Question

Difficulty: HardLinear Functions and Graphs

In the xyxy-plane, the graph of the linear function ff is perpendicular to the line with equation 3x+4y=483x + 4y = 48. The graph of ff intersects the xx-axis at the point (p,0)(p, 0) and the yy-axis at the point (0,q)(0, q), where q>0q > 0. If the distance between the points (p,0)(p, 0) and (0,q)(0, q) is 1515, what is the value of qq?

Answer: 12

Answer

12
First, find the slope of the given line by rewriting 3x+4y=483x + 4y = 48 in slope-intercept form: y=34x+12y = -\frac{3}{4}x + 12. The slope is 34-\frac{3}{4}. The graph of the linear function ff is perpendicular to this line, so its slope is the negative reciprocal, 43\frac{4}{3}. With a yy-intercept of (0,q)(0, q), the equation of ff is y=43x+qy = \frac{4}{3}x + q. Setting y=0y = 0 gives the xx-intercept (p,0)=(34q,0)(p, 0) = (-\frac{3}{4}q, 0). The distance between these intercepts is (34q)2+q2=2516q2=54q\sqrt{(-\frac{3}{4}q)^2 + q^2} = \sqrt{\frac{25}{16}q^2} = \frac{5}{4}q (since q>0q > 0). Given that the distance is 1515, we solve 54q=15\frac{5}{4}q = 15 to find q=12q = 12.

Step-by-Step Solution

1
Find the slope of the line 3x+4y=483x + 4y = 48.
The slope of the line is 34-\frac{3}{4}.
To find the slope of the perpendicular line ff, we first need the slope of the given line. Rewriting 3x+4y=483x + 4y = 48 in slope-intercept form gives y=34x+12y = -\frac{3}{4}x + 12.
2
Determine the slope of ff.
The slope of ff is 43\frac{4}{3}.
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Find the xx-intercept of ff in terms of qq.
The xx-intercept of ff is (34q,0)(-\frac{3}{4}q, 0).
Since the yy-intercept of ff is (0,q)(0, q), the equation of ff is y=43x+qy = \frac{4}{3}x + q. Setting y=0y = 0 gives 0=43x+q0 = \frac{4}{3}x + q, which simplifies to x=34qx = -\frac{3}{4}q.
4
Use the distance formula between the intercepts to solve for qq.
The value of qq is 1212.
The distance between (34q,0)(-\frac{3}{4}q, 0) and (0,q)(0, q) is (34q)2+q2=2516q2=54q\sqrt{(-\frac{3}{4}q)^2 + q^2} = \sqrt{\frac{25}{16}q^2} = \frac{5}{4}q since q>0q > 0. Setting this distance to 1515 gives 54q=15\frac{5}{4}q = 15, which yields q=12q = 12.

Key Concept

Properties of perpendicular lines, finding intercepts, and utilizing the distance formula in coordinate geometry.
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