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# Challenging Coordinate Geometry Practice Questions

Earlier, we featured a blog of Coordinate Geometry practice questions.  Here are eight more questions, some of which are challenging.

1) Graph G has a line of symmetry of x = –5/2.  Graph G passes through the point (3, 3).  What is the x-coordinate of another point that must have a y-coordinate of 3?

(A) –8

(B) –7

(C) –5

(D) –4

(E) 2

2) In the figure above, the point on segment JK that is four times as far from K as it is from J is:

3) Which point is the reflection of the point (–7, 5) over the line y = –x?

(A) (–7, 5)

(B) (–5, 7)

(C) (5, –7)

(D) (7, –5)

(E) (7, 5)

4) In a coordinate system, P = (2, 7) and Q = (2, –3).  Which could be the coordinates of R if PQR is an isosceles triangle?

I. (12, –3)

II. (–6, –9)

III. (–117, 2)

(A) I only

(B) I and II only

(C) I and III only

(D) II and III only

(E) I, II, and III

5) Point W = (5, 3).  Circle J has a center at point W and radius of r = 5.  This circle intersects the y-axis at one intercept and the x-axis at two intercepts.  What is the area of the triangle formed by these three intercepts?

(A) 7.5

(B) 12

(C) 15

(D) 24

(E) 30

6) Line M has a y-intercept of –4, and its slope must be an integer-multiple of 1/7.  Given that Line M passes below (4, –1) and above (5, –6), how many possible slopes could Line M have?

(A) 6

(B) 7

(C) 8

(D) 9

(E) 10

7) Line Q has the equation 5y – 3x = 45.  If Line S is perpendicular to Q, has an integer for its y-intercept, and intersects Q in the second quadrant, then how many possible Line S’s exist?  (Note: Intersections on one of the axes do not count.)

(A) 25

(B) 33

(C) 36

(D) 41

(E) 58

8) In the x-y plane, point (p, q) is a lattice point if both p and q are integers.  Circle C has a center at (–2, 1) and a radius of 6.  Some points, such as the center (–2, 1), are inside the circle, but a point such as (4, 1) is on the circle but not in the circle.  How many lattice points are in circle C?

(A) 36

(B) 72

(C) 89

(D) 96

(E) 109

## Some relevant blogs

Here are some blogs that you may find germane.  If you look at one of these and have an “aha!”, then you might want to look over these problems again.

4) Slopes

## Explanations to the practice questions

1) The line of symmetry is x = –2.5.  The point (3, 3) is 3 – (–2.5) = 5.5 to the right of this line of symmetry.  It’s reflection must be 5.5 units to the left of the line of symmetry, so (–2.5) – (5.5) = –8 is the x-coordinate.  Answer = (A)

2) Call the point P.  Then, PK = 4*JP.  Of course, JP + PK = JK, so JP + 4*JP = 5*JP = JK, and JP = (1/5)*JK.  Point P should be one fifth of the segment away from J.  This would be the point (–1, 3).  Answer = (A)

3) The point (–7, 5) is in the second quadrant, relatively close to the line y = –x, so the reflection would have to be another point in the second quadrant, and the only answer choice in the second quadrant not equal to the original point is (B), (–5, 7).  This question is very easy to solve by visual/spatial reasoning.

For those who want to know the “rule”: when a point is reflected over the line y = –x, the coordinates are reversed, and each one takes on its opposite sign.  The –7 becomes +7, and the +5 becomes –5, and they switch places, which also results in (B).

4) Notice that points P & Q are separated by 10 units vertically.

I. R = (12, –3)

Then, point R is 10 units to the right of Q, so we get a big 45-45-90 Right Isosceles Triangle:

Option I works.

II.  R = (–6, –9)

This is tricky.  This left 8 and down 6 from Q, so the triangle formed by the x & y separations between Q and this R form a 6-8-10 right triangle, and the distance QR = 10, making this an isosceles triangle as well.  This little 6-8-10 triangle on the QR connection is show in dashed lines:

Option II works.

III.  R = (–117, 2)

It’s a mistake to do any calculations at all with this one.  The line y = 2 is the perpendicular bisector of PQ:

This is an important Geometry idea: any point on the perpendicular bisector of a segment is automatically equidistant from the two endpoints of the segment.  This means, we could pick absolutely any point on the line y = 2, call it R, and this point R would be equidistant from P & Q, making PQR an isosceles triangle.  The point R = (–117, 2) is on this perpendicular bisector, so it is equidistant from P & Q, and PQR must therefore be isosceles.

Option III works.

Each of the three options works.  Answer = (E).

5) Well, if we go 5 units to the left of (5, 3), we’re at (0, 3): that’s the single y-intercept of the circle.

Now, think about the two x-intercepts.  Each one is a diagonal distance of r = 5 from (5, 3), and if we may a right triangle on either side, the vertical leg is the distance from (5, 3) straight down to the x-axis, which of course is 3.

Those two purple triangles must be 3-4-5 triangles, which means each one has a base of 4, and the distance between the two of them is 8.  One is at (1, 0) and the other is at (9, 0).

Now, think about the triangle formed by these three intercepts.  The base, from (1,0) to (9, 0) is 8, and while the third vertex is off-center, that doesn’t matter — the height is h = 3.  A = (0.5)bh = (0.5)(8)(3) = 12.  Answer = (B)

6) Well, for starters, zero is a multiple of every number, and a line with slope zero, the horizontal line y = –4 passes below (4, –1) and above (5, –6).  That’s horizontal line is our starting point.

The point (4, –1) is over 4, up 3 from the y-intercept (0, –4).  A line with a slope of +3/4 would go straight from (0, –4) to (4, –1).  Thus, we need a slope that is less than +3/4.  Notice that 3/4 = 21/28.  Notice that 5/7 = 20/28, so this would be less than 3/4.  Therefore, +1/7 through +5/7 will all slope up, obviously above (5, –6), and all will pass below (4, –1).  That’s five upward sloping lines.

The point (5, –6) is over 5, down 2, from the y-intercept (0, –4).  A line with a slope of –2/5 would go straight from (0, –4) to (5, –6).  Thus, we need a slope that is more than –2/5; another way to say that is, we need a negative slope whose absolute value is less than +2/5.  Well, 2/5 = 14/35, while 2/7 = 10/35 and 3/7 = 15/35, so (2/7) < (2/5) < (3, 7).  The negatively sloping lines obviously pass below (4, –1), but only two of them, –1/7 and –2/7, pass above (5, –6).

That’s one horizontal line, five upward sloping lines, and two downward sloping lines, for a total of eight.  Answer = (C).

7)  First of all, Line Q 5y – 3x = 45 re-written in slope-intercept form is y = (3/5)x + 9.  Line Q has a y-intercept of +9, so if Line S also has a y-intercept of  +9, they would intersect on the y-axis, not in the second quadrant.  Therefore, the y-intercept of Line S must be less than 9, and the highest value it could have is a y-intercept of 8.

Now, let’s think about the bottom end.  Line Q has a y-intercept of +9 and an x-intercept of –15.  Line S, perpendicular to Line Q, must have a slope of –5/3, the negative reciprocal of Line Q’s slope.  If we start at the x-intercept of point Q, the point (–15, 0), we would follow the slope of –5/3 over 15 and down 25 to (0, –25).  If Line S had a y-intercept of –25, it would intersect Line Q at (–15, 0), on the x-axis, and not in the second quadrant.  Therefore, a y-intercept of –25 doesn’t work, and the lowest value that would work is one above it, y-intercept of –24.

Any y-intercept less than or equal to +8 and greater than or equal to –24 would work.  That’s 8 positive y-intercepts, the intercept of zero at the origin, and 24 negative intercepts, for a total of 8 + 1 + 24 = 33.  Answer = (B)

8) This is a tough question.  First of all, obviously, the center a lattice point in the circle.  If we move horizontally or vertically, the first 5 lattice points will be inside the circle, and the sixth one will be on the circle, so the points on the circle don’t count as being “in” the circle.

Now, we just are going to focus on one quadrant of the circle, the upper right quadrant.  Suppose we go over one unit and up the circle.

Well, that’s a right triangle, with hypotenuse of r = 6, and horizontal leg of 1, so  if the vertical leg is x, then

Well, the square root of 35 is bigger than 5, so five points in that column are inside the circle.  Now, move two units to the right.  Again, hypotenuse r = 6, short leg = 2, vertical leg = x, so

Still bigger than 5, so there five points in this second column.  Now, move three units to the right.  Again, hypotenuse r = 6, short leg = 3, vertical leg = x, so

Still bigger than 5, so there five points in this third column.  Here’s where we are so far:

Clearly, by symmetry, we can reflect this array over the 45° line, to get more points in the circle:

Because that row at a height of 5 only is 3 points wide, we know that the column 5 units to the right can only be 3 points height.  We  have to check that highest point in the column 4 units to the right.  Again, hypotenuse r = 6, short leg = 4, vertical leg = x, so

Clearly, this is bigger than 4, so that fourth column can be four points high.

That completes the quadrant.  Within the quadrant, that’s 5 + 5 + 5 + 4 + 3 = 22 points.  Multiply that by four, and that’s 88 points within the four quadrants.  Add the 20 points on the vertical & horizontal segments, and the one point at the center, and we have 88 + 20 + 1 = 109.  Answer = (E).

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### 14 Responses to Challenging Coordinate Geometry Practice Questions

1. D September 14, 2016 at 9:41 am #

Are these GMAT level questions? The calculations involved in some of the questions are very lengthy. The questions are of average level difficulty, but I am not able to apply any shortcuts

• Magoosh Test Prep Expert September 21, 2016 at 6:19 am #

Hi D,

We think they are GMAT-level, but extra hard ones. If this type of GMAT question appeared, it would be a rarity, not the norm. These are specifically designed to be the top tier of challenging questions. I hope that clarifies! 🙂

2. abdul June 5, 2016 at 8:07 pm #

Question 5 was euphoric. You guys have extremely high quality problems.does magoosh have quant question bank for more practice? .No fancy math required to solve such problems. Just pure mental strength, calmness, and FOCUS. Sometimes, I feel like my brain is exploding,but when I get the solution, the endorphins rush in.

• Magoosh Test Prep Expert June 6, 2016 at 2:26 pm #

What a beautifully written comment, Abdul. Looks like you’ve really been doing your homework not just for GMAT math, but for Verbal and AWA vocabulary. 😉 Magoosh does indeed have a quant question bank, with hundreds of questions. Those questions are available to our GMAT Premium subscribers. (And as I type this, our GMAT Premium memberships are on sale.)

3. Aniket September 16, 2015 at 7:05 am #

Hi Mike,

Let me begin by thanking you for this set of challenging problems.

I had a doubt in the 7th Q; how did we reach (0, -25) starting from point (-15, 0) with a slope of -5/3?

4. Michael May 22, 2015 at 1:07 pm #

Hello, is there a faster way to do Question 4? Doing it by drawing everything out and then calculating the distance will surely take more than 2 minutes?

• Mike McGarry May 23, 2015 at 12:49 pm #

Dear Michael,
I drew everything out in the solution, just to help people visualize, but when you develop your intuition for the coordinate plane, you really could do that entire problem without even touching pencil to paper. For example, you should be able to see that (2,7) and (2, -3) are 10 units apart, and that any other 10 unit distance would give you an isosceles. You should recognize that a point is 10 units away horizontally without a calculation. You should recognize, with a calculation, that a right triangle with legs 6 & 8 will have a hypotenuse of 10. You should recognize the perpendicular bisector quality. It’s not a matter of doing a different calculation. It’s a matter of being fluent in the geometry of the coordinate plan. This problem has three important pieces of insight that, if you remember them, will save you time in a host of other problems.
Does all this make sense?
Mike 🙂

5. Ash October 22, 2014 at 4:48 am #

Hi Mike,

i have a question on the first problem.

Your solution of (–2.5) – (–5.5) = –8 is the x-coordinate doesn’t make sense to me. Do you mean -2.5-5.5=-8?

Can you please clarify. Thank you!

• Mike October 22, 2014 at 12:01 pm #

Dear Ash,
I’m sorry: that was a typo. There should have been a positive 5.5 in the parentheses. I just corrected that in the post. Thank you for pointing this out. Best of luck to you!
Mike 🙂

6. Tushar August 31, 2014 at 2:02 am #

Hi Mike!
is there any generalization we can draw from Q.8?
Any pattern that works..
Please also provide a link to count similar points inside any triangle or similar questions.
Thanks..

• Mike August 31, 2014 at 10:26 am #

Dear Tushar,
I’m happy to respond. 🙂 The short answer is “no.” Like many harder GMAT problems, this one is a little unlike any other. Many of the most difficult GMAT questions are difficult, in part, because there is nothing else like them — when we see the problem, we have never seen anything else like it. That’s part of what’s challenging about these questions.
Having said that, I would say: study the explanation carefully. It may be that some method, or some angle of thinking, discussed here may be helpful in a completely different context. As always in studying solutions, don’t just fix on the concrete steps, what to do in this particular problem. Think about the choices I made in the solution, the ways of framing and organizing the information. Those perceptual choices are what would be most transferable to other situations.
Does all this make sense?
Mike 🙂

• Sid August 13, 2015 at 9:36 am #

Hi Mike,

Thanks a lot for all the solutions. I was wondering if there was a shorter way for the last problem. Here is how I tried it:

From the co-ordinate data, it can be ascertained that the boundary co-ordinates of the circle are: (-8,1) (4,1) (-2,7) and (-2,-5), which form a 12 X 12 square.

As p,q are lattice points and are integers, the lattice points internal to it (excluding the boundary points) would be expressed by the formula n^2 – (n+1). (fiddling with a few squares, this formula can be arrived at).

Now as we have to exclude boundary points on the circle, then the size of the square would be shrunk by one, i.e. 11 X 11. Hence all the internal points within the circle, excluding points on circumference are 11^2 – (11 + 1) = 109

• Mike McGarry August 13, 2015 at 1:46 pm #

Dear Sid,
I’m happy to respond. 🙂 What you have done is essentially an approximation method, and because the answer choices are spread out, estimation is not a bad way to attack this problem. What you have pursued is not an exact solution, and it is only by blind luck that your approach produced the exact OA. Among other things, the square that would include those four boundary points would be a 13 x 13 square, not a 12 x 12 square. You got lucky that this mistake put you more in the vicinity of the correct answer.
Yes, it is always important to be aware of approximation methods to attack any difficult problem. It’s also very important to be precise in distinguishing what’s an approximation method and what’s an exact calculation.
Does all this make sense?
Mike 🙂

• Sid August 13, 2015 at 9:22 pm #

Hi MIke,

Thanks a lot for the reply. Ya i realized the mistake later on…

I actually wanted to try a smaller solution…. 🙂

Hope you could please post some more challenging problems. They seem to be real fun.

Thanks,
Sid

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