|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 2 x 1) and (y 2 y 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
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)²
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.
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
N.B.: To find the x-intercept of a line, set y = 0 and solve for x.
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.
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|
EXERCISE 4 Solutions
|2) a) (10, 11)||b) ( 3, 4)||c) (6, 6)|
|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)|
|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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