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Chapter 2: Problem 42

Find a number \(t\) such that the line in the \(x y\) -plane containing the points \((-3, t)\) and (4,3) is perpendicular to the line \(y=-5 x+999\).

Short Answer

Expert verified
\(t = \frac{28}{5}\)

Step by step solution

01

Find the slope of the line through the points (-3, t) and (4,3)

First, we need to use the slope formula, which is given by: \(m = \frac{y_2 - y_1}{x_2 - x_1}\) In this case, we have the points (-3, t) and (4,3). Plugging these into the formula, we get: \(m = \frac{3 - t}{4 - (-3)}\)
02

Find the slope of the given line y = -5x + 999

The slope of the line y = -5x + 999 is -5, since that is the coefficient of the x-term.
03

Use the fact that the product of the slopes of perpendicular lines is -1 to create an equation

Since the lines are perpendicular, the product of their slopes should be -1. So, we have: \(-5 \times \frac{3 - t}{4 - (-3)} = -1\)
04

Solve the equation for t

First, we will simplify the equation: \(-5 \times \frac{3 - t}{7} = -1\) We can now cross-multiply: \(-35 + 5t = -7\) Now, we will solve for t: \(5t = 28\) \(t = \frac{28}{5}\) So, the value of t that makes the line through the points (-3, t) and (4,3) perpendicular to the given line y = -5x + 999 is \(t = \frac{28}{5}\).

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