CAL-PREP ANALYTIC GEOMETRY REVIEW

D/ ANALYTIC GEOMETRY

Introduction: Formulas and Equations (Rules of Correspondence)

When we do Analytic Geometry, we use the coordinates of the Cartesian plane in 2 types of mathematical expressions. We use formulas to calculate the distance between 2 points, the midpoint of a line segment, a particular point of division, etc., and we use equations or rules of correspondence to describe loci or paths such as lines, circles, ellipses etc.

A FORMULA is a set of instructions we use to find a particular point, distance, or area etc. For instance, the distance formula we use to find the length of a line segment given its endpoints, is simply a restatement of the Pythagorean theorem where (x 2x 1) and (y 2y 1) represent the lengths of the legs of the right triangle formed using the line segment as the hypotenuse. The midpoint formula simply says to average the coordinates -- find their mean -- to get the middle point.

An EQUATION or RULE OF CORRESPONDENCE describes a locus, path, or set of points that satisfy a given pre-condition. It is an algebraic statement that defines the relation between the x and y-values of the points that make up the curve of the locus. For example, the equation of a line is an algebraic statement that says: "the change in the y-values divided by the change in the x-values is equal to the slope of the line since a line's slope is constant.

There are situations, such as when writing the equation of a circle, that we alter the defining statement somewhat. Because the distance formula involves a big old ugly square root sign, we write the equation of a circle to describe the relation between the points on the curve and the center point by the SQUARE OF THE RADIUS. A circle is the path of all points located exactly one radius distance from the center point.

So, the equation of a circle centered at the origin with a radius of 5 cm. would be
: which says that the distance between any point (x, y) on the circumference and the center (0, 0) is 5 units. Since (just as in politics) we want to eliminate the radical, we square both sides of this statement.

The equation of a circle states that the square of the distance between any point (x, y) on the circumference and the center is the square of the radius.

So, the circle above is x² + y² = 25.
With the center at (1, – 3), the equation would read ( x – 1 )² + ( y + 3 )² = 25

1/ Midpoint and Distance Formula

MID-POINT FORMULA:

Average of the x and y coordinates of the endpoints. Add them and take ½.

example: the midpoint of (–1, 3) and (7, 9) is ½ ( [–1 + 7 ], [3 + 9] ) = (3, 6)

DISTANCE FORMULA: is just Pythagoras. Take square root of (delta x)² + (delta y

example: the distance between (–1, 3) and (7, 9) is:

2/ Equations of Lines

To write the equation of a line we need 2 things: any point and the slope of the line.

When we have 2 points, we can find the slope between them.

Slope is the ratio of rise to run. The slope of a straight line is a constant.

The equation of a line is an algebraic statement that says:

The slope between any point (x, y) and a given point P
is equal to the slope of the line
.

The equation of the line with slope = m through given point P(a, b) is

Which says that the slope between any point (x, y) and given point P = m.

Example: the equation of the line through (–2, 1) and slope = 5 is

which becomes y = 5(x + 2) + 1 or y = 5x + 11

Example: the equation of the line through (–2, 1) and (6, – 3) is
.

This is the most efficient or " generic " way to write the equation of a line.

The equation of a horizontal line through (a, b) is y = b . (y-value doesn't change)

The equation of a vertical line through (a, b) is x = a . (x-value doesn't change)

example: the equation of the horizontal line through (–3, 2) is y = 2.

the vertical line through this point has equation x = 3.

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Forms of the Linear Equation:

1) standard or function form:

y = mx + b : slope = m, y-int = b, x-int = – b/m

ex: the equation of the line with slope = – 3 and y-intercept = 5 is y = – 3x + 5.

2) general form:

Ax + By + C = 0: slope = –A/B, y-int = – C/B, x-int = – C/A.

ex: find the slope, y-int., and x-int. of 4x – 3y – 12 = 0

solution: A = 4, B = – 3, C = – 12, therefore slope = – 4/–3 = 4/3

the y-intercept = – (–12)/–3 = – 4, and the x-intercept = – (–12)/4 = 3

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N.B.: To find the x-intercept of a line, set y = 0 and solve for x.

To find the y-intercept of a line, set x = 0 and solve for y.

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3) symmetric form:

The symmetric form is ; x-int = a, y-int = b, slope = – b/a.

Parallel Lines: have the same slope.
example: y = 2x + 1, and y = 2x – 7 are parallel lines with different y-intercepts.

Perpendicular Lines: have negative reciprocal slopes.
example: if the slope of a line = 4, then a line perpendicular to it has a slope of – ¼.

Reminder: INTEGERS are fractions with denominator = 1!!

If the slope of a line is 3, to get from one point to another on the line,
we'd have to rise 3 units and run 1.

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EXERCISE 4:

1) Find the length of PQ:
 a) P(–2, – 5) Q(22, – 17) b) P(6, 1) Q(–12, 7) c) P(–1, – 1) Q(13, – 11)

2) Find M, the mid-point of each line segment PQ in question #1:

3) Write the equation of theses lines:
 a) Through (–2, 1) parallel to the x-axis. b) Through (–2, 1) parallel to the y-axis. c) Through (–2, 1) with slope = – 3. d) Through (–2, 1) and (6, – 5). e) Through the y-intercept of 2x – 3y + 6 = 0 and parallel to the line y = 6x + 7. f) Through the x-intercept of 2x – 3y + 6 = 0 and perpendicular to the line y = 6x + 7. g) The right bisector of the line segment joining (–3, 1) and (– 9, – 5).

4) List the slope, x-intercept, and y-intercept of these lines:
 a) 2x + 3y – 12 = 0 b) y = – 7x + 3 c) – 5x – 2y – 15 = 0

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EXERCISE 4 Solutions

1)

 2) a) (10, – 11) b) (– 3, 4) c) (6, – 6)

3)

 a) y = 1 b) x = – 2 c) y – 1 = – 3(x + 2) d) y – 1 = – 3/4 (x + 2) e) y = 6x + 2 f) y = – 1/6 (x + 3) g) y + 2 = – 1(x + 6)

4)

 a) slope = – 2/3 x-int = 6 y-int = 4 b) slope = – 7 x-int = 3/7 y-int = 3 c) slope = – 5/2 x-int = – 3 y-int = – 15/2

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(all content of the MathRoom Lessons © Tammy the Tutor; 2004 - ).