Lines and linear equations

Graphs of currently

Geometry taught us that precisely one line crosses through any type of two points. We have the right to use this fact in algebra as well. When drawing the graph that a line, us only require two points, and then usage a right edge to connect them. Remember, though, the lines room infinitely long: they carry out not start and stop in ~ the two points we supplied to attract them.

Lines have the right to be express algebraically as an equation the relates the $y$-values come the $x$-values. We deserve to use the same reality that we used earlier that 2 points are included in specifically one line. With just two points, we have the right to determine the equation that a line. Prior to we do this, let"s talk about some very important characteristics of lines: slope, $y$-intercept, and $x$-intercept.


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steep

Think that the slope of a line as its "steepness": how easily it rises or falls from left come right. This worth is suggested in the graph above as $fracDelta yDelta x$, i beg your pardon specifies exactly how much the heat rises or falls (change in $y$) as we relocate from left to right (change in $x$). It is vital to relate slope or steepness come the price of vertical change per horizontal change. A popular instance is that of speed, which steps the readjust in street per adjust in time. Whereby a line have the right to represent the street traveled at assorted points in time, the steep of the heat represents the speed. A steep line represents high speed, conversely, very small steepness to represent a lot slower rate of travel, or short speed. This is depicted in the graph below.


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Speed graph


The vertical axis represents distance, and also the horizontal axis to represent time. The red line is steeper than the blue and green lines. Notice the street traveled after one hour ~ above the red line is about 5 miles. That is much greater than the distance traveled ~ above the blue or green lines ~ one hour - around $1$ mile and also $frac15$, respectively. The steeper the line, the higher the street traveled every unit of time. In various other words, steepness or slope represents speed. The red present is the fastest, with the best slope, and also the environment-friendly line is the slowest, with the smallest slope.

Slope have the right to be share in 4 ways: positive, negative, zero, and also undefined slope. Optimistic slope method that as we relocate from left to right on the graph, the heat rises. An unfavorable slope means that together we relocate from left to ideal on the graph, the line falls. Zero slope way that the line is horizontal: it neither rises nor falls as we move from left to right. Vertical lines are claimed to have "undefined slope," as their slope shows up to be part infinitely large, undefined value. See the graphs below that show each that the 4 slope types.


positive slope: an unfavorable slope: Zero steep (Horizontal): unknown slope (Vertical):
$fracDelta yDelta x gt 0$ $fracDelta yDelta x lt 0$ $Delta y = 0$, $Delta x eq 0$, therefore $fracDelta yDelta x = 0$ $Delta x = 0$, so $fracDelta yDelta x$ is undefined
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Investigate the behavior of a heat by adjusting the steep via the "$m$-slider".

Watch this video on steep for an ext insight into the concept.


$y$-Intercept

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

it looks choose the $y$-intercepts room $(0, 1)$, $(0, 0)$, and $(0, 1)$ because that the very first three graphs. Over there is no $y$-intercept on the 4th graph - the line never ever crosses the $y$-axis. Investigate the habits of a line by adjusting the $y$-intercept via the "$b$-slider".

$x$-Intercept

The $x$-intercept is a similar concept as $y$-intercept: it is the allude where the line crosses the $x$-axis. This happens when $y = 0$. The $x$-intercept is not supplied as often as $y$-intercept, together we will certainly see once determing the equation of a line. What are the $x$-intercepts that the currently in the graphs above?

the looks prefer the $x$-intercepts are $(-frac12, 0)$ and also $(0, 0)$ because that the first two graphs. There is no $x$-intercept ~ above the 3rd graph. The fourth graph has an $x$-intercept at $(-1, 0)$.

Equations of lines

In order to create an equation the a line, us usually need to determine the slope of the heat first.

Calculating steep

Algebraically, slope is calculated as the ratio of the change in the $y$ value to the adjust in the $x$ value between any two clues on the line. If we have actually two points, $(x_1, y_1)$ and $(x_2, y_2)$, slope is expressed as:

$$box extslope = m = fracDelta yDelta x = fracy_2 - y_1x_2 - x_1.$$

keep in mind that we use the letter $m$ to represent slope. A line that is an extremely steep has actually $m$ values through very large magnitude, whereas together line the is no steep has $m$ values with very small magnitude. Because that example, slopes the $100$ and also $-1,000$ have much bigger magnitude than slopes the $-0.1$ or $1$.

Example:

uncover the steep of the line that passes v points $(-2, 1)$ and also $(5, 8)$.

making use of the formula because that slope, and letting suggest $(x_1, y_1) = (-2, 1)$ and suggest $(x_2, y_2) = (5, 8)$, $$eginalign* m &= fracDelta yDelta x = fracy_2 - y_1x_2 - x_1\<1ex> &= frac8 - 15 - (-2)\<1ex> &= frac75 + 2\<1ex> &= frac77\<1ex> &= 1 endalign*$$

keep in mind that us chose suggest $(-2, 1)$ together $(x_1, y_1)$ and allude $(5, 8)$ as $(x_2, y_2)$. This to be by choice, as we can have let point $(5, 8)$ it is in $(x_1, y_1)$ and allude $(-2, 1)$ be $(x_1, y_1)$. Just 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 view the slope is the very same either means we select the an initial and second points. We can now conclude the the slope of the line the passes through points $(-2, 1)$ and $(5, 8)$ is $1$.

Watch this video for much more examples on calculating slope.

now that we know what slope and $y$-intercepts are, we can determine the equation the a heat given any two points on the line. There space two primary ways to write the equation the a line: point-slope form and slope-intercept form. Us will very first look in ~ point-slope form.

Point-Slope kind

The point-slope type of an equation the passes v the point $(x_1, y_1)$ v slope $m$ is the following:

$$box extPoint-Slope form: y - y_1 = m(x - x_1).$$ Example:

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

using the formula for the point-slope form of the equation that the line, we deserve to just substitute the slope and allude coordinate values directly. In other words, $m = 2$ and also $(x_1, y_2) = (5, 4)$. So, the equation the the heat is $$y - 4 = 2(x - 5).$$

Example:

offered two points, $(-3, -5)$ and $(2, 5)$, create the point-slope equation that the line that passes v them.

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

Graphically, we can verify the steep by looking in ~ the adjust in $y$-values versus the change in $x$-values between the 2 points:


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Graph of line passing with $(2, 5)$ and also $(-3, -5)$.

You are watching: What is the slope of a vertical line


We deserve to now use one of the points together with the steep to write the equation the the line: $$eginalign* y - y_1 &= m(x - x_1) \ y - 5 &= 2(x - 2) quadcheckmark endalign*$$

us could also have used the other suggest to write the equation that the line: $$eginalign* y - y_1 &= m(x - x_1) \ y - (-5) &= 2(x - (-3)) \ y + 5 &= 2(x + 3) quadcheckmark endalign*$$

however wait! Those two equations look different. How deserve to they both define the very same line? If we leveling the equations, we view that lock are certainly the same. Let"s do just that: $$eginalign* y - 5 &= 2(x - 2) \ y - 5 &= 2x - 4 \ y - 5 + 5 &= 2x - 4 + 5 \ y &= 2x + 1 quadcheckmark endalign*$$ $$eginalign* y + 5 &= 2(x + 3) \ y + 5 &= 2x + 6 \ y + 5 - 5 &= 2x + 6 - 5 \ y &= 2x + 1 quadcheckmark endalign*$$

So, making use of either suggest to create the point-slope type of the equation outcomes in the exact same "simplified" equation. We will certainly see next that this streamlined equation is one more important type of direct equations.

Slope-Intercept form

Another means to to express the equation of a line is slope-intercept form.

$$box extSlope-Intercept form: y = mx + b.$$

In this equation, $m$ again is the steep of the line, and also $(0, b)$ is the $y$-intercept. Choose point-slope form, every we need are 2 points in order to create the equation that passes with them in slope-intercept form.

Constants vs. Variables

that is crucial to keep in mind that in the equation for slope-intercept form, the letter $a$ and $b$ room constant values, as opposed to the letter $x$ and $y$, which space variables. Remember, constants represent a "fixed" number - it does not change. A variable have the right to be one of many values - it have the right to change. A offered line includes many points, every of which has actually a distinctive $x$ and $y$ value, yet that heat only has actually one slope-intercept equation through one worth each for $m$ and $b$.


Example:

given the same two points above, $(-3, -5)$ and also $(2, 5)$, compose the slope-intercept kind of the equation that the line that passes with them.

We already calculated the slope, $m$, over to be $2$. We have the right to then use one of the points to fix for $b$. Utilizing $(2, 5)$, $$eginalign* y &= 2x + b \ 5 &= 2(2) + b \ 5 &= 4 + b \ 1 &= b. endalign*$$ So, the equation the the line in slope-intercept type is, $$y = 2x + 1.$$ The $y$-intercept of the line is $(0, b) = (0, 1)$. Look in ~ the graph over to verify this is the $y$-intercept. At what allude does the line cross the $y$-axis?

At first glance, it seems the point-slope and also slope-intercept equations the the line are different, however they really do explain the same line. We deserve to verify this through "simplifying" the point-slope kind 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 an ext examples on writing equations of lines in slope-intercept form.

Horizontal and Vertical lines

currently that we can write equations the lines, we require to think about two special situations of lines: horizontal and also vertical. Us claimed above that horizontal lines have actually slope $m = 0$, and that vertical lines have undefined slope. How can we use this to identify the equations of horizontal and vertical lines?

vertical lines
Facts about vertical lines If 2 points have the exact same $x$-coordinates, just a vertical line deserve to pass v both points. Each point on a vertical line has actually the very same $x$-coordinate. If 2 points have the exact same $x$-coordinate, $c$, the equation that the line is $x = c$. The $x$-intercept of a vertical heat $x = c$ is the suggest $(c, 0)$. except for the line $x = 0$, vertical lines do not have actually a $y$-intercept.
Example:

consider two points, $(2, 0)$ and $(2, 1)$. What is the equation the the line that passes v them?


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Graph of heat passing through points $(2, 0)$ and also $(2, 1)$


First, keep in mind that the $x$-coordinate is the same for both points. In fact, if we plot any suggest from the line, we have the right to see the the $x$-coordinate will be $2$. We recognize that only a upright line can pass through the points, so the equation of that line must be $x = 2$.

But, 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 renders it a upright line.

Slope-intercept and also point-slope creates

in ~ this point, you could ask, "how deserve to I compose the equation that a vertical line in slope-intercept or point-slope form?" The price is the you really can only write the equation that a vertical heat one way. For vertical lines, $x$ is the same, or constant, because that all worths of $y$. Since $y$ can be any type of number because that vertical lines, the variable $y$ go not show up in the equation of a upright line.

Horizontal currently
Facts around horizontal present If 2 points have actually the very same $y$-coordinates, just a horizontal line can pass through both points. Each allude on a horizontal line has the same $y$-coordinate. If 2 points have actually the same $y$-coordinate, $b$, the equation of the line is $y = b$. The $y$-intercept the a horizontal heat $y = b$ is the allude $(0, b)$. except for the line $y = 0$, horizontal lines carry out not have actually an $x$-intercept.
Example:

think about two points, $(3, 4)$ and $(0, 4)$. What is the equation that the line that passes through them?


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Graph of line passing v points $(3, 4)$ and also $(0, 4)$


First, note that the $y$-coordinate is the very same for both points. In fact, if us plot any suggest on the line, we deserve to see the the $y$-coordinate is $4$. We understand that only a horizontal line have the right to pass through the points, for this reason the equation of the line must be $y = 4$.

How deserve to we verify this algebraically? First, calculation the slope: $$eginalign* m &= frac4 - 40 - 3 \<1ex> &= frac0-3 \<1ex> &= 0 endalign*$$ Then, making use of slope-intercept form, we can substitute $0$ because that $m$, and also solve because that $y$: $$eginalign* y &= (0)x + b \<1ex> &= b endalign*$$ This tells us that every allude on the line has $y$-coordinate $b.$ due to the fact that we understand two points on the line have actually $y$-coordinate $4$, $b$ need to be $4$, and so the equation the the heat is $y = 4$.

Slope-intercept and also Point-slope creates

similar to upright lines, the equation of a horizontal line deserve to only be composed one way. For horizontal lines, $y$ is the very same for all values of $x$. Due to the fact that $x$ might be any kind of number because that horizontal lines, the change $x$ does not show up in the equation the a horizontal line.

Parallel and Perpendicular currently

now that we know how to characterize lines by your slope, we deserve to identify if 2 lines room parallel or perpendicular by their slopes.

Parallel present

In geometry, we room told the two distinctive lines that execute not intersect room parallel. Looking in ~ the graph below, there room two lines the seem to never to intersect. What deserve to we say about their slopes?


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It appears that the lines over have the same slope, and that is correct. Non-vertical parallel lines have the same slope. Any kind of two vertical lines, however, are additionally parallel. It is necessary to note that vertical lines have actually undefined slope.

Perpendicular lines

We know from geometry the perpendicular lines form an angle of $90^circ$. The blue and red present in the graph listed below are perpendicular. What perform we an alert about their slopes?


also though this is one certain example, the relationship in between the slopes uses to every perpendicular lines. Ignoring the indicators for now, notice the vertical readjust in the blue line amounts to the horizontal adjust in the red line. Likewise, the the vertical readjust in the red line amounts to the horizontal readjust in the blue line. So, then, what are the slopes the these 2 lines? $$ extslope the blue line = frac-21 = -2$$ $$ extslope of red line = frac12$$

The other truth to notification is the the indications of the slopes that the lines room not the same. The blue line has actually a an adverse slope and also the red line has actually a positive slope. If we multiply the slopes, us get, $$-2 imes frac12 = -1.$$ This inverse and negative relationship between slopes is true for all perpendicular lines, except horizontal and also vertical lines.

right here is an additional example of 2 perpendicular lines:


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


two lines are perpendicular if among the following is true: The product of your slopes is $-1$. One line is vertical and also the various other is horizontal.

Exercises

Calculate the steep of the heat passing v the provided points.

1. $(2, 1)$ and $(6, 9)$ 2. $(-4, -2)$ and also $(2, -3)$ 3. $(3, 0)$ and $(6, 2)$
4. $(0, 9)$ and $(4, 7)$ 5. $(-2, frac12)$ and also $(-5, frac12)$ 6. $(-5, -1)$ and $(2, 3)$
7. $(-10, 3)$ and $(-10, 4)$ 8. $(-6, -4)$ and also $(6, 5)$ 9. $(5, -2)$ and $(-4, -2)$

Find the slope of every 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 kind of the equation the the line through the offered slope and containing the provided point.

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 kind of the equation the the heat passing through the offered pair of points.

31. $(1, 5)$ and also $(4, 2)$ 32. $(3, 7)$ and also $(4, 8)$ 33. $(-3, 1)$ and $(3, 5)$
34. $(-2, 3)$ and $(3, 5)$ 35. $(5, 0)$ and also $(0, -2)$ 36. $(-2, 0)$ and also $(0, 3)$
37. $(0, 0)$ and $(-1, 1)$ 38. $(1, 1)$ and $(3, 1)$ 39. $(3, 2)$ and also $(3, -2)$

Exercises 40-48: compose the slope-intercept form of the equation the the line through the provided slope and containing the given point in practice 22-30.

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Exercises 49-57: write the slope-intercept type of the equation the the heat passing through the provided pair of points in exercises 31-39.