Lines

The slope of the line which passes through the points $(x_1,y_1)$ and $(x_2,y_2)$ is

$$m = \dfrac{y_2 - y_1}{x_2 - x_1}.$$

The slope measures the rate at which a line goes up or down as you move to the right. For example, a line with slope 4 goes up 4 units for every 1 unit you move to the right. A line with slope -2 goes down 2 units for every 1 unit you move to the right.

$$\hbox{\epsfysize=2in \epsffile{lines1.eps}}$$


Example. Find the slope of the line which passes through $(1,4)$ and $(-5,3)$ .

$$\dfrac{3 - 4}{-5 - 1} = \dfrac{1}{6}.\quad\halmos$$


Example. A line has slope 3. If you move 2 units to the right, how far up or down does the line go?

$$\hbox{\epsfysize=2in \epsffile{lines2.eps}}$$

$$2 \cdot 3 = 6.\quad\halmos$$


Example. A line has slope $-\dfrac{2}{3}$ . Does it go from northwest to southeast or from southwest to northeast?

$$\hbox{\epsfysize=2in \epsffile{lines3.eps}}$$

It goes from northwest to southeast.


Example. Find the slope of the line which passes through $(0,2)$ and $(17,2)$ .

$$m = \dfrac{2 - 2}{17 - 0} = 0$$

A line with slope 0 is horizontal --- parallel to the x-axis.


Example. Find the slope of the line which passes through $(2,0)$ and $(2,17)$ .

If I use the slope formula, I get

$$m = \dfrac{17 - 0}{2 - 2} = \dfrac{17}{0}\quad (\hbox{undefined})$$

A vertical line has undefined slope.


A line may be represented by various equations. Here are a few important forms:

$$ax + by = c, \hbox{ where } a, b, \quad\quad\hbox{and}\quad\quad c \hbox{ are numbers}$$

$$y - y_0 = m(x - x_0), \hbox{ where } m, x_0, \quad\quad\hbox{and}\quad\quad y_0 \hbox{ are numbers}$$

$$y = mx + b, \hbox{ where } m \quad\quad\hbox{and}\quad\quad b \hbox{ are numbers}$$

You can get from one form to another using algebra. For example:

$$y - y_0 = m(x - x_0), \quad y - y_0 = mx - mx_0, \quad -mx + y = -mx_0 + y_0.$$

The last equation is in $ax + by = c$ form, with $a = -m$ , $b = 1$ , and $c = -mx_0 + y_0$ .


Example. Which of the following equations represent lines?

  1. $4x - 3y = 17$
  2. $\dfrac{1}{2}x = 5 + 3y$
  3. $x^2 + 3y = 42$
  4. $\dfrac{y}{x} = 6$

$4x - 3y = 17$ is a line, as is $\dfrac{1}{2}x = 5 + 3y$ .

$x^2 + 3y = 42$ is not a line, because of the $x^2$ term.

$\dfrac{y}{x} = 6$ is almost a line --- in fact, if I multiply both sides by x, I get $y = 6x$ , which is a line. But the original equation is a line minus one point, because plugging in $x = 0$ would cause division by 0.


Consider the form

$$y - y_0 = m(x - x_0), \hbox{ where } m, x_0, \quad\quad\hbox{and}\quad\quad y_0 \hbox{ are numbers}$$

This is called point-slope form. It is the equation of a line with slope m which passes through the point $(x_0,y_0)$ .


Example. Find the equation of the line with slope -17 which passes through the point $(3,5)$ .

$$y - 5 = -17(x - 3)\quad\halmos$$


Example. Find the slope of the line $y - 16 = 3(x + 4)$ . Find at least two different points on the line.

The slope is $m = 3$ . And from the fact that the equation is in point-slope form, I can see that the line passes through the point $(-4,16)$ .

To get another point on the line, plug in a random value for x and solve for y. For example, if I use $x =
   1$ , I get

$$y - 16 = 3(1 + 4), \quad y - 16 = 15, \quad y = 31.$$

Therefore, the point $(1,31)$ also lies on the line.


Example. Find the equation of the line which passes through the points $(4,2)$ and $(7,-10)$ .

First, I'll find the slope:

$$m = \dfrac{-10 - 2}{7 - 4} = -4.$$

Using the point $(4,2)$ and point-slope form, I get the equation

$$y - 2 = -4(x - 4).$$

If I used the point $(7,-10)$ , I'd get

$$y + 10 = -4(x - 7).$$

In fact, the two forms are equivalent:

$$y - 2 = -4(x - 4), \quad y - 2 = -4x + 16, \quad y = -4x + 18$$

$$y + 10 = -4(x - 7), \quad y + 10 = -4x + 28, \quad y = -4x + 18\quad\halmos$$


Example. Find the equation of the horizontal line which passes through the point $(3,42)$ . Find the equation of the vertical line which passes through the point $(3,42)$ .

A horizontal line has slope 0, so the equation of the horizontal line which passes through the point $(3,42)$ is

$$y - 42 = 0\cdot (x - 3), \quad y - 42 = 0, \quad y = 42.$$

A vertical line has undefined slope. But if you draw the picture, you can see that the vertical line passing through $(3,42)$ is $x = 3$ :

$$\hbox{\epsfysize=2in \epsffile{lines4.eps}}$$

This works in general. For example, the horizontal line passing through $(-17,8)$ is $y = 8$ . The vertical line passing through $(-17,8)$ is $x = -17$ .


The y-intercept of a line is the point where the line intersects the y-axis. Likewise, the x-intercept is the point where the line intersects the x-axis.

The form

$$y = mx + b, \hbox{ where } m \quad\quad\hbox{and}\quad\quad b \hbox{ are numbers}$$

is called slope-intercept form. In this equation, m is the slope of the line and b is the y-intercept.


Example. Find the equation of the line with slope 15 and y-intercept -32.

$$y = 15x + 32\quad\halmos$$


Example. Find the slope and y-intercept of the line $y = -17x$ .

The slope is -17, while the y-intercept is $y = 0$ .


Example. Find the slope and y-intercept of the line $3x + 4y = 24$ .

I use algebra to put the equation into slope-intercept form:

$$3x + 4y = 24, \quad 4y = -3x + 24, \quad y = -\dfrac{3}{4}x + 6.$$

The slope is $m = -\dfrac{3}{4}$ and the y-intercept is $y = 6$ .


Example. Find the point where the lines $3x + y = 10$ and $2x - y = 5$ intersect.

To find where two graphs intersect, solve their equations simultaneously.

From $3x + y = 10$ , I get $y = 10 - 3x$ . Plug this into $2x - y = 5$ :

$$2x - (10 - 3x) = 5, \quad 2x - 10 + 3x = 5, \quad 5x - 10 = 5, \quad 5x = 15, \quad x = 3.$$

Then $y = 10 - 3x = 10 - 3\cdot 3 = 1$ . The point of intersection is $(3,1)$ .


Example. Find the x-intercept of the line $2x - 5y = 10$ .

To find the x-intercept, set $y = 0$ and solve for x. The reason this works is that $y = 0$ is the x-axis, and the x-intercept is the place where the line intersects the x-axis. By the rule of thumb from the last problem, I therefore solve $2x - 5y =
   10$ and $y = 0$ simultaneously:

$$2x - 5y = 10, \quad 2x - 5\cdot 0 = 10, \quad 2x = 10, \quad x = 5.$$

The x-intercept is $x = 5$ .

In the same way, you can find the y-intercept by setting $x = 0$ and solving for y.


For example, 5 and $-\dfrac{1}{5}$ are negative reciprocals of one another. So are $\dfrac{3}{2}$ and $-\dfrac{2}{3}$ . Two numbers are negative reciprocals if their product is -1:

$$\dfrac{4}{22}\cdot \left(-\dfrac{11}{2}\right) = -\dfrac{44}{44} = -1 \quad \hbox{They're negative reciprocals}$$

$$\dfrac{36}{8}\cdot \left(-\dfrac{2}{3}\right) = -\dfrac{72}{24} = -3 \quad \hbox{They aren't negative reciprocals}$$

Of course, any horizontal line is perpendicular to any vertical line.


Example. Determine whether the following lines are parallel, perpendicular, or neither:

$$3x - 4y = 12 \quad\quad\hbox{and}\quad\quad 8x + 6y = 11$$

I find the slopes by putting the lines into slope-intercept form:

$$3x - 4y = 12, \quad -4y = -3x + 12, \quad y = \dfrac{3}{4}x - 3$$

$$8x + 6y = 11, \quad 6y = -8x + 11, \quad y = -\dfrac{8}{6}x + \dfrac{11}{6}$$

The slopes of the lines are $\dfrac{3}{4}$ and $-\dfrac{8}{6}$ . The slopes aren't equal, so the lines aren't parallel. Since

$$\dfrac{3}{4}\cdot -\dfrac{8}{6} = -1,$$

the slopes are negative reciprocals. Hence, the lines are perpendicular.


Example. Find the equation of the line which passes through the point $(3,-4)$ and is parallel to the line $2x - 5y = 20$ .

Find the slope of the given line:

$$2x - 5y = 20, \quad -5y = -2x + 20, \quad y = \dfrac{2}{5}x - 4.$$

The given line has slope $\dfrac{2}{5}$ . The line I want is parallel to the given line, so it also has slope $\dfrac{2}{5}$ . Since my line passes through the point $(3,-4)$ , it has point-slope form

$$y + 4 = \dfrac{2}{5}(x - 3).\quad\halmos$$


Example. Find the equation of the line which is perpendicular to the line $x - 12y =
   4$ and passes through the point $(7,5)$ .

Find the slope of the given line:

$$x - 12y = 4, \quad -12y = -x + 4, \quad y = \dfrac{1}{12}x - \dfrac{1}{3}.$$

The given line has slope $\dfrac{1}{12}$ . The line I want is perpendicular to the given line, so my line must have slope -12 (the negative reciprocal of $\dfrac{1}{12}$ ). Since my line passes through $(7,5)$ , its equation is

$$y - 5 = -12(x - 7.\quad\halmos$$


Example. Find the equation of the line which passes through the point $(3,-1)$ and is parallel to the line which passes through $(2,4)$ and $(5,-3)$ .

The slope of the line which passes through $(2,4)$ and $(5,-3)$ is

$$m = \dfrac{-3 - 4}{5 - 2} = -\dfrac{7}{3}.$$

The line I want is parallel to this line, so it also has slope $-\dfrac{7}{3}$ . Since my line passes through $(3,-1)$ , its point-slope form is

$$y + 1 = -\dfrac{7}{3}(x - 3).\quad\halmos$$


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