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Straight-Line Equations

3.4 Analyze linear functions as straight-line equations and interpret their graphs in contextual problems.

4.4 Solve contextual problems related to linear functions as straight-line equations.

Straight-line equations

After studying this material, students are expected to be able to:

  1. analyze linear functions and their graphs,
  2. determine the slope of a line,
  3. determine the relationship between two lines from their slopes,
  4. form a straight-line equation from the given information.

Observe the following graph.

Line graph

For the equations y=2xy = 2x and y=2x+1y = 2x + 1, the coefficient of xx is 2. This value describes the comparison between the change in yy and the change in xx. It is called the slope or gradient of the line.

If the slope is denoted by mm, then the general form of a straight-line equation is:

y=mx+c y = mx + c

where:

  • mm = slope,
  • cc = constant.

Examples of straight-line equations:

  1. y=3x4y = 3x - 4
  2. 2x3y5=02x - 3y - 5 = 0

To draw the graph of y=mx+cy = mx + c:

  1. find the intercept with the xx-axis,
  2. find the intercept with the yy-axis,
  3. connect the two points.

Draw the graph of:

y=3x+6 y = -3x + 6

Find the xx-intercept by setting y=0y = 0:

0=3x+63x=6x=2 \begin{aligned} 0 &= -3x + 6 \\ 3x &= 6 \\ x &= 2 \end{aligned}

So the xx-intercept is (2,0)(2,0).

Find the yy-intercept by setting x=0x = 0:

y=3(0)+6y=6 \begin{aligned} y &= -3(0) + 6 \\ y &= 6 \end{aligned}

So the yy-intercept is (0,6)(0,6).

Example graph

The slope shows the degree of inclination of a line relative to the xx-axis and is usually denoted by mm.

Line through the origin

Formula:

m=y1x1 m = \frac{y_1}{x_1}

Intercept form line

Formula:

m=ab m = \frac{-a}{b}

Formula:

m=y2y1x2x1 m = \frac{y_2 - y_1}{x_2 - x_1}

Find the slope of the line through A(1,3)A(-1,3) and B(7,5)B(7,-5).

mAB=537(1)=88=1 \begin{aligned} m_{AB} &= \frac{-5 - 3}{7 - (-1)} \\ &= \frac{-8}{8} \\ &= -1 \end{aligned}

Use the following simulation to better understand slope visually.

Two parallel lines have the same slope.

Example:

  1. The line through (1,6)(1,6) and (3,2)(3,2)
  2. The line through (2,9)(2,-9) and (3,1)(-3,1)

Parallel lines

First slope:

m1=2631=42=2 m_1 = \frac{2-6}{3-1} = \frac{-4}{2} = -2

Second slope:

m2=1(9)32=105=2 m_2 = \frac{1-(-9)}{-3-2} = \frac{10}{-5} = -2

Since m1=m2m_1 = m_2, the lines are parallel.

Two lines are perpendicular if:

m1×m2=1 m_1 \times m_2 = -1

Example:

  1. The line through (0,8)(0,-8) and (3,1)(3,1)
  2. The line through (3,3)(-3,3) and (6,0)(6,0)

Perpendicular lines

First slope:

m1=1(8)30=93=3 m_1 = \frac{1-(-8)}{3-0} = \frac{9}{3} = 3

Second slope:

m2=036(3)=39=13 m_2 = \frac{0-3}{6-(-3)} = \frac{-3}{9} = -\frac{1}{3}

Because:

3×(13)=1 3 \times \left(-\frac{1}{3}\right) = -1

the lines are perpendicular.

If a line passes through (x1,y1)(x_1,y_1) and has slope mm, then:

yy1=m(xx1) y - y_1 = m(x - x_1)

Find the equation of a line with slope 3 passing through (2,5)(-2,5).

y5=3(x(2))y5=3(x+2)y5=3x+6y=3x+11 \begin{aligned} y - 5 &= 3(x - (-2)) \\ y - 5 &= 3(x + 2) \\ y - 5 &= 3x + 6 \\ y &= 3x + 11 \end{aligned}

So the equation is:

y=3x+11 y = 3x + 11

If a line passes through (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2), then:

yy1y2y1=xx1x2x1 \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1}

or:

yy1=y2y1x2x1(xx1) y-y_1 = \frac{y_2-y_1}{x_2-x_1}(x-x_1)

Find the equation of the line through (10,3)(10,3) and (2,7)(2,7).

y373=x10210y34=x108y3=12(x10)y3=12x+5y=12x+8 \begin{aligned} \frac{y-3}{7-3} &= \frac{x-10}{2-10} \\ \frac{y-3}{4} &= \frac{x-10}{-8} \\ y-3 &= -\frac{1}{2}(x-10) \\ y-3 &= -\frac{1}{2}x + 5 \\ y &= -\frac{1}{2}x + 8 \end{aligned}

So the equation is:

y=12x+8 y = -\frac{1}{2}x + 8

Answer the following questions to test your understanding.

  1. Find the slope of the line in the figure below.

Slope question

  1. The line through P(2,y)P(2,y) and Q(5,7)Q(5,7) has slope 1-1. Find yy.
  2. Line gg is perpendicular to the line through P(0,7)P(0,7) and Q(2,1)Q(2,1). Find the slope of line gg.
  3. Find the equation of the line through (1,2)(1,2) and perpendicular to 3x+4y=83x+4y=8.
  4. Find the equation of a line with slope 23\frac{2}{3} passing through (6,5)(-6,-5).

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