{"id":4443,"date":"2024-06-19T12:00:14","date_gmt":"2024-06-19T19:00:14","guid":{"rendered":"https:\/\/magoosh.com\/gmat\/?p=4443"},"modified":"2024-06-03T16:45:21","modified_gmt":"2024-06-03T23:45:21","slug":"challenging-gmat-coordinate-geometry-practice-questions","status":"publish","type":"post","link":"https:\/\/magoosh.com\/gmat\/challenging-gmat-coordinate-geometry-practice-questions\/","title":{"rendered":"Challenging GMAT Coordinate Geometry Practice Questions"},"content":{"rendered":"

\"\"<\/p>\n

This post was updated in 2024 for the new GMAT.<\/strong><\/p>\n

Earlier, we featured a blog of Coordinate Geometry practice questions<\/a>.\u00a0 Here are eight more questions, some of which are challenging.<\/p>\n

1) Graph G has a line of symmetry of x = \u20135\/2.\u00a0 Graph G passes through the point (3, 3).\u00a0 What is the x-coordinate of another point that must have a y-coordinate of 3?<\/p>\n

(A) \u20138<\/p>\n

(B) \u20137<\/p>\n

(C) \u20135<\/p>\n

(D) \u20134<\/p>\n

(E) 2<\/p>\n

\"cgpq_img1\"<\/p>\n

2) In the figure above, the point on segment JK that is four times as far from K as it is from J is:<\/p>\n

\"cgpq_img2\"<\/p>\n

3) Which point is the reflection of the point (\u20137, 5) over the line y = \u2013x?<\/p>\n

(A) (\u20137, 5)<\/p>\n

(B) (\u20135, 7)<\/p>\n

(C) (5, \u20137)<\/p>\n

(D) (7, \u20135)<\/p>\n

(E) (7, 5)<\/p>\n

4) In a coordinate system, P = (2, 7) and Q = (2, \u20133).\u00a0 Which could be the coordinates of R if PQR is an isosceles triangle?<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 I. (12, \u20133)<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 II. (\u20136, \u20139)<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 III. (\u2013117, 2)<\/p>\n

(A) I only<\/p>\n

(B) I and II only<\/p>\n

(C) I and III only<\/p>\n

(D) II and III only<\/p>\n

(E) I, II, and III<\/p>\n

5) Point W = (5, 3).\u00a0 Circle J has a center at point W and radius of r = 5.\u00a0 This circle intersects the y-axis at one intercept and the x-axis at two intercepts.\u00a0 What is the area of the triangle formed by these three intercepts?<\/p>\n

(A) 7.5<\/p>\n

(B) 12<\/p>\n

(C) 15<\/p>\n

(D) 24<\/p>\n

(E) 30<\/p>\n

6) Line M has a y-intercept of \u20134, and its slope must be an integer-multiple of 1\/7.\u00a0 Given that Line M passes below (4, \u20131) and above (5, \u20136), how many possible slopes could Line M have?<\/p>\n

(A) 6<\/p>\n

(B) 7<\/p>\n

(C) 8<\/p>\n

(D) 9<\/p>\n

(E) 10<\/p>\n

7) Line Q has the equation 5y \u2013 3x = 45.\u00a0 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?\u00a0 (Note: Intersections on one of the axes do not count.)<\/p>\n

(A) 25<\/p>\n

(B) 33<\/p>\n

(C) 36<\/p>\n

(D) 41<\/p>\n

(E) 58<\/p>\n

8) In the x-y plane, point (p, q) is a lattice point if both p and q are integers.\u00a0 Circle C has a center at (\u20132, 1) and a radius of 6.\u00a0 Some points, such as the center (\u20132, 1), are inside the circle, but a point such as (4, 1) is on<\/i> the circle but not in<\/i> the circle.\u00a0 How many lattice points are in circle C?<\/p>\n

(A) 36<\/p>\n

(B) 72<\/p>\n

(C) 89<\/p>\n

(D) 96<\/p>\n

(E) 109<\/p>\n

Some relevant blogs<\/h2>\n

Here are some blogs that you may find germane.\u00a0 If you look at one of these and have an “aha!”, then you might want to look over these problems again.<\/p>\n

1) Quadrants in the x-y plane<\/a><\/p>\n

2) Special Properties of the line y = x<\/a><\/p>\n

3) Distance between two points<\/a><\/p>\n

4) Slopes<\/a><\/p>\n

5) Midpoints and Parallel & Perpendicular lines<\/a><\/p>\n

6) The Pythagorean Theorem<\/a><\/p>\n

7) Pythagorean Triplets<\/a><\/p>\n

If you found this problems helpful or have a question about any after reading the TE, please let us know in the comments section.<\/p>\n

\"cgpq_img3\"<\/p>\n

 <\/p>\n

Explanations to the practice questions<\/h2>\n

1) The line of symmetry is x = \u20132.5.\u00a0 The point (3, 3) is 3 \u2013 (\u20132.5) = 5.5 to the right of this line of symmetry.\u00a0 It’s reflection must be 5.5 units to the left of the line of symmetry, so (\u20132.5) \u2013 (5.5) = \u20138 is the x-coordinate.\u00a0 Answer = (A)<\/b><\/p>\n

2) Call the point P.\u00a0 Then, PK = 4*JP.\u00a0 Of course, JP + PK = JK, so JP + 4*JP = 5*JP = JK, and JP = (1\/5)*JK. \u00a0Point P should be one fifth of the segment away from J.\u00a0 This would be the point (\u20131, 3).\u00a0 Answer = (A)<\/b><\/p>\n

3) The point (\u20137, 5) is in the second quadrant, relatively close to the line y = \u2013x, 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)<\/b>, (\u20135, 7).\u00a0 This question is very easy to solve by visual\/spatial reasoning.<\/p>\n

For those who want to know the “rule”: when a point is reflected over the line y = \u2013x, the coordinates are reversed, and each one takes on its opposite sign.\u00a0 The \u20137 becomes +7, and the +5 becomes \u20135, and they switch places, which also results in (B)<\/b>.<\/p>\n

4) Notice that points P & Q are separated by 10 units vertically.<\/p>\n

I. R = (12, \u20133)<\/p>\n

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

\"cgpq_img4\"<\/p>\n

Option I works.<\/p>\n

II.\u00a0 R = (\u20136, \u20139)<\/p>\n

This is tricky.\u00a0 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.\u00a0 This little 6-8-10 triangle on the QR connection is show in dashed lines:<\/p>\n

\"cgpq_img5\"<\/p>\n

Option II works.<\/p>\n

III.\u00a0 R = (\u2013117, 2)<\/p>\n

It’s a mistake to do any calculations at all with this one.\u00a0 The line y = 2 is the perpendicular bisector of PQ:<\/p>\n

\"cgpq_img6\"<\/p>\n

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.\u00a0 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.\u00a0 The point R = (\u2013117, 2) is on this perpendicular bisector, so it is equidistant from P & Q, and PQR must therefore be isosceles.<\/p>\n

Option III works.<\/p>\n

Each of the three options works.\u00a0 Answer = (E)<\/b>.<\/p>\n

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.<\/p>\n

Now, think about the two x-intercepts.\u00a0 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.<\/p>\n

\"cgpq_img7\"<\/p>\n

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.\u00a0 One is at (1, 0) and the other is at (9, 0).<\/p>\n

Now, think about the triangle formed by these three intercepts.\u00a0 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.\u00a0 A = (0.5)bh = (0.5)(8)(3) = 12.\u00a0 Answer = (B)<\/b><\/p>\n

6) Well, for starters, zero is a multiple of every number, and a line with slope zero, the horizontal line y = \u20134 passes below (4, \u20131) and above (5, \u20136).\u00a0 That’s horizontal line is our starting point.<\/p>\n

The point (4, \u20131) is over 4, up 3 from the y-intercept (0, \u20134).\u00a0 A line with a slope of +3\/4 would go straight from (0, \u20134) to (4, \u20131).\u00a0 Thus, we need a slope that is less than +3\/4.\u00a0 Notice that 3\/4 = 21\/28.\u00a0 Notice that 5\/7 = 20\/28, so this would be less than 3\/4.\u00a0 Therefore, +1\/7 through +5\/7 will all slope up, obviously above (5, \u20136), and all will pass below (4, \u20131).\u00a0 That’s five upward sloping lines.<\/p>\n

The point (5, \u20136) is over 5, down 2, from the y-intercept (0, \u20134).\u00a0 A line with a slope of \u20132\/5 would go straight from (0, \u20134) to (5, \u20136).\u00a0 Thus, we need a slope that is more than \u20132\/5; another way to say that is, we need a negative slope whose absolute value is less than +2\/5.\u00a0 Well, 2\/5 = 14\/35, while 2\/7 = 10\/35 and 3\/7 = 15\/35, so (2\/7) < (2\/5) < (3, 7).\u00a0 The negatively sloping lines obviously pass below (4, \u20131), but only two of them, \u20131\/7 and \u20132\/7, pass above (5, \u20136).<\/p>\n

That’s one horizontal line, five upward sloping lines, and two downward sloping lines, for a total of eight.\u00a0 Answer = (C)<\/b>.<\/p>\n

\"cgpq_img8\"<\/p>\n

7)\u00a0 First of all, Line Q 5y \u2013 3x = 45 re-written in slope-intercept form is y = (3\/5)x + 9.\u00a0 Line Q has a y-intercept of +9, so if Line S also has a y-intercept of\u00a0 +9, they would intersect on the y-axis, not in the second quadrant.\u00a0 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.<\/p>\n

Now, let’s think about the bottom end.\u00a0 Line Q has a y-intercept of +9 and an x-intercept of \u201315.\u00a0 Line S, perpendicular to Line Q, must have a slope of \u20135\/3, the negative reciprocal of Line Q’s slope.\u00a0 If we start at the x-intercept of point Q, the point (\u201315, 0), we would follow the slope of \u20135\/3 over 15 and down 25 to (0, \u201325).\u00a0 If Line S had a y-intercept of \u201325, it would intersect Line Q at (\u201315, 0), on the x-axis, and not in the second quadrant.\u00a0 Therefore, a y-intercept of \u201325 doesn’t work, and the lowest value that would work is one above it, y-intercept of \u201324.<\/p>\n

Any y-intercept less than or equal to +8 and greater than or equal to \u201324 would work.\u00a0 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.\u00a0 Answer = (B)<\/b><\/p>\n

8) This is a tough question.\u00a0 First of all, obviously, the center a lattice point in the circle.\u00a0 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<\/i><\/b> the circle don’t count as being “in” the circle.<\/p>\n

Now, we just are going to focus on one quadrant of the circle, the upper right quadrant.\u00a0 Suppose we go over one unit and up the circle.<\/p>\n

\"cgpq_img9\"<\/p>\n

Well, that’s a right triangle, with hypotenuse of r = 6, and horizontal leg of 1, so\u00a0 if the vertical leg is x, then<\/p>\n

\"cgpq_img10\"<\/p>\n

Well, the square root of 35 is bigger than 5, so five points in that column are inside the circle.\u00a0 Now, move two units to the right.\u00a0 Again, hypotenuse r = 6, short leg = 2, vertical leg = x, so<\/p>\n

\"cgpq_img11\"<\/p>\n

Still bigger than 5, so there five points in this second column.\u00a0 Now, move three units to the right.\u00a0 Again, hypotenuse r = 6, short leg = 3, vertical leg = x, so<\/p>\n

\"cgpq_img12\"<\/p>\n

Still bigger than 5, so there five points in this third column.\u00a0 Here’s where we are so far:<\/p>\n

\"cgpq_img13\"<\/p>\n

Clearly, by symmetry, we can reflect this array over the 45\u00b0 line, to get more points in the circle:<\/p>\n

\"cgpq_img14\"<\/p>\n

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.\u00a0 We\u00a0 have to check that highest point in the column 4 units to the right.\u00a0 Again, hypotenuse r = 6, short leg = 4, vertical leg = x, so<\/p>\n

\"AAAA\"<\/p>\n

Clearly, this is bigger than 4, so that fourth column can be four points high.<\/p>\n

\"cgpq_img16\"<\/p>\n

That completes the quadrant.\u00a0 Within the quadrant, that\u2019s 5 + 5 + 5 + 4 + 3 = 22 points.\u00a0 Multiply that by four, and that’s 88 points within the four quadrants.\u00a0 Add the 20 points on the vertical & horizontal segments, and the one point at the center, and we have 88 + 20 + 1 = 109.\u00a0 Answer = (E)<\/b>.<\/p>\n

\"cgpq_img17\"<\/p>\n","protected":false},"excerpt":{"rendered":"

This post was updated in 2024 for the new GMAT. Earlier, we featured a blog of Coordinate Geometry practice questions.\u00a0 Here are eight more questions, some of which are challenging. 1) Graph G has a line of symmetry of x = \u20135\/2.\u00a0 Graph G passes through the point (3, 3).\u00a0 What is the x-coordinate of […]<\/p>\n","protected":false},"author":26,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[124],"tags":[],"ppma_author":[13209],"class_list":["post-4443","post","type-post","status-publish","format-standard","hentry","category-geometry"],"acf":[],"yoast_head":"\nGMAT Coordinate Geometry Practice | Magoosh Study Resources<\/title>\n<meta name=\"description\" content=\"GMAT coordinate geometry problems stumping you? 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