Let’s talk a bit about lines. Like everything else on the SAT, questions about lines will require some very basic knowledge of a few math rules, but you don’t need to know everything you ever learned about Cartesian coordinate planes. It’s important to be able to differentiate what’s important from what’s not.

We’ll get into mathy math in a minute, but first take a moment to try to remember the first time a math teacher introduced the concept of slope to you. You’ll often need math, but I encourage you to think about whether a line question can be solved simply by counting over and up from point to point using “rise over run” before you get too involved in equations. You might save yourself a lot of time and aggravation that way.

##### Keep it simple if you can.
1. Line j has a slope of ⅓ and passes through the point (1, 1). Which of the following points is NOT on line j?

(A) (–5, –1)
(B) (–2, 0)
(C) (0, –3)
(D) (4, 2)
(E) (7, 3)

Don’t even think of writing an equation here! Just apply the first thing you ever learned about slope: that it’s the rise (up and down) over the run (left and right). You can easily list other points on the line simply by adding 1 to the y-value for every 3 you add to the x-value. This is easier to show than it is to say. Click this drawing to blow it up if it’s too small:

That’s it! (C) is your answer; all of the other points clearly fall on the graph!

##### OK, on to the mathy math.

If you’re going to do algebra, you’re going to want to use slope-intercept form whenever possible. If you’re given the equation of a line and it’s not in slope-intercept form already, your first step is to get it there. Many questions will consist of nothing more than comparing the slopes of numerous lines, so make sure you know this cold:

SLOPE-INTERCEPT FORM OF A LINE
y = mx + b

Some times you won’t be given a line equation at all, just two points. In that case, you can calculate the slope using this formula:

SLOPE FORMULA
$\tiny&space;\dpi{300}&space;\fn_jvn&space;m=\frac{rise}{run}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Know the slope formula and the slope-intercept form of a line, and you’re well on your way.

##### Other facts to know cold.
1. Parallel lines have the same slope.
2. Perpendicular lines have negative reciprocal slopes (so if one line has a slope of 2, a perpendicular line has a slope of –1/2; if one has a slope of –16/5, the other has a slope of 5/16).
3. The x-intercept of a line is the x-value of the equation when y = 0.
4. The y-intercept of a line (b in the slope-intercept form) is the y-value of the equation when x = 0.
##### One other reminder that deserves its own heading.

When you are told that a particular point is on a line, that’s the same as being told that the equation of the line works out when that point is plugged into the equation for x and y. In other words, (4,6) is on the line y = x + 2 because 6 = 4 + 2. When a question gives you a point and an equation, PUT THE POINT INTO THE EQUATION.

##### Let’s wrassle with a tough example.

I’ve seen variants of the following question appear on multiple tests. Let’s see if, given what we know about lines, we can figure it out:

1. Line n has a slope of –1/2 and a positive y-intercept. Line m passes through the origin, is perpendicular to line n, and intersects line n at the point (a, a+2). What is the value of a?

(A) 0.5
(B) 1
(C) 1.75
(D) 2
(E) 2.5

With only a cursory glance, it seems like we don’t have much to work with here. Look closer. Note that the question states that line m passes through the origin. This is very important. It’s common for the SAT writers to tell you that, and it’s also common for students to completely breeze by it. When a line passes through the origin, that means we have the y-intercept (zero). It also, more generally, means we have a POINT, which isn’t super-important here, but will be on lots of questions.

Since we know the y-intercept of line m, and we can easily calculate the slope (line m is perpendicular to line n, so the slope’s gotta be the negative reciprocal of n‘s slope of -1/2, remember?):

b = 0
m = 2

we can write the equation of the line:

y = mx + b
y = 2x + 0
y = 2x

plug our point (a, a+2) into our equation:

a+2 = 2a

and solve:

a = 2.