Lines and also linear equations

Graphs of lines

Geometry taught us that precisely one line crosses through any two points. We deserve to use this reality in algebra as well. When drawing the graph of a line, us only require two points, and also then usage a straight edge to affix them. Remember, though, that lines room infinitely long: they execute not start and stop in ~ the 2 points we supplied to attract them.

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


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steep

Think of the slope of a line as its "steepness": how conveniently it rises or drops from left to right. This worth is indicated in the graph above as $fracDelta yDelta x$, i m sorry specifies just how much the heat rises or falls (change in $y$) together we relocate from left to appropriate (change in $x$). The is crucial to relate steep or steepness to the rate of vertical adjust per horizontal change. A popular example is the of speed, which measures the change in distance per readjust in time. Whereby a line have the right to represent the distance traveled at assorted points in time, the slope of the heat represents the speed. A steep heat represents high speed, whereas very small steepness represents a lot slower price 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 heat is steeper than the blue and also green lines. An alert the distance traveled ~ one hour ~ above the red line is about 5 miles. The is much higher than the street traveled top top the blue or eco-friendly lines after one hour - about $1$ mile and also $frac15$, respectively. The steeper the line, the better the street traveled every unit that time. In various other words, steepness or slope to represent speed. The red present is the fastest, v the greatest slope, and also the eco-friendly line is the slowest, v the the smallest slope.

Slope deserve to be divide in 4 ways: positive, negative, zero, and also undefined slope. Hopeful slope method that together we relocate from left to appropriate on the graph, the heat rises. An adverse slope way that together we relocate from left to ideal on the graph, the heat falls. Zero slope way that the heat is horizontal: it neither rises nor falls as we relocate from left to right. Vertical lines are said to have actually "undefined slope," together their slope appears to be some infinitely large, unknown value. See the graphs listed below that display each that the four slope types.


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

Watch this video on slope for much more insight into the concept.


$y$-Intercept

The $y$-intercept that a line is the suggest where the line crosses the $y$-axis. Note that this happens as soon as $x = 0$. What space the $y$-intercepts the the present in the graphs above?

it looks like the $y$-intercepts space $(0, 1)$, $(0, 0)$, and $(0, 1)$ because that the an initial three graphs. Over there is no $y$-intercept ~ above the 4th graph - the line never crosses the $y$-axis. Investigate the behavior of a line by adjusting the $y$-intercept via the "$b$-slider".

$x$-Intercept

The $x$-intercept is a comparable concept together $y$-intercept: that is the suggest where the line crosses the $x$-axis. This happens when $y = 0$. The $x$-intercept is not offered as frequently as $y$-intercept, together we will certainly see when determing the equation the a line. What room the $x$-intercepts of the currently in the graphs above?

the looks choose the $x$-intercepts space $(-frac12, 0)$ and $(0, 0)$ for the very first two graphs. There is no $x$-intercept top top the third graph. The fourth graph has an $x$-intercept in ~ $(-1, 0)$.

Equations of lines

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

Calculating slope

Algebraically, slope is calculated as the proportion of the readjust in the $y$ worth to the change in the $x$ worth between any type of two point out on the line. If we have two points, $(x_1, y_1)$ and also $(x_2, y_2)$, slope is to express as:

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

keep in mind that we use the letter $m$ to denote slope. A line the is very steep has actually $m$ values through very large magnitude, whereas as line the is not steep has actually $m$ values through very small magnitude. Because that example, slopes that $100$ and also $-1,000$ have actually much bigger magnitude 보다 slopes that $-0.1$ or $1$.

Example:

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

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

note that us chose allude $(-2, 1)$ together $(x_1, y_1)$ and allude $(5, 8)$ as $(x_2, y_2)$. This to be by choice, together we can have let point $(5, 8)$ be $(x_1, y_1)$ and suggest $(-2, 1)$ it is in $(x_1, y_1)$. How does that affect 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 check out the steep is the exact same either means we choose the first and 2nd points. We have the right to now conclude the the slope of the line that passes v points $(-2, 1)$ and also $(5, 8)$ is $1$.

Watch this video for more examples on calculating slope.

currently that we recognize what slope and also $y$-intercepts are, we have the right to determine the equation of a line given any kind of two clues on the line. There room two main ways to create the equation of a line: point-slope kind and slope-intercept form. We will first look in ~ point-slope form.

Point-Slope type

The point-slope form of one equation that passes with the allude $(x_1, y_1)$ v slope $m$ is the following:

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

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

using the formula because that the point-slope type of the equation of the line, we have the right to just instead of the steep and point coordinate values directly. In other words, $m = 2$ and $(x_1, y_2) = (5, 4)$. So, the equation of the line is $$y - 4 = 2(x - 5).$$

Example:

offered two points, $(-3, -5)$ and also $(2, 5)$, compose the point-slope equation that the line that passes with 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 deserve to verify the slope by looking in ~ the adjust in $y$-values versus the adjust in $x$-values between the two points:


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Graph of heat passing v $(2, 5)$ and also $(-3, -5)$.

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


We have the right to now use among the points along with the steep to create the equation that the line: $$eginalign* y - y_1 &= m(x - x_1) \ y - 5 &= 2(x - 2) quadcheckmark endalign*$$

we could additionally have provided the other allude to create the equation the the line: $$eginalign* y - y_1 &= m(x - x_1) \ y - (-5) &= 2(x - (-3)) \ y + 5 &= 2(x + 3) quadcheckmark endalign*$$

yet wait! Those 2 equations look different. How have the right to they both explain the very same line? If we simplify the equations, we check out that castle are undoubtedly 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, using either point to compose the point-slope kind of the equation results in the same "simplified" equation. We will see following that this simplified equation is another important form of linear equations.

Slope-Intercept type

Another way to refer the equation the a heat is slope-intercept form.

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

In this equation, $m$ again is the slope of the line, and $(0, b)$ is the $y$-intercept. Favor point-slope form, every we need are two points in bespeak to write the equation the passes v them in slope-intercept form.

Constants vs. Variables

the is important to note that in the equation because that slope-intercept form, the letter $a$ and $b$ are constant values, as opposed come the letters $x$ and also $y$, which space variables. Remember, constants represent a "fixed" number - it does no change. A variable deserve to be one of numerous values - it have the right to change. A given line consists of many points, every of which has actually a distinct $x$ and also $y$ value, but that line only has one slope-intercept equation v one worth each for $m$ and also $b$.


Example:

offered the very same two clues above, $(-3, -5)$ and $(2, 5)$, write the slope-intercept form of the equation of the line the passes with them.

We currently calculated the slope, $m$, above to be $2$. We have the right to then use one of the points to settle for $b$. Making use of $(2, 5)$, $$eginalign* y &= 2x + b \ 5 &= 2(2) + b \ 5 &= 4 + b \ 1 &= b. endalign*$$ So, the equation that the heat in slope-intercept type is, $$y = 2x + 1.$$ The $y$-intercept that the heat is $(0, b) = (0, 1)$. Look at the graph over to verify this is the $y$-intercept. In ~ what point does the heat cross the $y$-axis?

At very first glance, it appears the point-slope and also slope-intercept equations of the line are different, yet they really do describe the very same line. We have the right to verify this by "simplifying" the point-slope form 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 much more examples on composing equations of lines in slope-intercept form.

Horizontal and also Vertical currently

currently that we can write equations of lines, we require to take into consideration two special cases of lines: horizontal and vertical. We claimed above that horizontal lines have actually slope $m = 0$, and also that vertical lines have undefined slope. How deserve to we use this to determine the equations that horizontal and vertical lines?

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

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


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


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

But, how can we verify this algebraically? very first off, what is the slope? we calculate slope together $$eginalign* m &= frac1 - 02 - 2 \<1ex> &= frac10 \<1ex> &= extundefined endalign*$$ In this case, the slope worth is undefined, which renders it a upright line.

Slope-intercept and also point-slope forms

at this point, you can ask, "how deserve to I create the equation the a vertical line in slope-intercept or point-slope form?" The answer is that you really can only write the equation the a vertical line one way. Because that vertical lines, $x$ is the same, or constant, because that all worths of $y$. Because $y$ could be any number for vertical lines, the variable $y$ go not appear in the equation the a vertical line.

Horizontal lines
Facts around horizontal lines If two points have the very same $y$-coordinates, only a horizontal line have the right to pass through both points. Each point on a horizontal line has the very same $y$-coordinate. If 2 points have actually the very same $y$-coordinate, $b$, the equation that the heat is $y = b$. The $y$-intercept of a horizontal heat $y = b$ is the allude $(0, b)$. except for the heat $y = 0$, horizontal lines perform not have actually an $x$-intercept.
Example:

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


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


First, note that the $y$-coordinate is the exact same for both points. In fact, if we plot any point on the line, we have the right to see the the $y$-coordinate is $4$. We understand that just a horizontal line can pass v the points, for this reason the equation of that line have to be $y = 4$.

How have the right 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 solve for $y$: $$eginalign* y &= (0)x + b \<1ex> &= b endalign*$$ This tells united state that every allude on the line has actually $y$-coordinate $b.$ since we know two points on the line have $y$-coordinate $4$, $b$ have to be $4$, and so the equation the the heat is $y = 4$.

Slope-intercept and also Point-slope develops

comparable to upright lines, the equation the a horizontal line have the right to only be composed one way. For horizontal lines, $y$ is the exact same for all worths of $x$. Due to the fact that $x$ could be any type of number for horizontal lines, the change $x$ does not appear in the equation the a horizontal line.

Parallel and Perpendicular present

currently that us know just how to characterize present by your slope, we can identify if 2 lines room parallel or perpendicular by your slopes.

Parallel present

In geometry, we room told the two unique lines that execute not intersect room parallel. Looking at the graph below, there space two lines the seem to never ever to intersect. What have the right 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 actually the exact same slope. Any kind of two upright lines, however, are also parallel. The is important to keep in mind that vertical lines have actually undefined slope.

Perpendicular lines

We understand from geometry that perpendicular lines type an angle of $90^circ$. The blue and also red currently in the graph below are perpendicular. What do we an alert about their slopes?


even though this is one details example, the relationship in between the slopes applies to all perpendicular lines. Skip the indicators for now, notice the vertical change in the blue line amounts to the horizontal adjust in the red line. Likewise, the the vertical readjust in the red line equates to the horizontal adjust in the blue line. So, then, what are the slopes of these two lines? $$ extslope that blue line = frac-21 = -2$$ $$ extslope that red line = frac12$$

The other fact to an alert is that the indicators of the slopes that the lines room not the same. The blue line has actually a negative slope and the red line has actually a positive slope. If us multiply the slopes, we get, $$-2 imes frac12 = -1.$$ This station and an unfavorable relationship between slopes is true for every perpendicular lines, except horizontal and vertical lines.

here is one more example of 2 perpendicular lines:


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


2 lines space perpendicular if one of the following is true: The product of their slopes is $-1$. One line is vertical and also the various other is horizontal.

Exercises

Calculate the slope of the heat passing through the given points.

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

Find the steep 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 the the line through the given slope and containing the offered 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 form of the equation the the heat passing v the given pair the points.

31. $(1, 5)$ and also $(4, 2)$ 32. $(3, 7)$ and also $(4, 8)$ 33. $(-3, 1)$ and also $(3, 5)$
34. $(-2, 3)$ and also $(3, 5)$ 35. $(5, 0)$ and $(0, -2)$ 36. $(-2, 0)$ and $(0, 3)$
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 type of the equation the the line with the given slope and also containing the given allude in exercises 22-30.

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Exercises 49-57: write the slope-intercept kind of the equation that the heat passing through the given pair of point out in practice 31-39.