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Chapter 6: Problem 28

Find the inclination \(\theta\) (in radians and degrees) of the line passing through the points. $$(3, \sqrt{3}),(6,-2 \sqrt{3})$$

Short Answer

Expert verified
The inclination \(\theta\) of the given line is -1.25 radians or -71.57 degrees.

Step by step solution

01

Determine the slope

Use the formula for the slope of a line given two points, (x1,y1) and (x2,y2). The formula is \((y2-y1)/(x2-x1)\). Substituting the given points into this formula, \(((-2 \sqrt{3})-(\sqrt{3}))/((6)-(3))\), we find that the slope, m, equals -3.
02

Find the inclination in radians

To find the inclination \(\theta\) in radians of the line, we use the formula \(\theta=arctan(m)\), where m is the slope. Substituting the slope into the formula, \(\theta=arctan(-3)\), we get \(\theta=-1.25 \, rad\). Remember that the range of arctan is \(-\pi/2\) to \(\pi/2\), and the negative sign indicates the direction of inclination.
03

Convert radians to degrees

The conversion from radians to degrees involves multiplication by the conversion factor \(\frac{180}{\pi}\). Therefore, the inclination \(\theta\) in degrees is \(-1.25 \cdot \frac{180}{\pi} = -71.57^\circ \)

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Most popular questions from this chapter

In Exercises \(91-116\), convert the polar equation to rectangular form. $$r=\frac{6}{2 \cos \theta-3 \sin \theta}$$

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

True or False? Determine whether the statement is true or false. Justify your answer. The conic represented by the following equation is an ellipse. \(r^{2}=\frac{16}{9-4 \cos \left(\theta+\frac{\pi}{4}\right)}\)

Determine whether the statement is true or false. Justify your answer. It is possible for a parabola to intersect its directrix.

In Exercises \(71-90,\) convert the rectangular equation to polar form. Assume \(a > 0\). $$x^{2}+y^{2}=9 a^{2}$$
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