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Chapter 2: Problem 39

Find the equation of the line in the \(x y\) -plane that contains the point (4,1) and that is perpendicular to the line \(y=3 x+5\).

Short Answer

Expert verified
The equation of the line containing the point (4,1) and perpendicular to the line \(y = 3x + 5\) is \(y = -\frac{1}{3}x + \frac{7}{3}\).

Step by step solution

01

Find the slope of the given line

The given line equation is \(y = 3x + 5\). We can see that the slope of this line is the coefficient of \(x\), which is 3.
02

Find the slope of the perpendicular line

Two lines are perpendicular if the product of their slopes is -1. Let's denote the slope of the perpendicular line as \(m\). Since the slope of the given line is 3, we have: \(m \times 3 = -1\) Solve for \(m\): \(m = -\frac{1}{3}\) So the slope of the perpendicular line is -1/3.
03

Use point-slope form to find the equation of the line

The point-slope form of a linear equation is given by: \(y - y_1 = m(x - x_1)\) In our case, the point \((x_1, y_1) = (4,1)\) and the slope \(m = -\frac{1}{3}\). Substitute the values into the equation: \(y - 1 = -\frac{1}{3}(x - 4)\)
04

Simplify the equation

Distribute the slope and simplify the equation: \(y - 1 = -\frac{1}{3}x + \frac{4}{3}\) Add 1 to both sides: \(y = -\frac{1}{3}x + \frac{4}{3} + 1\) To combine the constants, write 1 as \(\frac{3}{3}\): \(y = -\frac{1}{3}x + \frac{4}{3} + \frac{3}{3}\) Finally, add the constants: \(y = -\frac{1}{3}x + \frac{7}{3}\) So the equation of the line is \(y = -\frac{1}{3}x + \frac{7}{3}\).

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