Lesson Objectives
• Learn how to find the inclination of a line
• Learn how to find the angle between two lines
• Learn how to find the distance between a point and a line

## How to Find the Inclination of a Line

### Definition of the Angle of Inclination of a Line

Let's now learn about a few trigonometric concepts that involve lines. Let's begin with a line in slope-intercept form: $$y=mx + b$$ Where m is the slope and (0, b) is the y-intercept.
Every nonhorizontal line must intersect the x-axis. The angle formed by the intersection determines what is known as the inclination of the line. The inclination of a nonhorizontal line is the positive angle θ measured counterclockwise from the x-axis to the line. The value of θ will always be less than 180°.

### Horizontal Line: θ = 0 For any horizontal line, the slope is zero.

### Vertical Line: θ = 90° For any vertical line, the slope is undefined.

### Positively-Sloped Line: 0° < θ < 90° For any positively-sloped line, θ will be greater than 0° and less than 90° (Acute Angle).

### Negatively-Sloped Line: 90° < θ < 180° For any negatively-sloped line, θ will be greater than 90° and less than 180° (Obtuse Angle).

### Finding the Inclination θ Given Slope m

Let's consider a line with a positive slope, where (x1, 0) will be the point that intersects the x-axis and (x2, y2) will be a second point on the line. As we can see from our image above, the tangent of θ is equal to m: $$m=\frac{y_{2}}{x_{2}- x_{1}}$$ $$\text{tan}\hspace{.1em}θ=\frac{y_{2}}{x_{2}- x_{1}}$$ If a nonvertical line has inclination θ and slope m, then: $$m=\text{tan}\hspace{.1em}θ$$ $$θ=\text{tan}^{-1}(m)$$ When our slope is negative, we will need to take the extra step of adding our result to 180°. This will give us θ in quadrant II.
Let's look at an example.
Example #1: Find the inclination of the line. $$4x + 5y=20$$ Let's begin by placing the line in slope-intercept form. $$y=-\frac{4}{5}x + 4$$ Now, let's use our formula: $$θ=\text{tan}^{-1}\left(-\frac{4}{5}\right) ≈ -38.66°$$ Let's add 180° to our result to get our angle θ in quadrant II: $$θ ≈ 180° - 38.66°$$ $$θ ≈ 141.34°$$ Example #2: Find the inclination of the line. $$-2x + 3y=6$$ Let's begin by placing the line in slope-intercept form. $$y=\frac{2}{3}x + 2$$ Now, let's use our formula: $$θ=\text{tan}^{-1}\left(\frac{2}{3}\right) ≈ 33.69°$$ In some cases, we may be asked to find the slope of the line with a given inclination θ. Let's look at an example.
Example #3: Find the equation of the line given the inclination θ and a point on the line. $$θ=135°$$ $$\text{Point}: (-7, 2)$$ First, let's consider: $$\text{tan}\hspace{.1em}θ=m$$ $$\text{tan}(135°)=-1$$ This tells us that our slope, m is -1.
Let's now use point-slope form to find the equation of our line: $$y - y_{1}=m(x - x_{1})$$ Plug in for x1 and y1 $$y - 2=-1(x - (-7))$$ Solve for y to obtain slope-intercept form: $$y - 2=-x - 7$$ $$y=-x - 5$$ ### How to Find the Angle Between Two Lines

When we have two distinct lines in a plane, they will either intersect or be parallel and never intersect. When they intersect and are nonperpendicular, their intersection forms two pairs of opposite angles. One pair will be acute, while the other will be obtuse. We consider the acute angle to be the angle between the two lines. To find our formula, we will start with the inclinations of the two lines. Let's suppose that our two lines have inclinations: $$θ_{1}, θ_{2}$$ When the two lines meet the following conditions: $$θ_{1}< θ_{2}$$ and $$θ_{2}- θ_{1}< 90°$$ Then the angle between the two lines is given by: $$θ=θ_{2}- θ_{1}$$ From here, we can use the difference identity for tangent to set up a formula. Let's take the tangent of each side: $$\text{tan}\hspace{.1em}θ=\text{tan}(θ_{2}- θ_{1})$$ $$\text{tan}\hspace{.1em}θ=\frac{\text{tan}\hspace{.1em}θ_{2}- \text{tan}\hspace{.1em}θ_{1}}{1 + \text{tan}\hspace{.1em}θ_{1}\hspace{.1em}\text{tan}\hspace{.1em}θ_{2}}$$ Earlier we stated the tangent of θ is equal to m. If we have two nonperpendicular lines with slopes m1 and m2, then the angle between the two lines is: $$\text{tan}\hspace{.1em}θ=\left|\frac{m_{2}- m_{1}}{1 + m_{1}\hspace{.1em}m_{2}}\right|$$ Since we are using the absolute value bars, the order in which we label the slopes does not change the answer. Let's look at an example.
Example #4: Find the angle between the two lines. $$3x - y=1$$ $$x - 3y=-6$$ Let's place each line in slope-intercept form: $$y=3x - 1$$ $$y=\frac{1}{3}x + 2$$ Let's grab our two slopes: $$m_{1}=3$$ $$m_{2}=\frac{1}{3}$$ Let's plug into the formula: $$\text{tan}\hspace{.1em}θ=\left|\frac{m_{2}- m_{1}}{1 + m_{1}\hspace{.1em}m_{2}}\right|$$ $$\text{tan}\hspace{.1em}θ=\left|\frac{\frac{1}{3}- 3}{1 + 3\cdot \frac{1}{3}}\right|$$ $$\text{tan}\hspace{.1em}θ=\left|\frac{-\frac{8}{3}}{2}\right|$$ $$\text{tan}\hspace{.1em}θ=\left|-\frac{8}{3}\cdot \frac{1}{2}\right|$$ $$\text{tan}\hspace{.1em}θ=\left|-\frac{4}{3}\right|$$ $$\text{tan}\hspace{.1em}θ=\frac{4}{3}$$ Let's use the inverse tangent function: $$θ=\text{tan}^{-1}\left(\frac{4}{3}\right)$$ $$θ ≈ 53.13°$$ Example #5: Find the angle between the two lines. $$3x + 4y=12$$ $$x - 4y=8$$ Let's place each line in slope-intercept form: $$y=-\frac{3}{4}x + 3$$ $$y=\frac{1}{4}x - 2$$ Let's grab our two slopes: $$m_{1}=-\frac{3}{4}$$ $$m_{2}=\frac{1}{4}$$ Let's plug into the formula: $$\text{tan}\hspace{.1em}θ=\left|\frac{m_{2}- m_{1}}{1 + m_{1}\hspace{.1em}m_{2}}\right|$$ $$\text{tan}\hspace{.1em}θ=\large{\left|\frac{\frac{1}{4}+ \frac{3}{4}}{1 - \frac{3}{4}\cdot \frac{1}{4}}\right|}$$ $$\text{tan}\hspace{.1em}θ=\large{\left|\frac{1}{1 - \frac{3}{16}}\right|}$$ $$\text{tan}\hspace{.1em}θ=\large{\left|\frac{1}{\frac{13}{16}}\right|}$$ $$\text{tan}\hspace{.1em}θ=\left|\frac{16}{13}\right|$$ $$\text{tan}\hspace{.1em}θ=\frac{16}{13}$$ Let's use the inverse tangent function: $$θ=\text{tan}^{-1}\left(\frac{16}{13}\right)$$ $$θ ≈ 50.91°$$ ### The Distance Between a Point and Line

In some cases, we may need to find the distance between a line and a point that is not on the line. This distance will be the length of the perpendicular line segment joining the point and the line.
Note: We will derive the following formula in the practice test section. A full step-by-step is provided in the practice test solution video.
The distance between the point (x1, y1) and the line ax + by + c = 0 is given by: $$d=\frac{|ax_{1}+ by_{1}+ c|}{\sqrt{a^{2}+ b^{2}}}$$ Let's look at an example.
Example #6: Find the distance between the point and line given. $$(5, 3)$$ $$y=3x + 4$$ Let's first convert the line into the form shown: $$ax + by + c=0$$ $$y=3x + 4$$ $$3x - y + 4=0$$ From here, we can plug into our formula: $$a=3, b=-1, c=4$$ $$x_{1}=5, y_{1}=3$$ $$d=\frac{|ax_{1}+ by_{1}+ c|}{\sqrt{a^{2}+ b^{2}}}$$ $$d=\frac{|3(5) + -1(3) + 4|}{\sqrt{3^{2}+ (-1)^2}}$$ $$d=\frac{|15 - 3 + 4|}{\sqrt{10}}$$ $$d=\frac{16}{\sqrt{10}}$$ Rationalize the denominator: $$d=\frac{16}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}}$$ $$d=\frac{16 \cdot \sqrt{10}}{10}$$ $$d=\frac{8\sqrt{10}}{5}$$ How could we achieve this result without the formula? First let's graph our line and point on the coordinate plane. The shortest distance from the point to the line is the perpendicular line segment joining the point and the line. We know that perpendicular lines have slopes whose product is -1. $$y=3x + 4$$ The perpendicular line would have a slope of -1/3. Additionally, we know that (5, 3) is on the line. This allows us to use the point-slope form of a line: $$y - y_{1}=m(x - x_{1})$$ $$y - 3=-\frac{1}{3}(x - 5)$$ $$y=-\frac{1}{3}x + \frac{14}{3}$$ Let's graph the two lines on the coordinate plane. From here, we need to find the point at which these two lines intersect. We can easily do this by solving a system of equations. $$1) \hspace{.1em}y=3x + 4$$ $$2) \hspace{.1em}y=-\frac{1}{3}x + \frac{14}{3}$$ We will plug (3x + 4) in for y in our second equation: $$3x + 4=-\frac{1}{3}x + \frac{14}{3}$$ $$3x + \frac{1}{3}x=-4 + \frac{14}{3}$$ $$\frac{9}{3}x + \frac{1}{3}x=-\frac{12}{3}+ \frac{14}{3}$$ $$\frac{10}{3}x=\frac{2}{3}$$ $$x=\frac{2}{3}\cdot \frac{3}{10}$$ $$x=\frac{1}{5}$$ Plug back in for x in either original equation. $$y=3\left(\frac{1}{5}\right) + 4$$ $$y=\frac{3}{5}+ \frac{20}{5}$$ $$y=\frac{23}{5}$$ Now we can find the distance between this point and (5, 3) using the distance formula. $$d=\sqrt{\left(5 - \frac{1}{5}\right)^2 + \left(3 - \frac{23}{5}\right)^2}$$ $$d=\sqrt{\left(\frac{25}{5}- \frac{1}{5}\right)^2 + \left(\frac{15}{5}- \frac{23}{5}\right)^2}$$ $$d=\sqrt{\left(\frac{24}{5}\right)^2 + \left(-\frac{8}{5}\right)^2}$$ $$d=\sqrt{\frac{576}{25}+ \frac{64}{25}}$$ $$d=\sqrt{\frac{640}{25}}$$ $$d=\sqrt{\frac{128}{5}}$$ $$d=\frac{\sqrt{128}}{\sqrt{5}}$$ $$d=\frac{8\sqrt{2}}{\sqrt{5}}$$ Rationalize the denominator: $$d=\frac{8\sqrt{2}}{\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}$$ $$d=\frac{8\sqrt{10}}{5}$$ Example #7: Find the distance between the point and line given. $$(4, 1)$$ $$2x - y=-1$$ Let's first convert the line into the form shown: $$ax + by + c=0$$ $$2x - y=-1$$ $$2x - y + 1=0$$ From here, we can plug into our formula: $$a=2, b=-1, c=1$$ $$x_{1}=4, y_{1}=1$$ $$d=\frac{|ax_{1}+ by_{1}+ c|}{\sqrt{a^{2}+ b^{2}}}$$ $$d=\frac{|2(4) + -1(1) + 1|}{\sqrt{2^{2}+ (-1)^2}}$$ $$d=\frac{|8 - 1 + 1|}{\sqrt{5}}$$ $$d=\frac{8}{\sqrt{5}}$$ Rationalize the denominator: $$d=\frac{8}{\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}$$ $$d=\frac{8 \sqrt{5}}{5}$$

#### Skills Check:

Example #1

Find the inclination of the line θ $$x - 6y=-42$$

A
$$θ ≈ 17.51°$$
B
$$θ ≈ 39.37°$$
C
$$θ ≈ 170.54°$$
D
$$θ ≈ 18.23°$$
E
$$θ ≈ 9.46°$$

Example #2

Find the acute angle θ between the two lines. $$11x + 4y=12$$ $$x + 8y=-32$$

A
$$θ ≈ 55.71°$$
B
$$θ ≈ 33.95°$$
C
$$θ ≈ 62.89°$$
D
$$θ ≈ 81.07°$$
E
$$θ ≈ 67.92°$$

Example #3

Find the shortest distance between the line and the point given. $$2x - y=10$$ $$(5, 2)$$

A
$$\text{d}=\frac{\sqrt{2}}{5}$$
B
$$\text{d}=\frac{2\sqrt{5}}{5}$$
C
$$\text{d}=\frac{2\sqrt{2}}{5}$$
D
$$\text{d}=\frac{\sqrt{5}}{2}$$
E
$$\text{d}=\frac{2\sqrt{13}}{5}$$         