2017 AMC 10B Problem 10

Slopes first. The line $ax - 2y = c$ has slope $\frac{a}{2}$, and $2x + by = -c$ has slope $-\frac{2}{b}$. Perpendicular lines have slopes whose product is $-1$:
$$\frac{a}{2} \cdot \left(-\frac{2}{b}\right) = -\frac{a}{b} = -1 \quad\Longrightarrow\quad a = b.$$

Now the common point $(1, -5)$ lies on both lines:
$$a(1) - 2(-5) = c \Rightarrow c = a + 10, \qquad 2(1) + b(-5) = -c \Rightarrow c = 5b - 2.$$
With $a = b$, set $a + 10 = 5a - 2$, so $4a = 12$ and $a = 3$. Then $c = 3 + 10 = 13$.

Check: the lines are $3x - 2y = 13$ and $2x + 3y = -13$; both pass through $(1, -5)$ and their slopes $\frac32$ and $-\frac23$ multiply to $-1$.

The answer is $\boxed{\textbf{(E)}\ 13}$.

Three unknowns ($a$, $b$, $c$) need three equations, and the problem supplies exactly three facts: perpendicularity and two point-on-line conditions. Translate each fact into an equation before solving anything; the perpendicular condition happens to be the cleanest and reduces the system immediately.

• Using slope -2/a for the first line (forgetting to solve for y) or mishandling the sign when substituting y = -5.
• Solving $a + 10 = 5a - 2$ correctly but reporting $a = 3$ or $-13$ (the value of $-c$) instead of $c$.