## Lines and linear equations

### Graphs of currently

Geometry taught united state that precisely one line the cross through any type of two points. We have the right to use this reality in algebra as well. When drawing the graph the a line, us only require two points, and then usage a right edge to attach them. Remember, though, the lines space infinitely long: they execute not start and also stop in ~ the two points we offered to draw them.

Lines can be expressed algebraically as an equation the relates the \$y\$-values come the \$x\$-values. We can use the same fact that us used previously that two points are had in precisely one line. With only two points, we can determine the equation that a line. Before we execute this, let"s talk about some really important qualities of lines: slope, \$y\$-intercept, and also \$x\$-intercept.

steep

Think that the slope of a line as its "steepness": how easily it rises or drops from left come right. This value is shown in the graph above as \$fracDelta yDelta x\$, which specifies just how much the line rises or falls (change in \$y\$) as we relocate from left to best (change in \$x\$). The is necessary to relate slope or steepness to the rate of vertical readjust per horizontal change. A popular instance is the of speed, which measures the readjust in street per change in time. Wherein a line have the right to represent the distance traveled at various points in time, the steep of the line represents the speed. A steep line represents high speed, whereas very tiny steepness represents a lot slower price of travel, or short speed. This is portrayed in the graph below.

Speed graph

The vertical axis to represent distance, and also the horizontal axis to represent time. The red heat is steeper than the blue and also green lines. Notification the distance traveled after ~ one hour top top the red line is about 5 miles. That is much better than the street traveled on the blue or eco-friendly lines ~ one hour - around \$1\$ mile and also \$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 present is the fastest, v the biggest slope, and the eco-friendly line is the slowest, through the smallest slope.

Slope can be share in four ways: positive, negative, zero, and undefined slope. Positive slope way that as we relocate from left to appropriate on the graph, the heat rises. An adverse slope means that as we move from left to right on the graph, the heat falls. Zero slope method that the heat is horizontal: it no rises nor drops as we relocate from left come right. Upright lines are claimed to have "undefined slope," as their slope appears to be some infinitely large, undefined value. Check out the graphs listed below that show each of the 4 slope types.

 hopeful slope: negative slope: Zero steep (Horizontal): undefined 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 unknown

Investigate the actions of a heat by adjusting the slope via the "\$m\$-slider".

Watch this video on steep for more insight into the concept.

\$y\$-Intercept

The \$y\$-intercept that a heat is the suggest where the line crosses the \$y\$-axis. Keep in mind that this happens as soon as \$x = 0\$. What space the \$y\$-intercepts of the currently in the graphs above?

the looks like the \$y\$-intercepts space \$(0, 1)\$, \$(0, 0)\$, and \$(0, 1)\$ for the first three graphs. There is no \$y\$-intercept top top the 4th graph - the line never ever crosses the \$y\$-axis. Investigate the actions of a line by adjusting the \$y\$-intercept via the "\$b\$-slider".

\$x\$-Intercept

The \$x\$-intercept is a similar concept together \$y\$-intercept: the is the point where the line crosses the \$x\$-axis. This happens as soon as \$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 are the \$x\$-intercepts the the lines in the graphs above?

it looks prefer the \$x\$-intercepts room \$(-frac12, 0)\$ and also \$(0, 0)\$ for the an initial two graphs. Over there is no \$x\$-intercept ~ above the third graph. The 4th graph has an \$x\$-intercept at \$(-1, 0)\$.

### Equations of present

In stimulate to create an equation of a line, we usually have to determine the steep of the heat first.

Calculating steep

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

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

keep in mind that we usage the letter \$m\$ to represent slope. A line the is very steep has \$m\$ values through very large magnitude, whereas together line that is no steep has \$m\$ values with very small magnitude. For example, slopes that \$100\$ and \$-1,000\$ have actually much larger magnitude 보다 slopes of \$-0.1\$ or \$1\$.

Example:

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

using the formula because that 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*\$\$

keep in mind that we chose suggest \$(-2, 1)\$ as \$(x_1, y_1)\$ and suggest \$(5, 8)\$ as \$(x_2, y_2)\$. This to be by choice, together we can have let suggest \$(5, 8)\$ be \$(x_1, y_1)\$ and allude \$(-2, 1)\$ be \$(x_1, y_1)\$. How does that affect the calculate 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 steep is the very same either means we select the very first and 2nd points. We deserve to now conclude the the slope of the line the passes v points \$(-2, 1)\$ and also \$(5, 8)\$ is \$1\$.

Watch this video for an ext examples ~ above calculating slope.

currently that we know what slope and \$y\$-intercepts are, we have the right to determine the equation the a line given any type of two points on the line. There space two primary ways to compose the equation of a line: point-slope type and slope-intercept form. Us will an initial look at point-slope form.

Point-Slope kind

The point-slope kind of one equation that passes v the suggest \$(x_1, y_1)\$ with 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 slope \$m = 2\$ and also passes with the allude \$(5, 4)\$ in point-slope form?

using the formula for the point-slope type of the equation that the line, we can just substitute the steep and allude coordinate worths 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 that the line the 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 have the right to verify the slope by looking in ~ the readjust in \$y\$-values matches the adjust in \$x\$-values between the two points:

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

You are watching: What is the y intercept of a vertical line

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

us could likewise have supplied the other point to create 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 2 equations watch different. How can they both describe the same line? If we leveling the equations, we watch that lock are indeed the same. Let"s do simply 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 point to create the point-slope type of the equation results in the exact same "simplified" equation. We will certainly see following that this streamlined equation is one more important kind of linear equations.

Slope-Intercept kind

Another means to to express the equation the a heat 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. Prefer point-slope form, all we require are two points in bespeak to create the equation the passes v them in slope-intercept form.

Constants vs. Variables

that is essential to keep in mind that in the equation because that slope-intercept form, the letters \$a\$ and \$b\$ are constant values, together opposed to the letter \$x\$ and \$y\$, which are variables. Remember, constants stand for a "fixed" number - the does no change. A variable can be one of countless values - it have the right to change. A provided line consists of many points, each of which has actually a distinctive \$x\$ and also \$y\$ value, however that line only has actually one slope-intercept equation v one value each because that \$m\$ and also \$b\$.

Example:

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

We currently calculated the slope, \$m\$, above to it is in \$2\$. We have the right to then use one of the points to fix for \$b\$. Making use of \$(2, 5)\$, \$\$eginalign* y &= 2x + b \ 5 &= 2(2) + b \ 5 &= 4 + b \ 1 &= b. endalign*\$\$ So, the equation the the heat 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. In ~ what suggest does the heat cross the \$y\$-axis?

At an initial glance, it appears the point-slope and slope-intercept equations that the line room different, but they yes, really do define 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 writing equations of present in slope-intercept form.

### Horizontal and Vertical present

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

upright lines
Facts around vertical present If 2 points have the exact same \$x\$-coordinates, just a vertical line have the right to pass through both points. Each suggest on a vertical line has actually the same \$x\$-coordinate. If two points have actually the very same \$x\$-coordinate, \$c\$, the equation that the heat is \$x = c\$. The \$x\$-intercept the a vertical heat \$x = c\$ is the allude \$(c, 0)\$. other than for the line \$x = 0\$, vertical lines do not have actually a \$y\$-intercept.
Example:

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

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

First, note that the \$x\$-coordinate is the very same for both points. In fact, if we plot any allude from the line, we can see that the \$x\$-coordinate will certainly be \$2\$. We know that only a upright line can pass v the points, so the equation of that line have to be \$x = 2\$.

But, how can we verify this algebraically? an initial 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 also point-slope creates

in ~ this point, you could ask, "how have the right to I write the equation that a vertical heat in slope-intercept or point-slope form?" The price is the you really can only create the equation that a vertical heat one way. Because that vertical lines, \$x\$ is the same, or constant, because that all worths of \$y\$. Due to the fact that \$y\$ can be any number because that vertical lines, the change \$y\$ does not appear in the equation the a vertical line.

Horizontal lines
Facts around horizontal lines If 2 points have the very same \$y\$-coordinates, just a horizontal line have the right to pass through both points. Each suggest on a horizontal line has the same \$y\$-coordinate. If 2 points have the exact same \$y\$-coordinate, \$b\$, the equation of the line is \$y = b\$. The \$y\$-intercept of a horizontal heat \$y = b\$ is the point \$(0, b)\$. other than for the heat \$y = 0\$, horizontal lines carry out not have actually an \$x\$-intercept.
Example:

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

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

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

How can we verify this algebraically? First, calculate the slope: \$\$eginalign* m &= frac4 - 40 - 3 \<1ex> &= frac0-3 \<1ex> &= 0 endalign*\$\$ Then, utilizing slope-intercept form, we deserve to substitute \$0\$ for \$m\$, and also solve for \$y\$: \$\$eginalign* y &= (0)x + b \<1ex> &= b endalign*\$\$ This tells united state that every point on the line has \$y\$-coordinate \$b.\$ since we recognize two point out on the line have \$y\$-coordinate \$4\$, \$b\$ need to be \$4\$, and also so the equation the the heat is \$y = 4\$.

Slope-intercept and also Point-slope develops

similar to upright lines, the equation that a horizontal line have the right to only be created one way. For horizontal lines, \$y\$ is the same for all worths of \$x\$. Due to the fact that \$x\$ can be any number because that horizontal lines, the change \$x\$ go not show up in the equation of a horizontal line.

### Parallel and also Perpendicular currently

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

Parallel currently

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

It appears that the lines over have the exact same slope, and that is correct. Non-vertical parallel lines have the exact same slope. Any type of two vertical lines, however, are additionally parallel. The is essential to keep in mind that vertical lines have undefined slope.

Perpendicular currently

We recognize from geometry that perpendicular lines kind an edge of \$90^circ\$. The blue and red lines in the graph below are perpendicular. What perform we notification about their slopes?

even though this is one particular example, the relationship between the slopes uses to every perpendicular lines. Skip the indicators for now, an alert the vertical adjust in the blue line amounts to the horizontal adjust in the red line. Likewise, the the vertical change in the red line equals the horizontal readjust in the blue line. So, then, what are the slopes that these 2 lines? \$\$ extslope the blue line = frac-21 = -2\$\$ \$\$ extslope the red line = frac12\$\$

The other truth to an alert is that the signs of the slopes the the lines room not the same. The blue line has a an unfavorable slope and also the red line has a confident slope. If we multiply the slopes, us get, \$\$-2 imes frac12 = -1.\$\$ This train station and an adverse relationship between slopes is true for every perpendicular lines, except horizontal and vertical lines.

here is another 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 currently are an unfavorable reciprocals, and therefore, your product is \$-1\$. Recall that the reciprocal that a number is \$1\$ split by the number. Let"s verify this through the examples above: The an adverse reciprocal the \$-2\$ is \$-frac1-2 = frac12 checkmark\$. The an adverse reciprocal the \$frac12\$ is \$-frac1frac12 = -2 checkmark\$. The negative reciprocal of \$-frac23\$ is \$-frac1-frac23 = frac32 checkmark\$. The an adverse reciprocal the \$frac32\$ is \$-frac1frac32 = -frac23 checkmark\$.

two lines are perpendicular if one of the following is true: The product of their slopes is \$-1\$. One line is vertical and the other is horizontal.

### Exercises

Calculate the steep of the line passing with the provided 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 \$(-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 \$(-4, -2)\$

Find the steep of each of the complying with 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 that the line v the offered slope and also containing the given 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 with the provided pair the 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 \$(3, -2)\$

Exercises 40-48: compose the slope-intercept kind of the equation of the line through the offered slope and also containing the given allude in practice 22-30.

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Exercises 49-57: create the slope-intercept form of the equation the the line passing v the offered pair of points in exercises 31-39.