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GMAT Practice Problems on Coordinate Geometry


1) The line shown passes through the point (A, 30).  Which of the following is closest to the value of A?

    (A) 58
    (B) 59
    (C) 60
    (D) 61
    (E) 62

2) Line A has a slope of -{5/3} and passes through the point (–2, 7).  What is the x-intercept of Line A?



3) Line J passes through the points (– 5, – 1), (2, 2), and (5, Q).  What is the value of Q?

AAA new answer choices

4) A circle has a center of (1, 2) and passes through (1, –3).  The circle passes through all of the following EXCEPT:

    (A) (–4, 2)
    (B) (–3, 5)
    (C) (0, 6)
    (D) (4, –2)
    (E) (5, 5)

5) Points J (3, 1) and K (– 1, – 3) are two vertices of an isosceles triangle.  If L is the third vertex and has a y-coordinate of 6, what is the x-coordinate of L?

    (A) –3
    (B) –4
    (C) –5
    (D) –6
    (E) –7

6) A parabola has one x-intercept at (– 4, 0).  If the vertex is at (2, 5), find the other x-intercept.

    (A) 5
    (B) 6
    (C) 7
    (D) 8
    (E) 9


Coordinate Geometry

The coordinate plane is also known as the x-y plane and the Cartesian plane, so named after its discoverer, Mr. Rene Descartes (1596 – 1650), the mathematician & philosopher who also said “I think therefore I am.”  Of course, it consists of two perpendicular numbers lines, the x- and y-axis, which define a grid that covers the entire infinite plane.   Here are some previous blogs on this most remarkable mathematical object, the coordinate plane:

1) Quadrants in the x-y plane

2) Lines & Slopes in the x-y plane

3) Midpoints and Parallel & Perpendicular Lines

4) Distance in the x-y plane

5) Special Properties of the line y = x

6) an earlier set of Coordinate Geometry practice questions

7) a set of challenging Coordinate Geometry practice questions

If you get some insights from some of those blogs, you might give the problems above a second look before reading the solutions below.



Practice problem explanations

1) The line appears to have a slope of ½ and y-intercept of 1, so its equation would be

y = 0.5x + 1

Plug in A for x and 30 for y.

30 = 0.5A + 1

29 = 0.5A

58 = A

Answer = (A)

2) Think about this visually.  A slope of – 5/3 means, among other things, left 3 spaces, down 5.  If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the x-intercept.  Let’s think about this in the vicinity of the x-intercept.


Obviously, (1, 2) is at a height of 2 above the x-axis.  Let b be the distance from (1, 0) to the x-intercept.   We know –h/b must equal the slope.


Now, we just have to add one to that to get the horizontal distance from the origin.


Answer = (B)

3) Think about this visually.  From (–5, –1) to (2, 2), the line moves right 7 and up 3, so that’s a slope of 3/7.

From the point (2, 2) to the point (5, Q), there’s a horizontal distance of 3, an unknown vertical distance — call it h, so that h = Q – 2, or Q = h + 2.  The ratio of this vertical and horizontal distance must equal the slope.


That’s h.  Now, add 2 to get Q.


Answer = (C)

4) The radius is r = 5.  First of all, 5 above the center, the circle goes through (1, 7) on the top, and 5 to the left & right of the center, the circle goes through (–4, 2) and (6, 2) on the same horizontal line as the center.  That first point is choice (A).

That length of 5 can also be the hypotenuse of a 3-4-5 slope triangle, so starting from the center (1, 2), we could go over ±3 and up ±4, or over ±4 and up ±5.  This means the circle must go through

right 3, up 4 = (4, 6)

right 4, up 3 = (5, 5) = option (E)

right 3, down 4 = (4, –2) = option (D)

right 4, down 3 = (5, –1)

left 3, up 4 = (–2, 6)

left 4, up 3 = (–3, 5) = option (D)

left 3, down 4 = (–2, –2)

left 4, down 3 = (–3, –1)

That’s all the points other than option (C).  Notice that (–2, 6) and (4, 6) are on the circle, so another point between them, on the same horizontal line, (0, 6), could not be on the circle.


Answer = (C)

5) This one looks like it could involve a very complicated calculation, but there’s a very elegant way to do it.  This involves special properties of the line y = – x.  Notice that the coordinates of J & K have been switched, x for y and vice versa, and they have the opposite ± signs.  This means that J and K are reflections of each other over the mirror line y = – x.  As is always true of any reflection, every point on the mirror line is equidistant from a point and its reflection.   Thus, we just need any point on the line y = – x, for example, (–6, 6).

Answer = (D)

6) The line of symmetry of a parabola always passes through the vertex, so the equation of the line of symmetry is the vertical line x = 2.  The two x-intercepts are symmetrical around this line.  The point (– 4, 0) is six units to the left of the symmetry line, so the other should be six units to right, at (8, 0).


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6 Responses to GMAT Practice Problems on Coordinate Geometry

  1. Emily December 9, 2015 at 9:29 am #

    Hi Mike,

    I was wondering in Problem 4 how the radius of 5 could be the hypotenuse of a 3-4-5 triangle? I see how one could form a right triangle but isn’t it isosceles with legs of 5 and a hypotenuse of 5 square root 2? Thanks so much for your help!


  2. Jo May 23, 2015 at 1:22 pm #

    For question 2, why wouldn’t you just solve for b to find the equation for the line and then plug in zero for y to find value of x? It seems like a more straightforward approach.

    • Mike MᶜGarry
      Mike McGarry May 26, 2015 at 4:55 pm #

      Dear Jo:
      I’m happy to respond. 🙂 That solution is what I would call the “algebra junkie” solution. Yes, it may be about as quick here, although don’t underestimate how quick the visual reasoning is once you get used to it. Part of the reason I showed it is that all the “algebra junkies” out there tend to want to solve absolutely everything related to the coordinate plane by solving some equation. I am trying to give students the idea that the “algebra junkie” approach is not the only way, and not always the best way, but the visual reasoning approach won’t be quick if folks never practice it. That’s why I showed the solution in this latter method. The key to success in GMAT math is being fluent in multiple approaches. Does all this make sense?
      Mike 🙂

  3. Sirisha February 5, 2015 at 4:14 am #


    In prob 3, why can’t I take h=Q+1, instead of h=Q-2?

    Moreover, I did not understand how h=q-2.

    Please explain

    Thank you.

  4. Arun Panda December 8, 2014 at 9:39 pm #

    The blog provides an excellent opportunity to revise concepts on coordinate geometry, a frequently tested topic in GMAT, and the topic ranges from simple line to complex parabola. Although a line is simplest of all geometrical figures, a solid foundation is required to understand relationship between slope and intercept of line in order to save time by solving the problem efficiently. Like the problem solving questions, the DS questions, which tests fundamentals as well as advance concepts, are big challenge to solve within 2 minutes. Thanks to you- I could quickly revise concepts on this topic through these six questions.

    Kindly note that there is a typo. The answer to Q#3 is C instead of D.



    • Mike MᶜGarry
      Mike December 8, 2014 at 10:05 pm #

      Dear Arun,
      Thank you for you kind words, my friend, and thank you very much for pointing out the typo. I just correct that! I am very glad you found this helpful, and I hope you find some of the linked blogs also helpful. Best of luck, my friend!
      Mike 🙂

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