SAT 8

94	PART II / MATH REVIEW

If the scale factor of two similar solids is m:n, then the following ratios are true.

1.The ratio of their corresponding base perimeters or circumferences is also m:n.

2.The ratio of their base areas, total surface areas, and lateral surface areas is m2:n2.

3.The ratio of their volumes is m3:n3.

E X A M P L E :

The lateral surface areas of two similar right cylinders are 80π in3 and 120π in3, respectively. Find the ratio of their volumes.

First determine the scale factor of the similar cylinders by comparing their lateral surface areas.

80π:120π simplifies to 8:12, which simplifies to 2:3. Therefore, the scale factor is 2:3.

Now, find the volume ratio by cubing the scale factor:

23:33.

8:27. (Answer)

*COORDINATES IN THREE DIMENSIONS

Similar to graphing points (x, y) in a two-dimensional xy-coordinate plane, the point (x, y, z) can be graphed in a three-dimensional coordinate system having an x-axis, y-axis and z-axis. For example, the point (2, −3, 1) is a solution to the equation 2x + y − z = 0. The graph of the ordered triple is as shown:

z

(2,–3,1)

y

x

The distance between ordered triples (x1, y1, z1) and (x2, y2, z2) is given by the formula:

Distance = ( x2 − x1)2 + ( y2 − y1)2 + ( z2 − z1)2 .

CHAPTER 5 / SOLID GEOMETRY	95
E X A M P L E :

What is the distance between (0, 0, 7) and (4, 1, 0)?

The problem is asking for the length of the segment connecting the two points.

Simply apply the distance formula to get:

(0,0,7)

(4,1,0)

d= (4 − 0)2 + (1 − 0)2 + (0 − 7)2 .

d= 42 + 12 + (− 7)2 .

d= 16 + 1 + 49.

d = 66 ≈ 8.12. (Answer)

E X A M P L E :

A right circular cylinder has a height of 10 and a radius of 4. If X and Y are two points on the surface of the cylinder, what is the maximum possible length of XY?

Although this problem doesn’t ask about points specifically in the form (x, y, z), it does ask you to visualize points in three-dimensions. Think of X and Y as vertices of the right triangle shown below.

4	x

10

y

8

XY = 82 + 102

XY = 64 + 100

XY = 164 = 2 41 ≈ 12.81

(Answer)

CHAPTER 6

COORDINATE GEOMETRY

This chapter provides a review of coordinate geometry principles. On the Level 2 test, 10–14% of the questions relate specifically to coordinate geometry. Coordinate geometry focuses on graphing in an xy-coordinate plane. The pie chart shows approximately how much of the Level 2 test is related to coordinate geometry:

Numbers and

Operations

12% Algebra

20%

Data Analysis,

Statistics, and

Probability

8%

Solid

Geometry

4%

Coordinate

Geometry

12%

Functions

30%

Trigonometry

14%

The following geometry topics are covered in this chapter:

1.Plotting Points

2.Midpoint

3.Distance

4.Slope

5.Slope of Parallel and Perpendicular Lines

6.Equations of Lines

a.Horizontal and Vertical Lines

b.Standard Form

c.Point-Slope Form

d.Slope-Intercept Form

e.Determining x and y Intercepts

7.Circles

8.Parabolas

96

CHAPTER 6 / COORDINATE GEOMETRY

97

9.*Ellipses

10.*Hyperbolas

11.Graphing Inequalities

12.Graphing Absolute Value

13.Symmetry

14.Transformations

15.*Polar Coordinates

PLOTTING POINTS

A plane rectangular coordinate system is created by drawing two axes that intersect at right angles at an origin. The horizontal axis is called the x-axis and the vertical axis is called the y-axis. The axes create 4 quadrants in the plane, which are numbered using Roman numerals as shown:

y

II

I

x

III IV

An ordered pair (x, y) denotes a unique point in the plane. To plot a point, simply graph the location of its x and y coordinates. The x-coordinate of a point is also referred to as the abscissa, while the y-coordinate is also referred to as the ordinate.

E X A M P L E :

Graph the following points in the same coordinate plane: A(1, 4), B(−2, −3), C(−5, 1), D(4, 0) (see graph on page 98).

*Denotes concepts that are on the Level 2 test only.

98															PART II			/	MATH REVIEW
										y
								12
								11
								10
								9
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								5			A
								4
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				C				2
								1
														D
x																					x
									0
–12 –11 –10 –9	–8	–7	–6	–5	–4	–3	–2	–1		1	2	3	4	5	6	7	8	9	10	11	12
								–1
								–2
							B –3
								–4
								–5
								–6
								–7
								–8
								–9
								–10
								–11
								–12
										y

E X A M P L E :

Find three solutions of 2x + 3y = 1.

The graph of an equation in two variables is the set of all points whose x and y coordinates satisfy the equation. To find three solutions to this equation, simply choose 3 values for x and solve for the corresponding value of y. Write y in terms of x, and create an xy-table to determine solutions of the equation.

x	y =		1 − 2x						y
					3
0	y =	1			− 2(0)			1
					3				3
2	y =	1			− 2(2)			−1
					3
−1	y =		1		− 2(−1)			1
					3

	CHAPTER 6 / COORDINATE GEOMETRY			99
	The graph of 2x + 3y = 1 is a line, so there are infinitely many points that
		will satisfy the equation. Here three possible points are given.
		1
	1,		, (2, −1), (−1, 1) are three possible solutions.	(Answer)
		3

MIDPOINT

The midpoint of a segment with endpoints (x1, y1) and (x2, y2) is the point:

	x1 + x2		,	y1 + y2
		2		2

Finding the midpoint of a line segment can also be thought of as finding the average of the x and y coordinates.

E X A M P L E :

___

Find the midpoint AB given the endpoints A(1, 6) and B(−3, −7).

Substitute the given coordinates into the midpoint formula to get:

	1 + −3				6 + −7
				,			.
	2				2
	−2	,	−1
	2			.
			2
	−1,	−1				(Answer)
			.
		2

E X A M P L E :

___

M is the midpoint of AB . Find the coordinates of B if A has coordinates (3, 8) and M has coordinates (−4, 0).

Let the coordinates of B be (x2, y2), then:

	3 + x2	= −4	and	8 + y2	= 0.
	2			2
	3 + x2 = −8			8 + y2 = 0.
	x2	= −11		y2 = −8.
	B(−11,−8).		(Answer)

DISTANCE

The distance between any two points (x1, y1) and (x2, y2) is:

d = (x2 − x1 )2 + ( y2 − y1 )2 .

Some aspects of coordinate geometry, simply involve solving a plane geometry question by using an xy-coordinate plane. Here’s an example of one.

100 PART II / MATH REVIEW

E X A M P L E :

Find the perimeter of ABC given its vertices are A(2, 2), B(−1, 5), and C(−5, 2).

This problem requires applying the distance formula three times to find the length of the three sides of the triangle. Solve for AB, BC, and AC

as follows:
AB =	(2 − −1)2 + (2 − 5)2	=	(3)2 + (−3)2	=	9 + 9 =	18 = 3 2.
BC =	(−1 − −5)2 + (5 − 2)2	=	(4)2 + (3)2	=	16 + 9 =	25 = 5.
AC =	(2 − −5)2 + (2 − 2)2	=	(7)2 + (0)2	=	49 = 7.

___

Notice that AC is a horizontal line segment, so the distance from A to C can be easily found be determining the change in the x coordinates.2 − −5 = 7. The perimeter of the triangle is:

7 + 5 + 3	2.
12 + 3 2.	(Answer)

The concept of distance can also be applied to finding the area of figures in a coordinate plane. Use the distance formula to find the length of the base and/or height of a figure, and substitute these values into the appropriate area formula. (See Chapter 5 for a review of area formulas.)

E X A M P L E :

Find the area of parallelogram ABCD given its vertices are A(3, 1), B(2, −1), C(−1, −1), and D(0, 1).

Sketch a diagram to help picture the parallelogram.

										y
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								2		D			A
								1
x									0												x
–12 –11 –10 –9	–8	–7	–6	–5	–4	–3	–2	–1		1	2	3	4	5	6	7	8	9	10	11	12
								–1				B
								C –2
								–3
								–4
								–5
								–6
								–7
								–8
								–9
								–10
								–11
								–12

y

CHAPTER 6 / COORDINATE GEOMETRY

101

___

Assume BC is the base of the parallelogram, and the height is the length of the perpendicular from vertex D to side ___BC. Notice that there is no change in the y-coordinate in the segment BC, so the distance from B to C can be found simply by using absolute value.

BC = 2 − −1 = 3.

The height is 1 − −1 = 2.

The area of the parallelogram is therefore:

A = bh = 3(2) = 6

6 units2.

(Answer)

SLOPE

Slope is the measure of the steepness of a line. The slope of a line containing the points (x1, y1) and (x2, y2) is:

slope =	rise	=	change in y	=	y2	− y1	.
	run		change in x x2			− x1

Horizontal lines have no change in y, so the slope of a horizontal line is zero.

m =	0	= 0.

change in x

Vertical lines have no change in x. Because you cannot divide by zero, vertical lines have an undefined slope.

= change in y =

m undefined. 0

A line having positive slope rises from left to right, and a line having negative slope falls from left to right.

y		y
12		12
11		11
10		10
9		9
8		8
7		7
6		6
5		5
4		4
3		3
2		2
1		1
x –12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0	1 2 3 4 5 6 7 8 9 10 11 12	x x –12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0	1 2 3 4 5 6 7 8 9 10 11 12 x
–1		–1
–2		–2
–3		–3
–4		–4
–5		–5
–6		–6
–7		–7
–8		–8
–9		–9
–10		–10
–11		–11
–12		–12
y		y
Positive slope		Negative slope

102	PART II / MATH REVIEW

E X A M P L E :

Find the slope of the line containing the points S(6, −9) and R(−2, −5).

m = y2 − y1 . x2 − x1

m = −5 − −9 . −2 − 6

m = −	4 .
	8
m = −	1	.	(Answer)
	2

In the above example, we subtracted the coordinates of S from R. Order doesn’t matter, however, as long as you choose the same order in the numerator and the denominator. We could have found the slope by subtracting the coordinates of R from S to get the same answer:

	m =	−9 − −5	= −	4	= −	1	.
		6 − −2		8		2

SLOPE OF PARALLEL AND PERPENDICULAR LINES

You can determine whether or not lines are parallel (never intersect) or perpendicular (intersect at a right angle) by examining their slope:

1.Parallel lines have the same slope. m1 = m2.

2.Perpendicular lines have slopes that are opposite reciprocals. m1 m2 = −1.

E X A M P L E :

___

Find the slope of the perpendicular bisector of AB given A(0, −2) and

B(3, 3).

___

The slope of AB is:

m = y2 − y1 . x2 − x1

m = 3 − −2 = 5 . 3 − 0 3

___

The slope of any line perpendicular to AB, including the perpendicular

5 bisector, is the opposite reciprocal of 3 .

m = −	3 .	(Answer)
	5

EQUATIONS OF LINES

Linear equations are equations whose graphs are straight lines. Equations containing two variables x and y raised to the first power are linear. By definition, linear equations have a constant slope.

CHAPTER 6 / COORDINATE GEOMETRY

103

The following equations are NOT linear:
3x2 + 2y2 = 1	3	+ 4 y = 5.
	x
The following equations are linear:
y = mx + b.			y − y1 = m(x − x1 ).	Ax + By = C.
Slope-Intercept Form			Point-Slope Form	Standard Form

The three forms of the equation of a line are further explained in the succeeding paragraphs.

Horizontal and Vertical Lines

The most basic forms of linear equations are horizontal and vertical lines. Horizontal lines are written in the form y = a, where a is any constant. Vertical lines are written in the form x = a, where, again, a is any constant.

E X A M P L E :

Find the point of intersection of the lines y = 7 and x = 5. The graph of the two lines clearly shows their intersection.

										y
									12
									11
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									9
									8
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									5
									4
									3
									2
x									1														x
	–12 –11 –10 –9	–8	–7	–6	–5	–4	–3	–2	–1	0	1	2	3	4	5	6	7	8	9	10	11	12
									–1
									–2
									–3
									–4
									–5
									–6
									–7
									–8
									–9
									–10
									–11
									–12
										y

(5, 7). (Answer)

Standard Form

The standard form of the equation of a line is:

Ax + By = C (where A and B are both ≠ 0).

The slope of a line in standard form is − BA (B ≠ 0).