E-Liberary Pakistan: Straight-Line Equations: Slope-Intercept Form

Straight-line equations, or "linear" equations,graph as straight lines, and have simple variableexpressions 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.

I think 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.)
I like slope-intercept form the best. It is in the form "y=", which makes it easiest to plug into, either for graphing or doing word problems. Just plug in your x-value; the equation is already solved for y. Also, this is the only format you can plug into your (nowadays obligatory) graphing calculator; you have to have a "y=" format to use a graphing utility. But the best part about the slope-intercept form is that you can read off the slope and the intercept right from the equation. This is great for graphing, and can be quite useful for word problems. C

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

-value and a

-value for this line:

x = –1

and

y = –6

In the slope-intercept form of a straight line, I have

, and

. So the only thing I don't have so far is a value for is

(which gives me the

-intercept). Then all I need to do is plug in what they gave me for the slope and the

and

from this particular point, and then solve for

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 + b4 = (–²/₃)(–2) + b4 = ⁴/₃ + b4 – ⁴/₃ = b12/3 – 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 + b2 = (–²/₃)(1) + b2 = –²/₃ + b2 + ²/₃ = b⁶/₃ + 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