GRE_Math_Bible_eBook

Coordinate Geometry 211

Problem Set K:

Easy

1.The slope of the line 2x + y = 3 is NOT the same as the slope of which one of the following lines?

(A)2x + y = 5

(B)x + y/2 = 3

(C)x = –y/2 – 3

(D)y = 7 – 2x

(E)x + 2y = 9

4.In the coordinate system shown, if (b, a) lies in Quadrant III, then in which quadrant can the point (a, b) lie?

(A)I only

(B)II only

(C)III only

(D)IV only

(E)In any one of the four quadrants.

Coordinate Geometry 213

Hard

7.In the figure, the horizontal and vertical lines divide the square ABCD into 16 equal squares as shown. What is the area of the shaded region?

(A)4

(B)4.5

(C)5

(D)6.5

(E)7

y-axis

D(7, 5)

C(4, b)

G(6, 3)

H(2, 3)

8.All the lines in the rectangular coordinate system shown in the figure are either horizontal or vertical with respect to the x-axis. What is the area of the figure ABCDEFGH ?

(A)15

(B)15.5

(C)16

(D)16.5

(E)17

Coordinate Geometry 215

Answers and Solutions to Problem Set K

Easy

1. The slope of a line expressed as y = mx + b is m. Expressing the given line 2x + y = 3 in that format yields y = –2x + 3. Hence, the slope is –2, the coefficient of x. Let’s express each line in the form y = mx + b and pick the line whose slope is not –2.

Choice (A): 2x + y = 5; y = –2x + 5, slope is –2. Reject.

Choice (B): x + y/2 = 3; y = –2x + 6, slope is –2. Reject.

Choice (C): x = –y/2 – 3; y = –2x – 6, slope is –2. Reject.

Choice (D): y = 7 – 2x; y = – 2x + 7, slope is –2. Reject.

Choice (E): x + 2y = 9; y = − 12 x + 92 , slope is − 12 ≠ −2. Accept the choice.

The answer is (E).

2. Any two points and their midpoint always lie on the same line. So, the slope of AC equals the slope of BC. Hence, the columns are equal, and the answer is (C).

Medium

3.The formula for the slope of a line passing through two points (x1, y1) and (x2, y2) is y2 − y1 . Using this

x2 − x1

formula to calculate the slope in each column yields

Since 4/3 > 6/5, Column A is greater than Column B. The answer is (A).

4.We are given that the point (b, a) lies in Quadrant III. In this quadrant, both x- and y-coordinates are negative. So, both b and a are negative. So, the point (a, b) also lies in the same quadrant (since both x- and y-coordinates are again negative). Hence, the answer is (C).

5.Since point A(m, n) is in Quadrant II, the x-coordinate m is negative and the y-coordinate n is positive.

Hence, in point B(n, m), the x-coordinate is positive, and the y-coordinate is negative. So, point B must be in Quadrant IV. The answer is (D).

Coordinate Geometry 217

Hard

7. Since each side of the larger square measures 4 units and is divided by the four horizontal lines, the distance between any two adjacent horizontal lines must be 1 unit (= 4 units/4). Similarly, the larger square is divided into four vertical lines in the figure and any two adjacent lines are separated by 1 unit (= 4 units/4).

4

D Q C

The shaded region can be divided into two sub-regions by the line FE, forming ∆FPE and ∆FQE.

Now, the line-segment FE is spread across the two opposite sides of the larger square and measures 4 units. The altitude to it, shown in the bottom triangle, is spread vertically across the three horizontal segments and therefore measures 3 units.

Hence, the area of the lower triangle = 1/2 × base × height = 1/2 × 4 × 3 = 6.

The shaded region above the line FE is one of the four sections formed by the two diagonals of the rectangle ABEF and therefore its area must equal one-fourth the area of rectangle ABEF. The area of rectangle ABEF is length × width = FE × AF = 4 × 1 = 4. Hence, the area of the shaded region above the line FE is 1/4 × 4 = 1.

Summing the areas of the two shaded regions yields the area of the total shaded region: 1 + 6 = 7. The answer is (E).

218 GRE Math Bible

8. Drop a vertical line from C on to the side GH (to meet at, say, P), and draw a horizontal line from G on to the side DE (to meet at, say, Q). The resultant figure is as follows:

Since the point P is horizontal to the point H, its y-coordinate equals the y-coordinate of H, which is 3. Also, since P is vertical to the point C, its x-coordinate equals the x-coordinate of the point C, which is 4. Hence, the point P is (4, 3).

Similarly, the point Q is vertical to the point D and therefore its x-coordinate equals the x-coordinate of D, which is 7. Also, the point Q is horizontal to the point H and therefore takes its y-coordinate. Hence, the point Q is (7, 3).

Now, the shaded region in the given figure is the sum of the three rectangles ABPH, CDQP, and GQEF.

The area of the rectangle ABPH is length width =

BP AB =

(Difference in y-coordinates of B and P) (Difference in x-coordinates of A and B) =

(7 – 3)(4 – 2) =

4 2 =

8

The area of the rectangle CDQP is length width =

CD DQ =

(Difference in x-coordinates of C and D) (Difference in y-coordinates of D and Q) =

(7 – 4)(5 – 3) =

3 2 =

6

The area of the rectangle GQEF is length width =

GQ QE =

(Difference in x-coordinates of G and Q) (Difference in y-coordinates of Q and E) =

(7 – 6)(3 – 0) =

1 3 =

3

Hence, the total area is 8 + 6 + 3 = 17 sq. units. The answer is (E).

Coordinate Geometry 219

9. From the figure, since EC is perpendicular to the x-axis, C is a point vertically below point B. Hence, both have the same x-coordinate.

The line AC is horizontal and therefore its length equals the x-coordinate difference of A and C, which equals 6 – 0 = 6.

The line DC is horizontal and therefore its length equals the x-coordinate difference of D and C, which is 6 – 2 = 4.

The length of the vertical line BC equals the y-coordinate difference of B and C, which is 3 – 0 = 3.

The length of the vertical line EC equals the y-coordinate difference of E and C.

Now, in ∆ABC, A = a°, B = 90° – a°, and C = 90°. The sides opposite angles A and B are in the ratio BC/AC = 3/6 = 1/2.

Similarly, in ∆DEC, E = a°, D = 90° – a°, and C = 90° (So, ABC and DEC are similar triangles and their corresponding sides are proportional). Hence, the sides opposite angles E and D are in the ratio DC/EC = 1/2. Hence, we have

DC/EC = 1/2

EC = 2DC EC = 2 4 = 8

Hence, the y-coordinate of point E is 8, and the answer is (E).

10. If the two graphs in figures 1 and 2 were represented on a single coordinate plane, the figure would look like this:

lines have same slope (1), they must be parallel.

Now, the point above the origin (x = 0) on the line f(x) is (0, 2) and is 2 units above the origin. But, the corresponding point (above origin, i.e., above x = 0) on g(x) is (0, 1) and is only 1 unit above the origin. This combined with the fact that the lines f(x) and g(x) are parallel shows that each point on f(x) for a given value of x is above the corresponding point on g(x). Hence, for any given value of x, f(x) is greater than g(x). Hence, f(10) must be greater than g(10). Hence, Column A > Column B, and the answer is (A).