Cracking the SAT subject test in math 2, 2nd edition

Cracking the SAT Subject Test in Math 2, 2nd Edition CHAPTER 8 COORDINATE GEOMETRY DRILL EXPLANATIONS Drill 1 The Coordinate Plane 1 Point E, quadrant II 2 Point A, quadrant I 3 Point C, quadrant IV 4[.]

CHAPTER 8: COORDINATE GEOMETRY DRILL EXPLANATIONS

Drill 1: The Coordinate Plane

1 Point E, quadrant II

2 Point A, quadrant I

3 Point C, quadrant IV

4 Point D, quadrant III

5 Point B, quadrant I

Drill 2: The Equation of a Line

1 A You can plug the line’s slope m and the given point (x, y) into

the slope-intercept equation to get 1 = 0.6(3) + b So b = −0.8.

So the equation is y = 0.6x − 0.8.

To find the point that is also on this line, go to each answer

choice and plug the x-coordinate into the formula You’ll have the right answer when the formula produces a y-coordinate that

matches the given one

2 E Once again, get the line into slope-intercept form, y = 5x − 4.

Then Plug In zero You get a y-value of −4 Notice that this is the y-intercept (the value of y when x = 0).

4 B Put the line into the slope-intercept formula by isolating y So y

= −3x + 4 and the slope is −3.

6 D You can figure out the line formula (y = mx + b) from the graph.

The line has a y-intercept (b) of −2, and it rises 6 as it runs 2, giving it a slope (m) of 3 Use those values of m and b to test the

statements in the answer choices

Drill 3: Slope

1 C Use the slope formula on the point (0, 0) and

5 D Draw it Remember that perpendicular lines have slopes that

are negative reciprocals of each other A line containing the origin and the point (2, −1) has a slope of − The perpendicular line must then have a slope of 2 Quickly move through the answer choices, determining the slope of a line passing through the given point and the origin The one that gives a slope of 2 is correct

13 A Once again, the slope-intercept formula is your most powerful

tool Isolate y, and you get y = −3x + 5 The line must then have

a slope of −3 and a y-intercept of 5 Only (A) and (D) show lines

with negative slope, and the line in (D) has a slope which is

between −1 and 0, because it forms an angle with the x-axis that

is less than 45°

37 D Remember that perpendicular lines have slopes that are

negative reciprocals of each other The slopes of x and y are therefore negative reciprocals—you can think of them as x and

− The difference between them will therefore be the sum of a

number and its reciprocal: If x = 1, then the sum of x and its reciprocal is 2; if x = 5, then the sum of x and

its reciprocal is 5.2; but no sum of a number and its reciprocal can be less than 2 A sum of 0.8 is impossible

Drill 4: Line Segments

2 D Use the distance formula on the points (−5, 9) and (0, 0)

6 E Drawing a rough sketch and approximating allows you to

eliminate (A) and (B) Then, plug the points you know into the

midpoint formula and PITA for the coordinates of B The average of −4 and the x-coordinate of B is 1 The average of 3 and the y-coordinate of B is −1 That makes B the point (6, −5).

12 D You’ll essentially be using the distance formula on (2, 2) and the

points in the answer choices (−5, −3) is the point at the greatest distance from (2, 2)

28 E To find the area of the circle, you need the radius Since the

given line segment goes through the center, that segment is a diameter Find the length of the diameter using the distance formula: Divide by 2 to find the radius, which is 11.102 Finally, use the formula for area of a circle: π(11.102)2 = 387.201, which is (E)

21 C This is a quadratic function, which always produces a parabola

If a parabola has a maximum or minimum, then that extreme

value is the parabola’s vertex Just find the vertex The

x-coordinate is , which is 3 in this case The y-coordinate will

be f(3), or −1.

22 B Use that vertex formula again The x-coordinate is , which

is −1 in this case That’s enough to get you the right answer (If

you needed the y-coordinate as well, you’d just Plug In x = −1 and solve for y.)

25 D At every point on the x-axis, y = 0 Plug each of the answer

choices in for x, and see which one gives you y = 0 You could also put in 0 for y and solve for x; the solutions are 1 and 5.

36 B The parabola opens downward, so you know in the standard

form of the equation y = a(x − h)2 + k that a must be negative; eliminate (E) The axis of symmetry is negative, so h must be

negative This means in the standard form of the equation the

parentheses must be (x + h); eliminate (A) and (C) Finally, the vertex of the parabola is positive, so k must be positive;

eliminate (D) and choose (B)

18 B Just plug each point into the equation The one that does not

make the equation true is not on the circle

20 E If S and T are the endpoints of a diameter, then the distance

between them is 8 If they are very close to each other on the circle, then the distance between them approaches zero The

distance between S and T cannot be determined.

45 C Notice that because the y’s equal 0, you can cancel out the y’s in

all the answer choices Plug In the points: (2, 0) works in (C), (D), and (E), but (10, 0) works in (A), (B), and (C) It must be (C)

50 A First, rewrite the equation into the Standard Form Begin by

rearranging the equation so your x-terms and y-terms are

together and the constants are on the other side of the equation:

x2 + 4x + y2 + 8y = −4

Next, complete the square for both the x-terms and the y-terms.

To do so, take the b coefficient, divide by 2, square the result,

and add that to each term and to the other side of the equation:

(x2 + 4x + 4)+ (y2 + 8y +16) = −4 + 4 + 16

You have just created two perfect squares Factor as follows:

(x + 2)2 + (y + 4)2 =16

Now the equation is in the standard form The center of the circle is at (−2, −4) and the circle has a radius of 4 You are looking for a circle with the center in the third quadrant and

tangent to the x-axis: (A) meets these criteria.

15 E Because a = 4 and b = 5, the minor axis is 2(4) = 8 and the

major axis is 2(5) = 10

40 C For an ellipse in its general form, the center is (h, k), which in

this case is (−5, 3)

45 E There is no denominator under the y2, so it is understood to be

1 Therefore, the form of this ellipse centered at the origin is

Because the a2 value (denominator of the x-term) is greater than the b2 value (denominator of the y-term), the

ellipse’s major axis is horizontal Eliminate (A), (B), and (D) You can now plug the points given into the original equation

Alternatively, if you can remember that if a2 = 16 then a = 4, then you know the distance along the horizontal axis is 2a = 8.

Therefore the distance from the origin is 4 Either way, the answer is (E)

Drill 8: General Equations (hyperbolas)

38 B Like circles and ellipses, hyperbolas in general form have their

centers at (h, k) This one is centered at (−4, −5).

45 A If the hyperbola is opening to the left and right, the form of the

equation is , where the center is (h, k).

Eliminate (B) and (D), since those hyperbolae open up and

down Eliminate (E) because it isn’t the equation of a hyperbola: rather, it is the equation of an ellipse The center of

the hyperbola shown is in the fourth quadrant; h should be positive and k should be negative, which means the sign in the

x-term should be negative and the sign in the y-term should be

positive Choice (A) fits these criteria

Drill 9: Triaxial Coordinates

14 C This is once again a job for the Super Pythagorean Theorem,

which is simply another version of the 3-D distance formula It’s just like finding the long diagonal of a box which is 5 by 6 by 7

Set up this equation: d2 = 52 + 62 + 72, and solve

19 B A point will be outside the sphere if the distance between it and

the origin is greater than 6 Use the Super Pythagorean Theorem to measure the distance of each point from the origin

37 D First, Point D is not on the y-axis, so its x-coordinate cannot be

0; eliminate (A) and (B) Point B shares x- and y-coordinates with point D directly below it Make the origin point O Triangle

OCD will have a right angle at and are each edges of the cube with length 5 This is an isosceles (45-45-90) right triangle; therefore, will have length 5 = 7.071 Point D will have a y-coordinate of half the length of , or 3.536; eliminate (C) and (E) and choose (D)

Comprehensive Coordinate Geometry Drill

1 A The x-intercept must have a y-value of 0; eliminate (E) Next,

use the slope formula to find the slope of the line:

Next, use this slope, one of the given points,

and the x-intercept coordinates (x, 0) in the slope formula to solve for the x-intercept: Cross-multiply: 44 +

11x = 414, so x = 33.636 The x-intercept is therefore (33.636,

0), which is (A)

2 C First, find the slope of the given line by solving for y:

4x + 7y = 23

7y = 4x + 23

y=− x +

The slope of this line is the x coefficient: − A perpendicular

line will have a slope which is the negative reciprocal of this line, , which is (C)

5 D You need an equation that results in a hyperbola that opens

vertically Choices (A) and (B) describe parabolas and would give you only half of what you need ((A) could be the part of the

graph above the x-axis, but not below; (B) could be the part of the graph below the x-axis, but not above); eliminate (A) and

(B) Choice (E) describes an ellipse, because you are adding the

two terms; eliminate it If the x-term is positive, the hyperbola

opens horizontally; eliminate (C) and choose (D)

12 B The x-value of the vertex of a parabola given in the general form

of the equation, y = ax2 + bx + c, is found by using This is

Eliminate (A), (D), and (E) To find the y-value, plug this x-value into the original equation: −3(0.833)2 +

5(0.833) −11 = −8.917 These x- and y-values match (B).

18 A Start by finding the center of the circle by finding the midpoint

of the diameter shown: Because the

standard form of the equation of a circle is (x − h)2 + (y − k)2 =

r2, the answer must have (x − 10)2 + (y + 6.5)2; eliminate (B),

(D), and (E) Next, find the radius by finding the distance between the center and one of the points on the circle:

The right side of the equation must be radius squared, or 42.25; choose (A)

Next, use the distance formula to find the distance between (−0.75, 8) and (7, −5):

, (D)

25 A The equation for an ellipse centered on the origin is

where 2a is the width and 2b is the height of the ellipse If the ellipse contains the point (4, 0) as its widest point on the x-axis,

then its width must be 8, which means a = 4 The equation

needs to include ; only (A) fits this criteria Alternatively, you

can plug the x- and y-values of the points given into each

equation and eliminate the equations that are not true with the given values

28 E First, because there is no negative sign before the (x − 2)2, the

parabola should open up; eliminate (A) and (B) Next, (x − 2)2 would move the vertex to the right of the y-axis; eliminate (C)

because the vertex is to the left Finally, ≤ means that the shaded region should be below the parabola; eliminate (D) and choose (E)

36 B If the point is on the y-axis, the x value must be 0; eliminate

(A) Next, you have a few options First, you can draw and ballpark; the only answer which is not obviously closer to (0, 2) than (6, 0) is (B) Or, you can Plug In The Answers; start with (C) or (D) and find the distance from (0, 2) and (6, 0) for each answer

Finally, mathematically, the distance from (0, 2) will be the

difference in y-coordinates, or y − 2 The distance from (6, 0)

can be found using the distance formula:

Set these two equations equal to each other and solve:

y2 − 4y + 4 = 36 + y2

y = −8, so (B)

42 E If point E is at (0, 3.5), then the y-coordinates of both point B

and point C must also be 3.5 Use the function given to find the

x-coordinate of points B and C by making f(x) = 3.5:

The distance from point B to point C will be the difference in their x-values, which is 3.606 The distance from point A to point B will be the y-value of 3.5 To find the area of the

rectangle, multiply length times width: 3.606 × 3.5 = 12.619, (E)

49 E Every point on a sphere with radius 5 will be 5 units away from

the center of the sphere You could use the 3D distance formula

, but there’s something else going on here Each point in the answers shares

at least one coordinate with the center of the sphere, so you can look at the other two points and use the simpler 2D distance

formula For (A) you don’t even need to go that far: the y- and

z-coordinates are the same as the center, and the x-coordinate

differs by 5 Similarly, (B) shares the x- and z-coordinates and

differs from the center by 5 along the y-axis: eliminate (A) and (B) Choices (C) and (D) both share an x-coordinate with the center, so you can find the difference between the y- and

z-coordinates of each point and the center of the circle In both cases, you create a 3:4:5 Pythagorean triple; these points are a distance of 5 from the center of the circle, so eliminate (C) and (D) and choose (E)