2017 AMC 10B Problem 10

There are three unknowns, $a$, $b$ and $c$, and the problem gives three facts: the lines are perpendicular, and the point $(1, -5)$ is on each line. Turn each fact into an equation.

1. Slope of the first line. Solve $ax - 2y = c$ for $y$: $y = \frac{a}{2}x - \frac{c}{2}$. The slope is $\frac{a}{2}$.
2. Slope of the second line. Solve $2x + by = -c$ for $y$: $y = -\frac{2}{b}x - \frac{c}{b}$. The slope is $-\frac{2}{b}$.
3. Perpendicular. Perpendicular lines have slopes that multiply to $-1$:
$$\frac{a}{2} \cdot \left(-\frac{2}{b}\right) = -\frac{a}{b} = -1,$$
so $a = b$.
4. The point is on the first line. Put $x = 1$, $y = -5$: $a + 10 = c$.
5. The point is on the second line. Put $x = 1$, $y = -5$: $2 - 5b = -c$, so $c = 5b - 2$.
6. Since $a = b$, the last two equations say $a + 10 = 5a - 2$. So $4a = 12$ and $a = 3$.
7. Then $c = 3 + 10 = 13$.
8. Check: the lines are $3x - 2y = 13$ and $2x + 3y = -13$. Both contain $(1, -5)$: $3 + 10 = 13$ and $2 - 15 = -13$. Their slopes $\frac32$ and $-\frac23$ multiply to $-1$.

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

Count the unknowns and count the facts. Three unknowns need three equations, and here each sentence of the problem gives one. Write every fact as an equation before solving; the perpendicular condition is the simplest one and cuts the work down right away.

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 $\boxed{\textbf{(E)}\ 13}$.