If one line has slope $m$, what is the slope of a line perpendicular to it? Think about taking the negative reciprocal (flip the fraction and change the sign).

Apply the slope formula $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ to the points $(a+1, 2a-3)$ and $(4a-5, -6)$. Be careful with parentheses when subtracting expressions that contain $a$.

Once you have an expression for the slope of line L and a numerical value for the perpendicular slope, set them equal to each other and solve the resulting equation for $a$ using cross-multiplication.

In Desmos, use x in place of $a$. Enter the equation y = (-2x - 3)/(3x - 6) to represent the slope of line L as x changes.

On a new line, type y = 2/5 to draw a horizontal line showing all points where the slope would be $\frac{2}{5}$.

Zoom or pan until you see where the two graphs intersect, then tap the intersection point. The x-coordinate of this intersection is the value of $a$ that makes the slope of line L equal to $\frac{2}{5}$; this x-value should match the correct answer choice.

Rewrite the equation of the given line in slope-intercept form (solve for y):

The slope of this line is $-\frac{5}{2}$.

If two lines are perpendicular, their slopes are negative reciprocals: their product is $-1$.

Here the original line has slope $-\frac{5}{2}$, so the slope of any line perpendicular to it must be

So line L must have slope $\frac{2}{5}$.

Line L passes through $(a+1,\;2a-3)$ and $(4a-5,\;-6)$. Use the slope formula $m=\dfrac{y_2-y_1}{x_2-x_1}$:

Simplify the numerator and denominator:

So the slope of line L is

Because line L is perpendicular to the given line, its slope must be $\frac{2}{5}$. Set the expression for $m_L$ equal to $\frac{2}{5}$:

Cross-multiply:

Simplify both sides:

Move all terms involving $a$ to one side and constants to the other:

Divide both sides by $-16$:

So the correct answer is $-\dfrac{3}{16}$, which corresponds to choice C.