In the xy-plane, the graph of ax + by = c, where a, b, and c... | SAT Linear equations in two variables Question, Answer Explained

Question bank / Linear equations in two variables

Math · AlgebraHard

In the $xy$ -plane, the graph of $ax + by = c$ , where $a$ , $b$ , and $c$ are nonzero constants, is a line with a positive slope and a negative $y$ -intercept. Which of the following must be true?

$ab > 0$

$bc > 0$

$ac > 0$

$abc > 0$

Show answer and explanation

Answer: C

Solving for $y$ gives $y = -\frac{a}{b}x + \frac{c}{b}$ . A positive slope means $-\frac{a}{b} > 0$ , so $a$ and $b$ have opposite signs. A negative $y$ -intercept means $\frac{c}{b} < 0$ , so $c$ and $b$ have opposite signs. Then $a$ and $c$ both have the sign opposite to that of $b$ , so they have the same sign and $ac > 0$ .

Choice A comes from reading the slope as $\frac{a}{b}$ . The slope is $-\frac{a}{b}$ , so $a$ and $b$ must have opposite signs.

Why the other choices are wrong

A. The slope is -a/b. For it to be positive, a and b must have opposite signs, so ab < 0.
B. The y-intercept is c/b. For it to be negative, b and c must have opposite signs, so bc < 0.
D. The sign of abc depends on the sign of b, which isn't fixed. For example, a = 1, b = -1, and c = 1 give abc < 0.

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