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

**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

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 *x*² + *y*² = 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*)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* = 5*x* + 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 =

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

2) **general form:**

**A***x*** + B***y*** + C = 0**: slope = –A/B, ** y**-int = – C/B,

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

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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* = 2*x* + 1, and *y* = 2*x* – 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 - ).*