For two perpendicular lines, how are their slopes related? If one line has slope $m$, what is the slope of a line perpendicular to it?

Once you know the slope of line $m$, plug the point $(6,-2)$ into the equation $y = mx + b$ to solve for $b$, the $y$-intercept.

After you find $b$ in $y = mx + b$, remember that $b$ is the $y$-intercept—the value of $y$ when $x=0$.

Type 4x+3y=12 into Desmos to graph line $\ell$. Then, separately, solve for $y$ by hand to get $y = -\tfrac{4}{3}x + 4$, and type y = (-4/3)x + 4 to confirm it overlaps the same line—this verifies the slope $-\tfrac{4}{3}$.

Use the negative reciprocal slope $\tfrac{3}{4}$. In Desmos, enter y + 2 = (3/4)(x - 6) (or equivalently y = (3/4)(x - 6) - 2) to graph line $m$ passing through $(6,-2)$ and perpendicular to $\ell$.

On the graph of line $m$, look at the point where it crosses the $y$-axis (where $x=0$). For a precise value, add another line in Desmos like m(x) = (3/4)(x - 6) - 2 and then enter m(0); the output is the $y$-intercept.

Start with the equation of line $\ell$:

Solve for $y$ to put it into slope-intercept form $y = mx + b$:

So the slope of line $\ell$ is $-\tfrac{4}{3}$.

If two lines are perpendicular, their slopes are negative reciprocals of each other. That means if one slope is $m$, the other is $-\tfrac{1}{m}$.

Line $\ell$ has slope $-\tfrac{4}{3}$, so the slope of line $m$ must be:

Thus, the slope of line $m$ is $\tfrac{3}{4}$.

Line $m$ has slope $\tfrac{3}{4}$ and passes through $(6,-2)$. Use point-slope form:

Substitute $m = \tfrac{3}{4}$ and $(x_1, y_1) = (6, -2)$:

This simplifies to:

Now expand the right side:

Reduce $\tfrac{18}{4}$ to $\tfrac{9}{2}$:

Solve for $y$ to get the equation in the form $y = mx + b$:

Write $2$ as $\tfrac{4}{2}$ so the fractions have a common denominator:

In $y = mx + b$, the $y$-intercept is $b$, so the $y$-intercept of line $m$ is $-\tfrac{13}{2}$.