Test your GMAT skills with this GMAT problem solving practice question!

The general equation for a line is: ax + by + c = 0 for all values (x,y) that are on that line. In this form, the slope is –a/b and the y-intercept is –c/b.

The most common equation for a line is called slope-intercept form: y = mx + b, for all values (x, y) on the line. Here m is the slope and b is the y-intercept.

A modified version of slope-intercept form is called point-slope form: y – y₁ = m (x – x₁) + b. This equation is helpful if you are given two points on the line, (x, y) and (x₁, y₁).

For most GMAT questions involving lines, you can use any of these three equations to represent the line. You may find that one is easier depending on the information you are given and what the question is asking, so it’s helpful to practice manipulating all three. Let’s look at an example GMAT problem from Grockit:

Does the line y = ax + b pass through the point (2,5)?

(1) When it is reflected around the x-axis, the line passes through the point (1,-6).

(2) When it is reflected around the y-axis, the line passes through the point (-3,4).

Notice how y = ax + b looks almost exactly like the slope-intercept form, except here a = slope. To know whether (2,5) is a possible (x,y) for this equation, we would need to know the slope AND the y-intercept.

To find the slope of a line two known points are required. Slope = (y₂ – y) / (x₂ – x). The y-intercept is the point at which the line crosses the y-axis. As a pair, the y-intercept can always be expressed as (0, b) since the x-coordinate when a line crosses the y-axis is always zero. If we know the slope AND one pair of coordinates, it is possible to solve for the y-intercept. To answer this yes/no Data Sufficiency question, we need to know two points.

(1) If (1,-6) is on the reflection around the x-axis, then the point (1,6) is on the original line. However, we would not be able to find the slope or the y-intercept with only one coordinate pair.

(2) If (-3,4) is on the reflection around the y-axis, then the point (3,4) is on the original line. This is not sufficient for the same reason as the first statement.

Combining the statements, we see that we have two points on our line. This will be sufficient. Two distinct points are always enough to find the equation of the line. Keep in mind that for DS, we don’t have to solve, but just to make sure:

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

We can then plug in either of the points to find the value of b.
4 = -(3) + b
4 = -3 + b
7 = b

The equation of the line is y = -x + 7

(5) = -(2) +7
5 = 5

The answer is that the line does pass through the point (2,5).

You can check out more articles on Coordinate Geometry on the Grockit blog or in the Beat the GMAT Library!

Try this GMAT geometry practice question and test your skills today!

Parallel lines can be tested in either Problem Solving or Data Sufficiency and though the concept itself is relatively simple, the test questions presented can appear somewhat complex. Don’t make assumptions about lines that look parallel but may in fact not be. Let’s look at an example of how the “eyes can deceive” from GMATPrep:

This question tells us that x < 90 and y < 90 and that PS and QR are parallel. We need to be able to answer whether or not PQ < SR. We cannot assume that PQ and SR are parallel simply because they look parallel, or that x = y. Statement (1) tell us that x > y. Since x is greater than y, then PQ will be slightly shorter (more vertical) than SR. Thus, (1) is sufficient. Statement (2) tells us that the sum of the angles is greater than 90, but does not tell us the value of each angle. It is insufficient.  Let’s try a question from Grockit:

If the two horizontal lines in the figure provided are parallel, what is (a + f + d) – (e + h)?

1)  a + f = 180

2)  a – f = 60

This question solely relies on our knowledge of angle measures and parallel lines. Angles a and b are complementary (add up to 180, a straight line), and angles b and f are congruent, so we already know that a + f = 180 without Statement (1) telling us. (a + f + d) – (e + h) = a + f + d – e – h = 180 – h. Since we do not know the exact value of h (the difference between the obtuse and acute angles), Statement (1) is not sufficient. Statement (2) gives us the difference between the obtuse and acute angle, and so it is sufficient.

Follow me on Beat the GMAT to read more of my posts, or message me via Grockit for more help with Parallel Lines questions!

Based on the figure, is this is the correct graph for the equation:  y = ^{(x3 + x)}/_{x ?}

For a question like this, don’t get nervous if the equation given does not match the standard equation for a parabola. This equation simplifies to y = x² + 1. We can see from the graph that the y-intercept is indeed 1. When we plug in the x-coordinates of the other two given points, we know this must be the correct equation. When in doubt, plug in!

The x-intercepts are also called the “roots” or “solutions” of a parabola. On the GMAT, parabolas will often be referred to as “functions” interchangeably. The x-intercepts can be found using the quadratic formula:

To find the number of x-intercepts a given parabola has, calculate what is called the discriminant:  or the information underneath the radical in the quadratic formula.

If the discriminant is positive, the parabola has two intercepts with x-axis; if it is negative there are no intercepts with the x-axis, and if the discriminant is equal to zero there is one intercept with the x-axis.

The vertex represents the maximum (or minimum) value of the function. Think of it as the starting point of the function.

The vertex of the parabola is located at point  for the standard equation. If you are given the standard equation, you can find the vertex and the x-intercepts.

The standard equation of a circle is (x – h)² + (y – k)² = r² where (h, k) is the center point of the circle and r is the radius. For example, on test day you might see a circle plotted on a graph like so:

The question will ask you to find the equation of the circle. All we have to find is the center point (0, 4) and the radius (4), and then plug it into our equation.

(x – h)² + (y – k)² = r²

(x – 0)² + (y – 4)² = 4²

(x)² + (y – 4)² = 16

Need more Coordinate Geometry help? Schedule a 1-hour lesson with one of Grockit’s GMAT tutors to focus only on coordinate geometry questions from the Grockit GMAT question bank, or check out the Coordinate Geometry questions in Grockit’s Academy, SAT, ACT, or GRE question banks for even more practice!

There are two main equations for straight lines. One form looks like:
For an equation that looks like this the slope is and the y intercept is

For example, in the equation 2x + 3y + 6 = 0, the slope is -2/3 and the y-intercept is -2. The second equation is called slope-intercept form and looks like:   Here m is the slope and b is the y-intercept.

Distance Formula =     Use this to find the distance between two points.

Midpoint Formula =

Use this to find the midpoint between two points (notice how you are essentially finding the average of the x-coordinates and the average of the y-coordinates).

Slope = Rise / Run = Change in y / Change in x

Slopes can be positive, negative, zero, or undefined. Positive slopes tilt to the right. Negative slopes lean to the left. A line with a slope of zero is exactly horizontal. The line neither goes up nor goes down as x increases, which is why it has a 0 slope. Vertical lines have undefined slopes. When the two x coordinates are the same, their difference is zero. The slope calculation would leave a 0 in the denominator, which is called “undefined.”

As long as you know any two points on a line, you can find the slope. A line that passes through the origin must have (0,0) as one of its points. Remember that parallel lines have the same slope, and perpendicular lines have negative reciprocal slopes.

Points of intersection can be found by setting two lines equal. What is the point of intersection between the lines y = 2x – 1 and y = -1/2x + 4? Here’s a quick three-step solution:

1. Put both lines in slope-intercept form. (These example equations are already in slope-intercept form).
2. Set them equal & solve for x.

2x – 1 = -1/2x + 4

2.5x – 1 = 4

2.5x = 5

x = 2

3.   Plug x back in to either equation to find y.

y = 2x – 1

y = 2(2) – 1

y = 4 – 1

y = 3

The point of intersection will be (2,3). Check out Part 2 for parabolas and circles!