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Chapter 6: Problem 29

Find the inclination \(\theta\) (in radians and degrees) of the line passing through the points. $$(6,1),(10,8)$$

Short Answer

Expert verified
Then, the inclination of the line passing through the points (6,1) and (10,8) is \(\arctan(\frac{7}{4})\) radians or the equivalent in degrees.

Step by step solution

01

Identify the coordinates

The coordinates provided are (6,1) and (10,8). Hence, \(x_1 = 6\), \(y_1 = 1\), \(x_2 = 10\) and \(y_2 = 8\).
02

Calculate the rise and the run

The 'rise' is the change in y, and the 'run' is the change in x. Rise = \(y_2 - y_1 = 8 - 1 = 7\). Run = \(x_2 - x_1 = 10 - 6 = 4\).
03

Substitute these values into the formula for inclination

The inclination or slope of a line is equal to \(\theta = \arctan(\frac{y_2 - y_1}{x_2 - x_1})\). We substitute our values and get \(\theta = \arctan(\frac{7}{4})\).
04

Convert to radians and degrees

To get the value in radians, use a calculator to solve \(\theta = \arctan(\frac{7}{4})\). For degrees, remember that \(180掳 = \pi \, rad\), so multiply the radians by \(\frac{180}{\pi}\) to get the answer in degrees. Thus, \(\theta\)= angle in rad * \(\frac{180}{\pi}\).

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Most popular questions from this chapter

In Exercises \(91-116\), convert the polar equation to rectangular form. $$r=-2 \cos \theta$$

(a) Show that the distance between the points \(\left(r_{1}, \theta_{1}\right)\) and \(\left(r_{2}, \theta_{2}\right)\) is \(\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)}\) (b) Simplify the Distance Formula for \(\theta_{1}=\theta_{2} .\) Is the simplification what you expected? Explain. (c) Simplify the Distance Formula for \(\theta_{1}-\theta_{2}=90^{\circ}\) Is the simplification what you expected? Explain.

In Exercises \(117-126\), convert the polar equation to rectangular form. Then sketch its graph. $$\theta=\pi / 6$$

In Exercises \(91-116\), convert the polar equation to rectangular form. $$r=-5 \sin \theta$$

Consider a line with slope \(m\) and \(y\) -intercept \((0,4)\) (a) Write the distance \(d\) between the point \((3,1)\) and the line as a function of \(m\) (b) Graph the function in part (a). (c) Find the slope that yields the maximum distance between the point and the line. (d) Is it possible for the distance to be \(0 ?\) If so, what is the slope of the line that yields a distance of \(0 ?\) (e) Find the asymptote of the graph in part (b) and interpret its meaning in the context of the problem.
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