## Lines and also direct equations

### Graphs of lines

Geomeattempt taught us that exactly one line crosses through any 2 points. We can use this fact in algebra as well. When drawing the graph of a line, we just require two points, and also then use a straight edge to attach them. Remember, though, that lines are infinitely long: they execute not start and also stop at the 2 points we supplied to draw them.

Lines can be expressed algebraically as an equation that relates the $y$-values to the $x$-values. We have the right to usage the exact same reality that we used previously that 2 points are had in exactly one line. With only two points, we have the right to determine the equation of a line. Before we do this, let"s discuss some extremely necessary characteristics of lines: slope, $y$-intercept, and also $x$-intercept.

Slope

Think of the slope of a line as its "steepness": just how conveniently it rises or drops from left to right. This value is indicated in the graph above as $fracDelta yDelta x$, which specifies how a lot the line rises or drops (readjust in $y$) as we move from left to best (change in $x$). It is necessary to relate slope or steepness to the rate of vertical readjust per horizontal change. A renowned instance is that of rate, which actions the adjust in distance per readjust in time. Wright here a line deserve to represent the distance traveled at various points in time, the slope of the line represents the speed. A steep line represents high rate, whereas very little steepness represents a a lot slower price of take a trip, or low rate. This is shown in the graph below.

**Speed graph**

The vertical axis represents distance, and the horizontal axis represents time. The red line is steeper than the blue and green lines. Notice the distance traveled after one hour on the red line is about 5 miles. It is much better than the distance traveled on the blue or green lines after one hour - around $1$ mile and $frac15$, respectively. The steeper the line, the higher the distance traveled per unit of time. In other words, steepness or slope represents speed. The red lines is the fastest, via the best slope, and the green line is the slowest, via the smallest slope.

Slope have the right to be classified in 4 ways: positive, negative, zero, and also uncharacterized slope. Optimistic slope implies that as we move from left to right on the graph, the line rises. Negative slope implies that as we relocate from left to ideal on the graph, the line falls. Zero slope suggests that the line is horizontal: it neither rises nor drops as we move from left to appropriate. Vertical lines are said to have "uncharacterized slope," as their slope shows up to be some infinitely big, uncharacterized worth. See the graphs listed below that present each of the four slope kinds.

Optimistic slope: | Negative slope: | Zero slope (Horizontal): | Unidentified slope (Vertical): |

$fracDelta yDelta x gt 0$ | $fracDelta yDelta x lt 0$ | $Delta y = 0$, $Delta x eq 0$, so $fracDelta yDelta x = 0$ | $Delta x = 0$, so $fracDelta yDelta x$ is uncharacterized |

**Investigate** the behavior of a line by adjusting the slope through the "$m$-slider".

**Watch this video** on slope for even more insight into the idea.

$y$-Intercept

The $y$-intercept of a line is the suggest where the line crosses the $y$-axis. Keep in mind that this happens once $x = 0$. What are the $y$-intercepts of the lines in the graphs above?

It looks favor the $y$-intercepts are $(0, 1)$, $(0, 0)$, and also $(0, 1)$ for the initially three graphs. Tbelow is no $y$-intercept on the fourth graph - the line never crosses the $y$-axis. **Investigate** the behavior of a line by adjusting the $y$-intercept via the "$b$-slider".

The $x$-intercept is a similar concept as $y$-intercept: it is the point wright here the line crosses the $x$-axis. This happens as soon as $y = 0$. The $x$-intercept is not offered as frequently as $y$-intercept, as we will watch as soon as determing the equation of a line. What are the $x$-intercepts of the lines in the graphs above?

It looks prefer the $x$-intercepts are $(-frac12, 0)$ and also $(0, 0)$ for the first 2 graphs. There is no $x$-intercept on the 3rd graph. The fourth graph has actually an $x$-intercept at $(-1, 0)$.

### Equations of lines

In order to write an equation of a line, we normally need to recognize the slope of the line initially.

Calculating slopeAlgebraically, slope is calculated as the ratio of the change in the $y$ worth to the change in the $x$ value between any type of 2 points on the line. If we have actually 2 points, $(x_1, y_1)$ and also $(x_2, y_2)$, slope is expressed as:

$$boxNote that we use the letter $m$ to denote slope. A line that is extremely steep has $m$ worths through extremely large magnitude, whereas as line that is not steep has actually $m$ values through extremely tiny magnitude. For example, slopes of $100$ and also $-1,000$ have actually a lot bigger magnitude than slopes of $-0.1$ or $1$.

Example:Find the slope of the line that passes with points $(-2, 1)$ and also $(5, 8)$.

Using the formula for slope, and also letting allude $(x_1, y_1) = (-2, 1)$ and also point $(x_2, y_2) = (5, 8)$, $$eginalign* m &= fracDelta yDelta x = fracy_2 - y_1x_2 - x_1\<1ex> &= frac8 - 15 - (-2)\<1ex> &= frac75 + 2\<1ex> &= frac77\<1ex> &= 1 endalign*$$

Keep in mind that we made a decision allude $(-2, 1)$ as $(x_1, y_1)$ and also allude $(5, 8)$ as $(x_2, y_2)$. This was by alternative, as we might have let suggest $(5, 8)$ be $(x_1, y_1)$ and point $(-2, 1)$ be $(x_1, y_1)$. How does that impact the calculation of slope?

$$eginalign* m &= fracDelta yDelta x = fracy_2 - y_1x_2 - x_1\<1ex> &= frac1 - 8-2 - 5\<1ex> &= frac-7-7\<1ex> &= 1 endalign*$$

We watch the slope is the same either method we pick the first and also second points. We have the right to now conclude that the slope of the line that passes with points $(-2, 1)$ and also $(5, 8)$ is $1$.

**Watch this video** for even more examples on calculating slope.

Now that we recognize what slope and $y$-intercepts are, we deserve to recognize the equation of a line offered any two points on the line. Tright here are 2 primary means to compose the equation of a line: **point-slope** form and **slope-intercept** create. We will certainly initially look at point-slope form.

The point-slope develop of an equation that passes through the allude $(x_1, y_1)$ with slope $m$ is the following:

$$boxWhat is the equation of the line has actually slope $m = 2$ and passes through the point $(5, 4)$ in point-slope form?

Using the formula for the point-slope create of the equation of the line, we have the right to simply substitute the slope and also suggest coordinate values directly. In various other words, $m = 2$ and also $(x_1, y_2) = (5, 4)$. So, the equation of the line is $$y - 4 = 2(x - 5).$$

Example:Given two points, $(-3, -5)$ and $(2, 5)$, compose the point-slope equation of the line that passes via them.

First, we calculate the slope: $$eginalign* m &= fracy_2 - y_1x_2 - x_1\<1ex> &= frac5 - (-5)2 - (-3)\<1ex> &= frac105\<1ex> &= 2 endalign*$$

Graphically, we deserve to verify the slope by looking at the adjust in $y$-worths versus the readjust in $x$-values between the two points:

**Graph of line passing via $(2, 5)$ and also $(-3, -5)$.You are watching: What does the distance between two white horizontal lines on this graph represent?**

We can currently usage among the points in addition to the slope to compose the equation of the line: $$eginalign* y - y_1 &= m(x - x_1) \ y - 5 &= 2(x - 2) quadcheckmark endalign*$$

We could additionally have actually offered the other suggest to write the equation of the line: $$eginalign* y - y_1 &= m(x - x_1) \ y - (-5) &= 2(x - (-3)) \ y + 5 &= 2(x + 3) quadchecknote endalign*$$

But wait! Those two equations look different. How can they both explain the very same line? If we simplify the equations, we view that they are indeed the very same. Let"s execute simply that: $$eginalign* y - 5 &= 2(x - 2) \ y - 5 &= 2x - 4 \ y - 5 + 5 &= 2x - 4 + 5 \ y &= 2x + 1 quadchecknote endalign*$$ $$eginalign* y + 5 &= 2(x + 3) \ y + 5 &= 2x + 6 \ y + 5 - 5 &= 2x + 6 - 5 \ y &= 2x + 1 quadchecknote endalign*$$

So, making use of either allude to write the point-slope create of the equation results in the very same "simplified" equation. We will check out next that this streamlined equation is another necessary create of direct equations.

Slope-Intercept develop Another way to express the equation of a line is **slope-intercept** create.

In this equation, $m$ again is the slope of the line, and also $(0, b)$ is the $y$-intercept. Like point-slope form, all we require are 2 points in order to compose the equation that passes via them in slope-intercept create.

Constants vs. Variables It is necessary to note that in the equation for slope-intercept develop, the letters $a$ and $b$ are **constant** worths, as opposed to the letters $x$ and $y$, which are **variables**. Remember, constants reexisting a "fixed" number - it does not adjust. A variable deserve to be one of many kind of values - it deserve to change. A offered line contains many points, each of which has a distinctive $x$ and also $y$ worth, yet that line only has one slope-intercept equation via one value each for $m$ and $b$.

Example:

Given the very same 2 points over, $(-3, -5)$ and $(2, 5)$, compose the slope-intercept develop of the equation of the line that passes through them.

We already calculated the slope, $m$, over to be $2$. We can then usage among the points to settle for $b$. Using $(2, 5)$, $$eginalign* y &= 2x + b \ 5 &= 2(2) + b \ 5 &= 4 + b \ 1 &= b. endalign*$$ So, the equation of the line in slope-intercept form is, $$y = 2x + 1.$$ The $y$-intercept of the line is $(0, b) = (0, 1)$. Look at the graph over to verify this is the $y$-intercept. At what suggest does the line cross the $y$-axis?

At initially glance, it seems the point-slope and also slope-intercept equations of the line are different, yet they really execute define the very same line. We can verify this by "simplifying" the point-slope create as such: $$eginalign* y - 5 &= 2(x - 2) \ y - 5 &= 2x - 4 \ y - 5 + 5 &= 2x - 4 + 5 \ y &= 2x + 1 \ endalign*$$

**Watch this video** for more examples on creating equations of lines in slope-intercept develop.

### Horizontal and Vertical Lines

Now that we can create equations of lines, we should think about two one-of-a-kind instances of lines: horizontal and also vertical. We declared above that horizontal lines have slope $m = 0$, and that vertical lines have actually uncharacterized slope. How deserve to we usage this to recognize the equations of horizontal and also vertical lines?

Vertical linesFacts around vertical lines If 2 points have the exact same $x$-works with, only a vertical line deserve to pass via both points. Each suggest on a vertical line has the very same $x$-coordinate. If two points have actually the exact same $x$-coordinate, $c$, the equation of the line is $x = c$. The $x$-intercept of a vertical line $x = c$ is the suggest $(c, 0)$. Except for the line $x = 0$, vertical lines

**perform not**have a $y$-intercept.

Example:

Consider 2 points, $(2, 0)$ and $(2, 1)$. What is the equation of the line that passes with them?

**Graph of line passing with points $(2, 0)$ and $(2, 1)$**

First, note that the $x$-coordinate is the exact same for both points. In reality, if we plot any kind of allude from the line, we deserve to see that the $x$-coordinate will be $2$. We recognize that only a vertical line deserve to pass with the points, so the equation of that line must be $x = 2$.

But, just how have the right to we verify this algebraically? First off, what is the slope? We calculate slope as $$eginalign* m &= frac1 - 02 - 2 \<1ex> &= frac10 \<1ex> &= extundefined endalign*$$ In this case, the slope value is undefined, which provides it a vertical line.

Slope-intercept and point-slope developsAt this suggest, you might ask, "exactly how deserve to I write the equation of a vertical line in slope-intercept or point-slope form?" The answer is that you really have the right to just write the equation of a vertical line one means. For vertical lines, $x$ is the exact same, or constant, for all worths of $y$. Due to the fact that $y$ might be any number for vertical lines, the variable $y$ does not show up in the equation of a vertical line.

Horizontal linesFacts around horizontal lines If two points have actually the very same $y$-coordinates, only a horizontal line have the right to pass with both points. Each allude on a horizontal line has actually the same $y$-coordinate. If two points have actually the very same $y$-coordinate, $b$, the equation of the line is $y = b$. The $y$-intercept of a horizontal line $y = b$ is the point $(0, b)$. Except for the line $y = 0$, horizontal lines perform not have actually an $x$-intercept.

Example:

Consider two points, $(3, 4)$ and $(0, 4)$. What is the equation of the line that passes with them?

**Graph of line passing through points $(3, 4)$ and $(0, 4)$**

First, note that the $y$-coordinate is the same for both points. In truth, if we plot any kind of suggest on the line, we deserve to watch that the $y$-coordinate is $4$. We understand that only a horizontal line deserve to pass with the points, so the equation of that line should be $y = 4$.

How have the right to we verify this algebraically? First, calculate the slope: $$eginalign* m &= frac4 - 40 - 3 \<1ex> &= frac0-3 \<1ex> &= 0 endalign*$$ Then, making use of slope-intercept form, we deserve to substitute $0$ for $m$, and also fix for $y$: $$eginalign* y &= (0)x + b \<1ex> &= b endalign*$$ This tells us that eexceptionally allude on the line has $y$-coordinate $b.$ Due to the fact that we know 2 points on the line have actually $y$-coordinate $4$, $b$ need to be $4$, and also so the equation of the line is $y = 4$.

Slope-intercept and Point-slope developsComparable to vertical lines, the equation of a horizontal line deserve to only be written one method. For horizontal lines, $y$ is the same for all worths of $x$. Due to the fact that $x$ can be any number for horizontal lines, the variable $x$ does not appear in the equation of a horizontal line.

### Parallel and also Perpendicular lines

Now that we understand exactly how to characterize lines by their slope, we have the right to identify if 2 lines are parallel or perpendicular by their slopes.

Parallel linesIn geomeattempt, we are told that two unique lines that execute not intersect are parallel. Looking at the graph listed below, tbelow are two lines that seem to never before to intersect. What deserve to we say around their slopes?

It appears that the lines above have the same slope, and that is correct. Non-vertical parallel lines have the same slope. Any 2 vertical lines, however, are likewise parallel. It is vital to note that vertical lines have actually uncharacterized slope.

Perpendicular linesWe understand from geometry that perpendicular lines develop an angle of $90^circ$. The blue and red lines in the graph listed below are perpendicular. What perform we notification around their slopes?

Even though this is one certain instance, the partnership between the slopes applies to all perpendicular lines. Ignoring the indicators for currently, alert the vertical change in the blue line equals the horizontal adjust in the red line. Likewise, the the vertical readjust in the red line equates to the horizontal change in the blue line. So, then, what are the slopes of these 2 lines? $$ extslope of blue line = frac-21 = -2$$ $$ extslope of red line = frac12$$

The other fact to notice is that the indicators of the slopes of the lines are not the very same. The blue line has actually an adverse slope and the red line has actually a positive slope. If we multiply the slopes, we get, $$-2 imes frac12 = -1.$$ This inverse and negative relationship between slopes is true for all perpendicular lines, other than horizontal and vertical lines.

Here is another instance of two perpendicular lines:

$$ extslope of blue line = frac-23$$ $$ extslope of red line = frac32$$ $$ extProduct of slopes = frac-23 cdot frac32 = -1$$ Again, we see that the slopes of two perpendicular lines are negative reciprocals, and therefore, their product is $-1$. Respeak to that the **reciprocal** of a number is $1$ separated by the number. Let"s verify this through the examples above: The negative reciprocal of $-2$ is $-frac1-2 = frac12 checkmark$. The negative reciprocal of $frac12$ is $-frac1frac12 = -2 checkmark$. The negative reciprocal of $-frac23$ is $-frac1-frac23 = frac32 checkmark$. The negative reciprocal of $frac32$ is $-frac1frac32 = -frac23 checkmark$.

Two lines are perpendicular if one of the complying with is true: The product of their slopes is $-1$. One line is vertical and also the other is horizontal.

### Exercises

Calculate the slope of the line passing through the provided points.

1. $(2, 1)$ and also $(6, 9)$ | 2. $(-4, -2)$ and $(2, -3)$ | 3. $(3, 0)$ and also $(6, 2)$ |

4. $(0, 9)$ and also $(4, 7)$ | 5. $(-2, frac12)$ and also $(-5, frac12)$ | 6. $(-5, -1)$ and also $(2, 3)$ |

7. $(-10, 3)$ and also $(-10, 4)$ | 8. $(-6, -4)$ and $(6, 5)$ | 9. $(5, -2)$ and also $(-4, -2)$ |

Find the slope of each of the following lines.

10. $y - 2 = frac12(x - 2)$ | 11. $y + 1 = x - 4$ | 12. $y - frac23 = 4(x + 7)$ |

13. $y = -(x + 2)$ | 14. $2x + 3y = 6$ | 15. $y = -2x$ |

16. $y = x$ | 17. $y = 4$ | 18. $x = -2$ |

19. $x = 0$ | 20. $y = -1$ | 21. $y = 0$ |

Write the point-slope form of the equation of the line with the given slope and also containing the offered suggest.

22. $m = 6$; $(2, 7)$ | 23. $m = frac35$; $(9, 2)$ | 24. $m = -5$; $(6, 2)$ |

25. $m = -2$; $(-4, -1)$ | 26. $m = 1$; $(-2, -8)$ | 27. $m = -1$; $(-3, 6)$ |

28. $m = frac43$; $(7, -1)$ | 29. $m = frac72$; $(-3, 4)$ | 30. $m = -1$; $(-1, -1)$ |

Write the point-slope create of the equation of the line passing via the given pair of points.

31. $(1, 5)$ and $(4, 2)$ | 32. $(3, 7)$ and $(4, 8)$ | 33. $(-3, 1)$ and $(3, 5)$ |

34. $(-2, 3)$ and $(3, 5)$ | 35. $(5, 0)$ and also $(0, -2)$ | 36. $(-2, 0)$ and $(0, 3)$ |

37. $(0, 0)$ and also $(-1, 1)$ | 38. $(1, 1)$ and also $(3, 1)$ | 39. $(3, 2)$ and $(3, -2)$ |

Exercises 40-48: Write the slope-intercept create of the equation of the line with the offered slope and also containing the given point in exercises 22-30.

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Exercises 49-57: Write the slope-intercept develop of the equation of the line passing with the offered pair of points in exercises 31-39.