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Difficulty: Very hardLinear Equations in Two Variables

In the xyxy-plane, the graph of the linear equation ax+by=cax + by = c, where aa, bb, and cc are non-zero constants, has a positive slope and a negative yy-intercept. If the line passes through the point (b,a)(b, a), which of the following inequalities must be true?

  1. A
    ab>0ab > 0
  2. ac>0ac > 0Answer
  3. C
    bc>0bc > 0
  4. D
    ac<0ac < 0

Answer

The inequality ac>0ac > 0 must be true.
The correct inequality is ac>0ac > 0. Rearranging ax+by=cax + by = c into slope-intercept form gives y=−abx+cby = -\frac{a}{b}x + \frac{c}{b}. A positive slope means −ab>0-\frac{a}{b} > 0, so ab<0\frac{a}{b} < 0, indicating that aa and bb have opposite signs (ab<0ab < 0). A negative yy-intercept means cb<0\frac{c}{b} < 0, indicating that bb and cc have opposite signs (bc<0bc < 0). Substituting the point (b,a)(b, a) into the line's equation gives 2ab=c2ab = c. Since ab<0ab < 0, it follows that c<0c < 0. Using the relation bc<0bc < 0 with c<0c < 0, we find b>0b > 0. Using ab<0ab < 0 with b>0b > 0, we find aa is negative. Since both aa and cc are negative, their product acac must be positive.

Step-by-Step Solution

1
Rearrange the standard form equation ax+by=cax + by = c into slope-intercept form.
y=−abx+cby = -\frac{a}{b}x + \frac{c}{b}
To identify the expressions for the slope and the yy-intercept in terms of the constants aa, bb, and cc.
2
Determine the sign relationships between the coefficients based on the given signs of the slope and yy-intercept.
Since the slope is positive, −ab>0  ⟹  ab<0-\frac{a}{b} > 0 \implies ab < 0. Since the yy-intercept is negative, cb<0  ⟹  bc<0\frac{c}{b} < 0 \implies bc < 0.
This establishes that aa and bb have opposite signs, and bb and cc have opposite signs.
3
Substitute the coordinates of the point (b,a)(b, a) into the original equation ax+by=cax + by = c.
a(b)+b(a)=c  ⟹  2ab=ca(b) + b(a) = c \implies 2ab = c
Since the point lies on the line, substituting its coordinates must satisfy the equation.
4
Analyze the sign of cc using the relation c=2abc = 2ab.
Since ab<0ab < 0, it follows that 2ab<02ab < 0, which means c<0c < 0.
To determine whether the constant cc is positive or negative.
5
Determine the individual signs of bb and aa.
Since bc<0bc < 0 and c<0c < 0, bb must be positive (b>0b > 0). Since ab<0ab < 0 and b>0b > 0, aa must be negative (a<0a < 0).
To find the sign of each coefficient individually.
6
Determine the sign of the product acac.
Since both a<0a < 0 and c<0c < 0, their product acac must be positive (ac>0ac > 0).
To evaluate which inequality must be true.

Key Concept

Analyzing slope, intercepts, and points on a linear equation in two variables using sign analysis of its constants.

Alternative Method

Instead of solving for the individual signs of all variables, we can choose test values that satisfy the conditions. For example, let a=−1a = -1 and b=2b = 2, which gives a positive slope of −ab=12-\frac{a}{b} = \frac{1}{2}. Since the line passes through (b,a)=(2,−1)(b, a) = (2, -1), we can find cc: ax+by=c  ⟹  (−1)(2)+(2)(−1)=−4ax + by = c \implies (-1)(2) + (2)(-1) = -4, so c=−4c = -4. The yy-intercept is cb=−42=−2\frac{c}{b} = \frac{-4}{2} = -2, which is negative, satisfying all conditions. Evaluating the product of aa and cc gives (−1)(−4)=4>0(-1)(-4) = 4 > 0, confirming that the product of these constants must be positive.
Estimated Time:3m 0s
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Topics

Linear Equations in Two Variables

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