Each answer choice is in the form $Ax+By=C$. The slope of such a line is $-\dfrac{A}{B}$.

Compute the slopes:

• $4x-3y=17$: slope $=-\dfrac{4}{-3}=\dfrac{4}{3}$
• $4x+3y=7$: slope $=-\dfrac{4}{3}$
• $3x-4y=5$: slope $=-\dfrac{3}{-4}=\dfrac{3}{4}$
• $3x+4y=5$: slope $=-\dfrac{3}{4}$

Only the last equation has slope $-\dfrac{3}{4}$, so line $n$ must be represented by the last choice if it also passes through $(3,-1)$.

Now plug $(3,-1)$ into the remaining candidate with slope $-\dfrac{3}{4}$:

For $3x+4y=5$:

This is a true statement ($5=5$), so $(3,-1)$ lies on this line.

Therefore, the equation of line $n$ is $3x+4y=5$.