Question

Difficulty: HardParallel and Perpendicular Lines

In the standard (x,y)(x, y) coordinate plane, a line L1L_1 is perpendicular to the line that contains the points (3,5)(3, 5) and (1,8)(-1, 8). If L1L_1 is also parallel to the line defined by the equation ax+6y=15ax + 6y = 15, what is the value of the constant aa?

Answer: -8

Answer

The value of the constant aa is 8-8.
The slope of the line containing (3,5)(3, 5) and (1,8)(-1, 8) is 34-\frac{3}{4}. The slope of a line perpendicular to it is the negative reciprocal, which is 43\frac{4}{3}. Because line L1L_1 is parallel to the line ax+6y=15ax + 6y = 15, they must have equal slopes. The slope of ax+6y=15ax + 6y = 15 is a6-\frac{a}{6}. Setting the two slopes equal gives a6=43-\frac{a}{6} = \frac{4}{3}, which yields a=8a = -8.

Step-by-Step Solution

1
Calculate the slope of the line containing the points (3,5)(3, 5) and (1,8)(-1, 8).
The slope is 34-\frac{3}{4}.
Using the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} for the points (3,5)(3, 5) and (1,8)(-1, 8), we get m=8513=34=34m = \frac{8 - 5}{-1 - 3} = \frac{3}{-4} = -\frac{3}{4}.
2
Determine the slope of line L1L_1 using the perpendicular relationship.
The slope of L1L_1 is 43\frac{4}{3}.
Since line L1L_1 is perpendicular to the line with slope 34-\frac{3}{4}, its slope must be the negative reciprocal, which is 13/4=43-\frac{1}{-3/4} = \frac{4}{3}.
3
Express the slope of the line ax+6y=15ax + 6y = 15 in terms of aa.
The slope is a6-\frac{a}{6}.
Rewriting the equation ax+6y=15ax + 6y = 15 in slope-intercept form (y=mx+by = mx + b) gives 6y=ax+156y = -ax + 15, which simplifies to y=a6x+52y = -\frac{a}{6}x + \frac{5}{2}. The slope is the coefficient of xx, which is a6-\frac{a}{6}.
4
Set the slope of L1L_1 equal to the slope of the parallel line to solve for aa.
a=8a = -8
Because line L1L_1 is parallel to the line ax+6y=15ax + 6y = 15, their slopes are equal: a6=43-\frac{a}{6} = \frac{4}{3}. Multiplying both sides by 6-6 gives a=8a = -8.

Key Concept

Parallel lines have equal slopes, and perpendicular lines have slopes that are negative reciprocals of each other.
Estimated Time:1m 30s
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