Precalculus by Richard Wright

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7-01 Lines

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.6.7

Conic Sections
Figure 1: credit (wikimedia/Seahen)

These graphs can all be generated from the same general equation. They are called conic sections and can be used in applications ranging as widely as the orbit of planets to the shape of the curve in a satellite dish.

Conic Sections

Conic sections are the intersections of a plane with a double cone.

  1. Parabola: The plane intersects the cone parallel to a side of the cone. This will only intersect one of the double cones.
  2. Ellipse: The plane intersects a single cone such that it goes completely through the cone. A circle is a special case of an ellipse where the plane is perpendicular to the axis.
  3. Hyperbola: The plane intersects both cones such, but does not go through the apex of the cones.
  4. Degenerate conic sections
    1. Point: The plane intersects only at the apex of the cones.
    2. Single Line: The plane intersects the cones tangent to the surface of the cone.
    3. Intersecting Lines: The plane intersects the cones in such a way that the axis of the cones is on the plane.

The point and lines are called degenerate conic sections because they do not produce curves. This unit deals mainly with the three major conic sections, but starts out by looking at the degenerate line.

Lines

Lines are described by the general equation Ax + By + C = 0 or, more commonly, by slope-intercept form y = mx + b where m is the slope and b is the y-intercept.

Inclination

The inclination describes the steepness of a line. It is the angle the line makes with the positive x-axis.

Figure 2: Inclinations of two lines.

Since both slope and inclination describe the steepness of a line, they must be related. If a right angle is drawn on the line as in figure 3, the sides are x and y.

Figure 3

$$m=\frac{rise}{run}=\frac{y}{x}$$

Also,

$$\tan θ = \frac{opposite}{adjacent} = \frac{y}{x}$$

Since they both equal \(\frac{y}{x}\), set the slope and tangent equal to each other. This can be used to calculate the inclination of the line.

tan θ = m

where 0° < θ < 180°.

Remember that inverse tangent gives an angle between −90° and 90°. If you get a negative number when you solve for the inclination, add 180° because tangent has a period of 180°.

Inclination of a Line

tan θ = m

where 0° < θ < 180° is the inclination and m = slope.

Example 1: Find the Inclination of a Line

Find the inclination of 4x − 2y + 5 = 0.

Solution

First, find the slope by rewriting the equation in slope-intercept form.

$$4x-2y+5=0$$

$$-2y=-4x-5$$

$$y=2x+\frac{5}{2}$$

So the slope is 2.

Now, find the inclination.

$$\tan θ = 2$$

$$θ = \tan^{-1} 2 ≈ 63.4°$$

Figure 4
Try It 1

Find the inclination of x + 3y + 1 = 0.

Answer

161.6°

Angle between Two Lines

If two intersecting lines are drawn, an angle θ is formed at the intersection. That angle θ can be found using the inclinations of the lines.

Figure 5: Angle between two lines, θ.

In figure 5, \(β + θ_2 = 180°\) because they are a linear pair, so \(β = 180° - θ_2\). The sum of the three angles in a triangle is 180°, so

$$θ_1 + α + β = 180°$$

$$α = 180° - θ_1 - β$$

$$α = 180° - θ_1 - \left(180° - θ_2\right)$$

$$α = θ_2 - θ_1$$

α and θ are vertical angle, so they are congruent and the angle between two lines is

$$θ = θ_2 - θ_1$$

where 0° < θ < 90°.

This can be written with slopes. Take the tangent of both sides.

$$\tan θ = \tan \left(θ_2 - θ_1\right)$$

Use the difference formula for tangent.

$$\tan θ = \frac{\tan θ_2 - \tan θ_1}{1 + \tan θ_2 \tan θ_1}$$

Because the angles are inclinations, \(\tan θ_1 = m_1\) and \(\tan θ_2 = m_2\).

$$\tan θ = \left\lvert \frac{m_2 - m_1}{1 + m_1 m_2} \right\rvert$$

Angle between Two Lines

$$\tan θ = \left\lvert \frac{m_2 - m_1}{1 + m_1 m_2} \right\rvert$$

where 0° < θ < 90° is the angle between the lines and m1 and m2 are the slopes of the lines.

Example 2: Calculate the Angle between Two Lines

Find the angle between x − 2y = 0 and 3x + 2y − 1 = 0.

Solution

Begin by finding the slopes of the lines by rewriting them in slope-intercept form.

$$y = \frac{1}{2} x \text{ and } y = -\frac{3}{2} x + \frac{1}{2}$$

So the slopes are \(m_1 = \frac{1}{2}\) and \(m_2 = -\frac{3}{2}\).

Fill in the formula and find the angle.

$$\tan θ = \left\lvert \frac{m_2 - m_1}{1 + m_1 m_2} \right\rvert$$

$$\tan θ = \left\lvert \frac{-\frac{3}{2} - \frac{1}{2}}{1+\left(\frac{1}{2}\right)\left(-\frac{3}{2}\right)} \right\rvert$$

$$\tan θ = \left\lvert \frac{-4}{\frac{1}{4}} \right\rvert$$

$$\tan θ = 16$$

$$θ = 86.42°$$

Try It 2

Find the angle between 2x − 2y + 1 = 0 and x + y = 0.

Answer

90°

Distance from a Point to a Line

Figure 6

To find the distance from P to line Ax + By + C = 0, we can use vectors. Let \(P\left(x_1, y_1\right)\) be a point not on line Ax + By + C = 0. Let \(\overset{\rightharpoonup}{n} = \langle A, B \rangle\) be a perpendicular vector from point Q(x, y) on the line. The distance, d, between P and the line is the length of the orthogonal projection of \(\overset{\rightharpoonup}{QP}\) onto \(\overset{\rightharpoonup}{n}\).

$$proj_\overset{\rightharpoonup}{n} \overset{\rightharpoonup}{QP} = \frac{\overset{\rightharpoonup}{n} \cdot \overset{\rightharpoonup}{QP}}{\lVert \overset{\rightharpoonup}{n} \rVert}$$

This is slightly different than the formula from lesson 6-05. In this situation, we only want the length of the projection, so the direction unit vector \(\left(\frac{\overset{\rightharpoonup}{n}}{\lVert \overset{\rightharpoonup}{n} \rVert}\right)\) has not been included. Now, \(\overset{\rightharpoonup}{QP} = \langle x_1 - x, y_1 - y \rangle\), so

$$d = \left\lvert proj_\overset{\rightharpoonup}{n} \overset{\rightharpoonup}{QP} \right\rvert = \left\lvert \frac{A\left(x_1-x\right) + B\left(y_1-y\right)}{\sqrt{A^2 + B^2}} \right\rvert$$

$$=\frac{\lvert Ax_1 - Ax + By_1 - By \rvert}{\sqrt{A^2 + B^2}}$$

Because point \(Q\left(x, y\right)\) is on the line, Ax + By + C = 0, so C = −Ax − By.

$$d = \frac{\lvert Ax_1 + By_1 + C \rvert}{\sqrt{A^2 + B^2}}$$

Distance from a Point to a Line

$$d = \frac{\lvert Ax_1 + By_1 + C \rvert}{\sqrt{A^2 + B^2}}$$

where the point is \(P\left(x_1, y_1\right)\) and the line is Ax + By + C = 0.

Example 3: Find the Distance between a Point and a Line

Find the distance between the point (2, 5) and the line x + 3y − 2 = 0.

Solution

Fill in the distance formula with \(x_1 = 2\), \(y_1 = 5\), A = 1, B = 3, C = -2.

$$d = \frac{\lvert Ax_1 + By_1 + C \rvert}{\sqrt{A^2 + B^2}}$$

$$d = \frac{\lvert 1\left(2\right) + 3\left(5\right) + \left(-2\right) \rvert}{\sqrt{1^2 + 3^2}}$$

$$d = \frac{15}{\sqrt{10}} = \frac{3\sqrt{10}}{2}$$

Try It 3

Find the distance between the point (1, −2) and the line y = 2x − 1.

Answer

\(\frac{3\sqrt{5}}{5}\)

Lesson Summary
Inclination of a Line

tan θ = m

where 0° < θ < 180° is the inclination and m = slope.

Angle between Two Lines

$$\tan θ = \left\lvert \frac{m_2 - m_1}{1 + m_1 m_2} \right\rvert$$

where 0° < θ < 90° is the angle between the lines and m1 and m2 are the slopes of the lines.

Distance from a Point to a Line

$$d = \frac{\lvert Ax_1 + By_1 + C \rvert}{\sqrt{A^2 + B^2}}$$

where the point is \(P\left(x_1, y_1\right)\) and the line is Ax + By + C = 0.

Helpful videos about this lesson.

Practice Exercises

    Find the slope of the line with the given inclination.

  2. θ = 120°

    3. Find the inclination of the line that goes through the points.

  3. (−2, 1) and (0, 4)
  4. (3, 2) and (5, −2)

    5. Find the inclination of the line with the given equation.

  5. y = x + 3
  6. 2x − 3y + 5 = 0

    7. Find the angle between the two given lines.

  7. \(y = \frac{1}{2}x - 3\) and \(y = -\frac{2}{3}x + 1\)
  8. \(y = -x - \frac{1}{2}\) and \(y = \frac{3}{2}x + \frac{1}{2}\)
  9. 2x − y = 0 and x + y = 2
  10. x + 2y + 1 = 0 and −3x − 2y + 1 = 0

    11. Find the distance from the point to the line.

  11. x − y = 1 and (1, 3)
  12. 2x + y − 2 = 0 and (−2, 1)
  13. \(y = \frac{1}{3}x - 1\) and (0, 3)
  14. \(y = -\frac{2}{5}x + \frac{3}{5}\) and (−1, −2)

    15. Find the area of the triangle by (a) graphing the three vertices (b) find the equation of side AB (c) find the altitude to side AB and (d) calculate the area of the triangle.

  15. A(0, 2), B(1, 3), C(−2, 5)

    16. Find the equation of the line with the given inclination and x-intercept.

  16. θ = 30° and (2, 0)

    17. Mixed Review: Let \(m = 2\left(\cos \frac{π}{3} + i \sin \frac{π}{3}\right)\) and \(n = 4\left(\cos \frac{5π}{3} + i \sin \frac{5π}{3}\right)\).

  17. (6-07) Find m2.
  18. (6-07) Find \(\frac{n}{m}\).
  19. (6-06) Write m and n in standard form.
  20. (6-05) Find \(\langle 1, \sqrt{3}\rangle \cdot \langle 2, -2\sqrt{3}\rangle\)

Answers

  1. \(\sqrt{3}\)
  2. \(-\sqrt{3}\)
  3. 56.31º
  4. 116.57º
  5. 45º
  6. 33.69º
  7. 60.26º
  8. 78.69º
  9. 71.57º
  10. 29.74º
  11. \(\frac{3\sqrt{2}}{2}\)
  12. \(\sqrt{5}\)
  13. \(\frac{6\sqrt{10}}{5}\)
  14. \(\frac{15\sqrt{29}}{29}\)
  15. ; x – y + 2 = 0; \(\frac{5\sqrt{2}}{2}\); \(\frac{5}{2}\)
  16. \(\sqrt{3}x - 3y - 2\sqrt{3} = 0\)
  17. \(4\left(\cos \frac{2π}{3} + i \sin \frac{2π}{3}\right)\)
  18. \(2\left(\cos \frac{4π}{3} + i \sin \frac{4π}{3}\right)\)
  19. \(m = 1 + \sqrt{3}i\); \(n = 2 - 2\sqrt{3}i\)
  20. −4