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Chapter 4: Problem 3

Find the equation of the line in the \(x y\) -plane that contains the point (3,2) and makes an angle of \(41^{\circ}\) with the positive \(x\) -axis.

Short Answer

Expert verified
The equation of the line in the xy-plane that contains the point (3, 2) and makes an angle of \(41^{\circ}\) with the positive x-axis is \(y \approx 0.8693x - 0.6079\).

Step by step solution

01

Find the slope of the line using the given angle.

In order to find the slope, we can use the tangent of the angle. The slope (m) of the line that makes an angle of \(41^{\circ}\) with the positive x-axis is: \(m = \tan(41^{\circ})\) Using a calculator, we get: \(m \approx 0.8693\)
02

Use the point-slope form of the line equation.

The point-slope form of the equation of a line is given by: \(y - y_1 = m(x - x_1)\) where (x_1, y_1) is a point on the line and m is the slope of the line. In this case, we have the point (3, 2) and a slope of approximately 0.8693. Plug in the given values: \(y - 2 = 0.8693(x - 3)\)
03

Write the equation of the line in slope-intercept form.

To write the equation in slope-intercept form (y = mx + b), we can solve for y: \begin{align*} y - 2 &= 0.8693(x - 3) \\ y - 2 &= 0.8693x - 0.8693 \times 3 \\ y - 2 &= 0.8693x - 2.6079 \\ y &= 0.8693x - 0.6079 \end{align*} So the equation of the line in the xy-plane that contains the point (3, 2) and makes an angle of \(41^{\circ}\) with the positive x-axis is: \(y \approx 0.8693x - 0.6079\)

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