Once you know the slope of $4x + 2y = 10$, remember that a line perpendicular to it must have a slope that is the negative reciprocal of that slope.

After you find the perpendicular slope, plug in the point $(6,-5)$ using point-slope form $y - y_1 = m(x - x_1)$, then rearrange to get $y =$ (slope)$x$ + (intercept).

Make sure your final equation both has the correct perpendicular slope and gives $y = -5$ when $x = 6$.

Type 4x + 2y = 10 into Desmos (or solve for $y$ as $y = -2x + 5$ and enter that). Look at the graph and note that the slope is $-2$ (it goes down 2 units for every 1 unit to the right).

Type (6, -5) into Desmos to plot the point that line $r$ must go through.

Enter each answer choice equation into Desmos (one at a time or all together). For each line, check two things: (1) Does it pass exactly through the point $(6,-5)$? (2) Does its slope appear to be the negative reciprocal of $-2$ (that is, rising where the original line falls)? The equation that satisfies both conditions is the correct one.

Rewrite $4x + 2y = 10$ in slope-intercept form $y = mx + b$.

Subtract $4x$ from both sides:

Divide everything by $2$:

So the slope of this line is $-2$.

Perpendicular lines have slopes that are negative reciprocals. That means if one slope is $m$, the perpendicular slope is $-\dfrac{1}{m}$.

Here the original slope is $-2$, so the perpendicular slope is:

So line $r$ must have slope $\frac{1}{2}$.

We know line $r$ has slope $\frac{1}{2}$ and passes through $(6,-5)$.

Use the point-slope formula $y - y_1 = m(x - x_1)$ with $(x_1,y_1) = (6,-5)$ and $m = \frac{1}{2}$:

This simplifies to:

Distribute the $\frac{1}{2}$ on the right side:

Subtract $5$ from both sides:

So the equation of line $r$ is $y = \dfrac{1}{2}x - 8$, which corresponds to choice D.