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Math  /  Algebra

Question5. Suppose a line has a slope of k5\frac{k}{5}. Find the value of kk so that each of the following is true: a. The line is parallel to a line with slope −12-\frac{1}{2} b. The line is perpendicular to a line with slope −12-\frac{1}{2}

Studdy Solution

STEP 1

1. We are given a line with slope k5 \frac{k}{5} .
2. We need to find the value of k k for two separate conditions: a. The line is parallel to another line with slope −12 -\frac{1}{2} . b. The line is perpendicular to another line with slope −12 -\frac{1}{2} .

STEP 2

1. Determine the value of k k for the line to be parallel.
2. Determine the value of k k for the line to be perpendicular.

STEP 3

To find k k when the line is parallel, set the slope of the line equal to the slope of the given line.
The slope of the line is k5 \frac{k}{5} and it must equal −12 -\frac{1}{2} .
k5=−12 \frac{k}{5} = -\frac{1}{2}

STEP 4

Solve for k k by multiplying both sides of the equation by 5.
k=−12×5 k = -\frac{1}{2} \times 5 k=−52 k = -\frac{5}{2}
The value of k k when the line is parallel is:
−52 \boxed{-\frac{5}{2}}

STEP 5

To find k k when the line is perpendicular, use the fact that the product of the slopes of two perpendicular lines is −1-1.
The slope of the line is k5 \frac{k}{5} and the slope of the given line is −12 -\frac{1}{2} .
(k5)×(−12)=−1 \left(\frac{k}{5}\right) \times \left(-\frac{1}{2}\right) = -1

STEP 6

Solve for k k by simplifying and isolating k k .
k5×−12=−1 \frac{k}{5} \times -\frac{1}{2} = -1 −k10=−1 \frac{-k}{10} = -1
Multiply both sides by −10-10 to solve for k k .
−k=−10 -k = -10 k=10 k = 10
The value of k k when the line is perpendicular is:
10 \boxed{10}

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