Forms of Equations of lines:

Straight-line equations, or "linear" equations, graph as straight lines, and have simple variable expressions with no exponents on them. If you see an equation with only x and y — as opposed to, say x2 or sqrt(y) — then you're dealing with a straight-line equation.

There are different types of "standard" formats for straight lines; the particular "standard" format your book refers to may differ from that used in some other books. (There is, ironically, no standard definition of "standard form".) The various "standard" forms are often holdovers from a few centuries ago, when mathematicians couldn't handle very complicated equations, so they tended to obsess about the simple cases. Nowadays, you likely needn't worry too much about the "standard" forms; this lesson will only cover the more-helpful forms.






The most useful form of straight-line equations is the "slope-intercept" form:

y = mx + b

This is called the slope-intercept form because "m" is the slope and "b" gives the y-intercept. (For a review of how this equation is used for graphing, look at slope and graphing.)


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Common exercises will give you some pieces of information about a line, and you will have to come up with the equation of the line. How do you do that? You plug in whatever they give you, and solve for whatever you need, like this:

Find the equation of the straight line that has slope m = 4
and passes through the point (–1, –6).
Okay, they've given me the value of the slope; in this case, m = 4. Also, in giving me a point on the line, they have given me an x-value and a y-value for this line: x = –1 and y = –6.

In the slope-intercept form of a straight line, I have y, m, x, and b. So the only thing I don't have so far is a value for is b (which gives me the y-intercept). Then all I need to do is plug in what they gave me for the slope and the x and y from this particular point, and then solve for b:

y = mx + b
(–6) = (4)(–1) + b
–6 = –4 + b
–2 = b

Then the line equation must be "y = 4x – 2".

What if they don't give you the slope?

Find the equation of the line that passes through the points (–2, 4) and (1, 2).
Well, if I have two points on a straight line, I can always find the slope; that's what the slope formula is for.



Now I have the slope and two points. I know I can find the equation (by solving first for "b") if I have a point and the slope. So I need to pick one of the points (it doesn't matter which one), and use it to solve for b. Using the point (–2, 4), I get:

y = mx + b
4 = (– 2/3)(–2) + b
4 = 4/3 + b
4 – 4/3 = b
12/3 – 4/3 = b
b = 8/3

...so y = ( – 2/3 ) x + 8/3.

On the other hand, if I use the point (1, 2), I get:

y = mx + b
2 = (– 2/3)(1) + b
2 = – 2/3 + b
2 + 2/3 = b
6/3 + 2/3 = b
b = 8/3

So it doesn't matter which point I choose. Either way, the answer is the same:

y = (– 2/3)x + 8/3






The other format for straight-line equations is called the "point-slope" form. For this one, they give you a point (x1, y1) and a slope m, and have you plug it into this formula:

y – y1 = m(x – x1)

Don't let the subscripts scare you. They are just intended to indicate the point they give you. You have the generic "x" and generic "y" that are always in your equation, and then you have the specific x and y from the point they gave you; the specific x and y are what is subscripted in the formula. Here's how you use the point-slope formula:

Find the equation of the straight line that has slope m = 4 and passes through
the point (–1, –6).
This is the same line that I found on the previous page, so I already know what the answer is (namely, y = 4x – 2). But let's see how the process works with the point-slope formula.

They've given me m = 4, x1 = –1, and y1 = –6. I'll plug these values into the point-slope form, and solve for "y=":

y – y1 = m(x – x1)
y – (–6) = (4)(x – (–1))
y + 6 = 4(x + 1)
y + 6 = 4x + 4
y = 4x + 4 – 6

y = 4x – 2




You can find the straight-line equation using the point-slope form if they just give you a couple points:

Find the equation of the line that passes through the points (–2, 4) and (1, 2).
I've already answered this one, but let's look at the process. I should get the same result (namely, y = ( – 2/3 ) x + 8/3 ).

Given two points, I can always find the slope:



Then I can use either point as my (x1, y1), along with this slope Ive just calculated, and plug in to the point-slope form. Using (–2, 4) as the (x1, y1), I get:

y – y1 = m(x – x1)
y – (4) = ( – 2/3 )(x – (–2))
y – 4 = ( – 2/3 )(x + 2)
y – 4 = ( – 2/3 ) x – 4/3
y = ( – 2/3 ) x – 4/3 + 4
y = ( – 2/3 ) x – 4/3 + 12/3

y = ( – 2/3 ) x + 8/3







There is one other consideration for straight-line equations: finding parallel and perpendicular lines. Here is the usual format for the question:

Given the line 2x – 3y = 9 and the point (4, –1), find lines through the point that
are (a) parallel to the given line and (b) perpendicular to it.
In other words, they've given me a reference line — 2x – 3y = 9 — that I'll be comparing to, and some point somewhere else on the plane — namely, (4, –1). Then they want me to find the line through (4, –1) that is parallel to (that has the same slope as) 2x – 3y = 9. On top of that, they then want me to find the line through (4, –1) that is perpendicular to (that has a slope that is the negative reciprocal of the slope of) 2x – 3y = 9.
solve "2x – 3y = 9" for "y=", so that I can find my reference slope:

2x – 3y = 9
–3y = –2x + 9
y = ( 2/3)x – 3

So the reference slope from the reference line is m = 2/3.

Since a parallel line has an identical slope, then the parallel line through (4, –1) will have slope m = 2/3. I'll use the point-slope form to find the line:

y – (–1) = ( 2/3 )(x – 4)
y + 1 = ( 2/3 ) x – 8/3
y = ( 2/3 ) x – 8/3 – 3/3

y = ( 2/3 ) x – 11/3

This is the parallel line that they asked for.

For the perpendicular line, you need to find the perpendicular slope. The reference slope is m = 2/3, and, for the perpendicular slope, I'll flip this slope and change the sign. Then the perpendicular slope is m = – 3/2. So now I can do the point-slope form. Note that the only change from the calculations I just did is that the slope is different now.

y – (–1) = ( – 3/2 )(x – 4)
y + 1 = ( – 3/2 ) x + 6

y = ( – 3/2 ) x + 5

Then the full solution to this exercise is:

parallel: y = ( 2/3 ) x – 11/3
perpendicular: y = ( – 3/2 ) x + 5


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Warning: If a question asks you whether two given lines are "parallel, perpendicular, or neither", you must answer that question by finding their slopes, not by drawing a picture! Pictures can only give you a rough idea of what is going on, but you cannot tell "by looking" that lines with slopes of, say, m1 = 1.00 and m2 = 0.99 are NOT parallel, because they'll sure look parallel on their graphs. But since 1.00 does not equal 0.99, the lines are not parallel. Find the slopes; don't just draw the pictures.