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Difficulty: Very hardParallel and Perpendicular Lines

A line segment in the standard (x,y)(x, y) coordinate plane has endpoints at (1,a)(1, a) and (5,a2)(5, a^2). The perpendicular bisector of this segment is parallel to the line defined by the equation x+3y=6x + 3y = 6. What is the sum of all possible values of the constant aa?

Answer: 1

Answer

The sum of all possible values of the constant aa is 1.
To find the sum of all possible values of the constant aa, we first find the slope of the line x+3y=6x + 3y = 6 by writing it in slope-intercept form: y=−13x+2y = -\frac{1}{3}x + 2. Since the perpendicular bisector is parallel to this line, its slope is also −13-\frac{1}{3}. The line segment is perpendicular to its perpendicular bisector, so the slope of the line segment is the negative reciprocal of −13-\frac{1}{3}, which is 33. Setting the slope of the segment a2−a5−1\frac{a^2 - a}{5 - 1} equal to 33 gives the equation a2−a4=3\frac{a^2 - a}{4} = 3, which simplifies to a2−a−12=0a^2 - a - 12 = 0. Solving this quadratic equation yields (a−4)(a+3)=0(a - 4)(a + 3) = 0, giving the values a=4a = 4 and a=−3a = -3. The sum of these possible values is 4+(−3)=14 + (-3) = 1.

Step-by-Step Solution

1
Find the slope of the given line x+3y=6x + 3y = 6.
The slope is −13-\frac{1}{3}.
Rewriting the equation in slope-intercept form (y=mx+by = mx + b) gives y=−13x+2y = -\frac{1}{3}x + 2, showing the slope is −13-\frac{1}{3}.
2
Determine the slope of the line segment.
The slope is 3.
The line segment is perpendicular to its perpendicular bisector. Because the perpendicular bisector is parallel to the reference line, its slope is also −13-\frac{1}{3}. The line segment's slope is the negative reciprocal of −13-\frac{1}{3}, which is 33.
3
Write the slope of the segment in terms of aa and set it equal to 3.
a2−a4=3\frac{a^2 - a}{4} = 3
Using the slope formula with endpoints (1,a)(1, a) and (5,a2)(5, a^2) gives the expression a2−a5−1\frac{a^2 - a}{5 - 1}.
4
Solve the quadratic equation for aa.
a=4a = 4 or a=−3a = -3
Multiplying both sides by 4 yields a2−a=12a^2 - a = 12, which simplifies to the quadratic a2−a−12=0a^2 - a - 12 = 0. Factoring gives (a−4)(a+3)=0(a - 4)(a + 3) = 0.
5
Sum all possible values of aa.
1
The sum of the values is 4+(−3)=14 + (-3) = 1.

Key Concept

Understanding that parallel lines have equal slopes, perpendicular lines have slopes that are negative reciprocals of each other, and applying the slope formula to solve for coordinate variables.
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Topics

Parallel and Perpendicular Lines

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