section 1.3 -- the coordinate plane

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MA107 PreCalculus Section 1.3 The Coordinate Plane

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Lecture on Section 1.3 in Faires and DeFranza.

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Page 1: Section 1.3 -- The Coordinate Plane

MA107 PreCalculusSection 1.3

The Coordinate Plane

Page 2: Section 1.3 -- The Coordinate Plane

The Coordinate PlaneIf two copies of the number line, one horizontal

and one vertical, are placed so that they intersect at the zero point of each line, a pair of axes is formed.The horizontal number line is called the x-axis and

the vertical number line is called the y-axis.The point where the lines intersect is called the

origin.

We call this a rectangular coordinate plane or a Cartesian coordinate system.

Page 3: Section 1.3 -- The Coordinate Plane

The Coordinate Plane

Page 4: Section 1.3 -- The Coordinate Plane

The Coordinate Plane

Page 5: Section 1.3 -- The Coordinate Plane

Inequalities in Two Dimensions

The graph of an inequality in two variables consists of all ordered pairs that make the inequality a true statement.

Example: Suppose we want to graph the inequality .

Procedure:• Graph the boundary curve

.• Draw a solid curve if

equality is included.• Draw a dashed curve if

equality is not included.• Determine which region(s)

formed by the curve makes the inequality true by testing with one point from inside each region.

• Shade the region(s) that make the inequality true.

Page 6: Section 1.3 -- The Coordinate Plane

Inequalities in Two Dimensions

Systems of two inequalities:

Idea: Graph both inequalitiesand the region that has been shaded in twice is the region we’re looking for.

Example at left: Graph the solution set of the following system of inequalities:

Click here to see a dynamicexample of linear inequalities.

Page 7: Section 1.3 -- The Coordinate Plane

DistanceSuppose two points P1 and P2 have coordinates

. What is the distance between P1 and P2?

(x1,y1) and (x2,y2)

• The distance from P1 to P2 isthe length of the hypotenuseof a right triangle.• The length of the bottom sideis the same as the distance between x1 and x2 on the x-axis, that is, . The length of the vertical side is the same as the distance between y1 and y2, that is .

Page 8: Section 1.3 -- The Coordinate Plane

DistanceSo if we let d be the distance between P1 and P2,

by the Pythagorean Theorem ….

Now we take the square rootof both sides. Since distance is positive:

Since we’re squaring in there,we can dispense with theabsolute values and get

Distance Formula

Page 9: Section 1.3 -- The Coordinate Plane

DistanceExample: Find the distance between the points

(4,-7) and (-1,3).

Page 10: Section 1.3 -- The Coordinate Plane

DistanceSee Mathematica Player demo on distance.

Page 11: Section 1.3 -- The Coordinate Plane

MidpointThe midpoint of the line segment connecting the points

P1(x1,y1) and P2(x2,y2) is computed by simply averaging the x- and y-coordinates separately.

Midpoint FormulaThe midpoint between (x1,y1) and (x2,y2) is

Take a moment to find the coordinates of the point half way between and .

Answer:

Page 12: Section 1.3 -- The Coordinate Plane

Circles in the PlaneA circle is defined as the set of all points that

are the same distance from a given point.The distance is called the radius.The given point is called the center of the circle.

Let (x, y) be any point on a circle with center (h, k) and radius r as shown at left.Since (x, y) must be r units from the center of the circle, the distance formula gives

Page 13: Section 1.3 -- The Coordinate Plane

Standard Form of the Equation of a Circle

The graph of

is a circle of radius r (r ≥ 0) with center at the point (h, k). If the circle has center at the origin, the equation becomes

The circle is called the unit circle.

Page 14: Section 1.3 -- The Coordinate Plane

Circles in the PlaneExample:

The center is at the point (2, 3) and the radius is 1 unit.

We can figure out some points on the circle by starting with the center point, (2, 3), and adding or subtracting 1 from each coordinate.

So four examples would be (3, 3), (1, 3), (2, 2), and (2, 4).

Try: Find an equation of a circle with center (-3, 6) and radius 4.

Answer:

Page 15: Section 1.3 -- The Coordinate Plane

Completing the squareExample: Find the center and radius of the

circle whose equation is .

Since there is an x2-term and an x-term, we have to combine them to form the term by completing the square:We first group the x-terms together:

To turn the thing in parentheses into a perfect square, divide the coefficient of the x-term by 2, square that number and add it to both sides.

Page 16: Section 1.3 -- The Coordinate Plane

Completing the SquareWe can then factor the x-

terms:

Or to put it in the form of a circle:

We see that h is 1 and k is 0. So the center of the circle is (1,0) and the radius is 2.

Page 17: Section 1.3 -- The Coordinate Plane

Completing the SquareTo try on your own: Graph the equation

Hint: Divide first by 2.

Circle is centered at (-2,3) and hasradius 1.

Page 18: Section 1.3 -- The Coordinate Plane

Homeworkpg. 19: 1-19 odd, 25-43 odd, 51, 55, 57