1. Calculus and analytic geometry
Lecture 02
CS001
Calculus and analytic geometry

2. Rectangular Coordinate Systems
We shall now show how to assign an ordered pair (a, b) of real numbers to each point in a plane. Although we have also used the notation (a, b) to denote an open interval, there is little chance for confusion, since it should always be clear from our discussion whether (a, b) represents a point or an interval.

3. Rectangular Coordinate Systems
We introduce a rectangular, or Cartesian, coordinate system in a plane by means of two perpendicular coordinate lines, called coordinate axes, that intersect at the origin O, as shown in Figure 1.
Figure 1

4. Rectangular Coordinate Systems
We often refer to the horizontal line as the x-axis and the vertical line as the y-axis and label them x and y, respectively. The plane is then a coordinate plane, or an xy-plane. The coordinate axes divide the plane into four parts called the first, second, third, and fourth quadrants, labeled I, II, III, and IV, respectively (see Figure 1). Points on the axes do not belong to any quadrant. Each point P in an xy-plane may be assigned an ordered pair (a, b), as shown in Figure 1.

5. Rectangular Coordinate Systems
We call a the x-coordinate (or abscissa) of P, and b the y-coordinate (or ordinate). We say that P has coordinates (a, b) and refer
to the point (a, b) or the point P (a, b).
Conversely, every orderedpair (a, b)
determines a point P with coordinates
a and b. We plot a point by using a dot, as illustrated in Figure 2.
Figure 2

6. Rectangular Coordinate Systems
Distance between two points
The length of a horizontal line segment is the abscissa (x coordinate) of the point on the right minus the abscissa (x coordinate) of the point on the left.
Distance D = X2 – X1

7. Rectangular Coordinate Systems
Distance between two points
The length of a vertical line segment is the ordinate (y coordinate) of the upper point minus the ordinate (y coordinate) of the lower point.
Distance D = y2 – y1

8. Rectangular Coordinate Systems
To determine the distance between two points of a slant line segment add the square of the difference of the abscissa to the square of the difference of the ordinates and take the positive square root of the sum.

9. Rectangular Coordinate Systems

10. Example 1 – Finding the distance between points
Plot the points A(–3, 6) and B(5, 1), and find the distance d (A, B).
Solution: The points are plotted in Figure 4.
Figure 4

11. By the distance formula,
d (A, B) = =
 9.43.

12. INCLINATION AND SLOPE OF A LINE
The inclination of the line, L, (not parallel to the x-axis) is defined as the smallest positive angle measured from the positive direction of the x-axis or the counterclockwise direction to L.
The slope of the line is defined as the tangent of the angle of inclination.

13. INCLINATION AND SLOPE OF A LINE

14. INCLINATION AND SLOPE OF A LINE

15. PARALLEL AND PERPENDICULAR LINES
If two lines are parallel their slope are equal (m1=m2).
If two lines are perpendicular the slope of one of the line is the negative reciprocal of the slope of the other line.If m1 is the slope of L1 and m2 is the slope of L2 then, or m1m2 = -1.

16. Example 1 – Finding slopes
Sketch the line through each pair of points, and find its slope m:
(a) A(–1, 4) and B(3, 2)
(b) A(2, 5) and B(–2, –1)
(c) A(4, 3) and B(–2, 3)
(d) A(4, –1) and B(4, 4)

18. The lines are sketched in Figure 3
The lines are sketched in Figure 3. We use the definition of slope to find the slope of each line.
m =
m =
Figure 3(a)
Figure 3(b)

19. m = 0
m undefined
Figure 3(c)
Figure 3(d)

20. (a)
(b)
(c)
(d) The slope is undefined because the line is parallel to the y- axis. Note that if the formula for m is used, the denominator is zero.

21. LINES
Let us next find an equation of a line l through a point P1(x1, y1) with slope m. If P (x, y) is any point with x  x1 (see Figure 7), then P is on l if and only if the slope of the line through P1 and P is m—that is, if
Figure 7

22. LINES This equation may be written in the form y – y1 = m(x – x1).
Note that (x1, y1) is a solution of the last equation, and hence the points on l are precisely the points that correspond to the solutions.
This equation for l is referred to as the point-slope form.

23. Example 4 – Finding an equation of a line through two points
Find an equation of the line through A(1, 7) and B(–3, 2).
Solution:
The line is sketched in Figure 8.
Figure 8

24. The formula for the slope m gives us
We may use the coordinates of either A or B for (x1, y1) in the point-slope form.
Using A(1, 7) gives us the following:
y – 7 = (x – 1)
4(y – 7) = 5(x – 1)
point-slope form
multiply by 4

25. 4y – 28 = 5x – 5
–5x + 4y = 23
5x – 4y = –23
The last equation is one of the desired forms for an equation of a line.
Another is 5x – 4y + 23 = 0.
multiply factors
subtract 5x and add 28
multiply by –1

26. LINES
The point-slope form for the equation of a line may be rewritten as y = mx – mx1 + y1, which is of the form
y = mx + b
with b = –mx1 + y1. The real number b is the y-intercept of the graph, as indicated in Figure 9.
Figure 9

27. Example 5 – Expressing an equation in slope-intercept form
Express the equation 2x – 5y = 8 in slope-intercept form.
Solution:
Our goal is to solve the given equation for y to obtain the form
y = mx + b.
We may proceed as follows:
2x – 5y = 8
given

28. subtract 2x divide by –5 m = and y-intercept b = . equivalent equation
–5y = –2x + 8
y =
The last equation is the slope-intercept form y = mx + b with slope
m = and y-intercept b =
subtract 2x
divide by –5
equivalent equation