1.3 STRAIGHT  LINES

   DEFINITION

           Let P (x0, y0) be a point and let m be a real number. The line through P with

           slope m is the set of all points Q(x, y) with

                              y- y0 = m(x-x0).

 

             This equation is called the point - slope equation of the line (see Figure 1.3.1.)

             The vertical line through P is the set of all points Q(x, y) with x= x0.

             Vertical lines do not have slopes.

 

 

          

 

 

 

 

 

Figure 1.3.1

 

The slope is a measure of the direction of the line. Figure 1.3.2 shows lines with zero, positive, and negative slopes.

 

The line that crosses the y-axis at the point (0, b) and has slope m has the simple equation.

                                  y= mx + b.

 

 

 

 

 

Figure 1.3.2

 

This is called the slope-intercept equation for the line. We can get it from the point-slope equation by setting x0=0 and y0=b.

 

EXAMPLE 1  the line through the point P(-1, 2) with slope m= __(Figure 1.3.3)

              has the point-slope equation

              y-2 = (x-(-1)) ·(____),  or   y-2 = ___(x+1)

 

The slope- intercept equation is

                          y=___ x + ___.

 

 

 

 

Figure 1.3.3

 

We now describe the functions whose graphs are nonvertical lines.

 

DEFINITION

 

         A linear function is a function f of the form

                         f(x) = mx + b,

Where m and b are constants.

The graph of a linear function is just the line with slope-intercept equation

                        y=mx+b.

This is the line through (0, b) with slope m.

 

If two points on a line are known, the slope can be found as follows.

 

THEOREM 1

Suppose a line Lpasses through two distinct points P(x1, y1) and Q(x2, y2). If x1=x2, then the line L is vertical. If x1≠x2, then the slope of the line Lis equal to the change in ydivided by the change in x.

 

 

 

PROOF  Suppose x1≠x2, so L is not vertical. Let mbe the slope of L. L has

         the point-slope formula

                             y-y1 = m(x-x1).

 

         Substituting y2for y and x2for x, we see that m= (y2-y1)/(x2-x1).

 

Theorem 1  shows why the slope of a line is a measure of its direction. Sometimes Δxis called the run and Δy the rise. Thus the slope is equal to the rise divided by the run. A large positive slope means that the line is rising steeply to the right, and a small positive slope means the line rises slowly to the right. A negative slope means that the line goes downward to the right. These cases are illustrated in Figure 1.3.4.

   

 

 

 

 

 

 

 

 

Figure 1.3.4

 

There is exactly one line Lpassing through two distinct points P(x1, y2) and Q(x2, y2). If x1≠x2, we see from Theorem 1 that Lhas the equation

                         

 

 

This is called the two-point equation for the line.

 

EXAMPLE 2  Given P(3,1) and Q(1,4), find the changes in xand y, the slope, and the equation of the line through Pand Q. (See Figure 1.3.5.)

 

                       x= 1-3 = -2,     y=4-1=3.

The line through Pand Q has slope y/x=__, and its equation is

                      y-1= ___(x-3).

               

 

 

 

 

 

 

 

Figure 1.3.5

EXAMPLE  3  Given P(1,-1) and Q(1,2), as in Figure 1.3.6,

                     Δx= 1-1 = 0,   Δy= 2 - (-1) = 3.

          The line through P and Q is the vertical line x=1.

 

                

 

 

 

 

 

 

 

 

 

 

Figure 1.3.6

EXAMPLE  4  A particle moves along the y-axis with constant velocity. At time t=0 sec,

               it is at the point y=3ft. At time t=2 sec, it is at the point y= 11ft.

              

         The velocity is defined as the distance moved divided by the time elapsed, so the 

         velocity is

                       

 

If the motion of the particle is plotted in the(t,y) plane as in Figure  1.3.7,

 

 

 

 

 

 

 

 

 

 

Figure  1.3.7

             The result is a line through the points P(0,3) and Q(2,1). The velocity, being

             the ratio of Δy to Δt, is just the slope of this line. The line has the equation

                                    y-3 = 4t.

Suppose a particle moving with constant velocity is at the point y=y1 at time t=t1, and at the point y=y2 at time t=t2. Then the velocity is v= y/t. The motion of the particle plotted on the (t, y) plane is the line passing through the two points (t1,y1) and (t2, y2), and the velocity is the slope of this line.

 

An equation of the form   

                    Ax +by + c =0

 

Where a and b are not both zero is called a linear equation. The reason for this name is explained by the next theorem.

 

THEOREM  2

                 Every linear equation determines a line.

PROOF

Case 1  B=0. The equation Ax+C = 0 can be solved for x, x= -C/A. This is a vertical line.

Case 2  B≠0. In this case, we can solve the given equation for y, and the result is

            

 

           This is a line with slope -A/B crossing the y-axis at -C/B.

 

EXAMPLE  5   Find the slope of the line 6x-2y+7 = 0.

                The answer is m= -A/B = -6/(-2)=3.

 

To draw the graph of a linear equation, find two points on the line and draw the line through them with a ruler.

 

EXAMPLE  6  Draw the graph of the line 4x+2y+3=0.

               First solve for y as a function of x:

                            y= -2x-___.

 Next select any two values for x, say x=0 and x=1, and compute the corresponding values of y.

When    x=0, y=____.

When    x=1, y=____.

Finally, plot the two points (0,__) and (1, __), and draw the line through them.

(see Figure 1.3.8)

 

 

 

 

 

 

 

 

 

Figure 1.3.8

 

 

PROBLEMS  FOR  SECTION  1.3

In Problems 1-8, find the slope and equation of the line through Pand Q.

1   P(1,2),   Q(3,4)                      2   P(1, -3),   Q(0,2)

3   P(-4,1),   Q(- 4,2)                    4   P(2,5),   Q(2,7)

5   P(3,0),   Q(0,1)                      6   P(0,0),   Q(10,4)

7   P(1,3),   Q(3,3)                      8   P(6, -2),   Q(1, -2)

 

In Problems 9-16, find the equation of the line with slope mthrough the point P.

9   m=2,     p(3,3)                      10   m=3,    p(-2, 1)

11  m=____,     p(1, -4)                  12   m= -1,   p(2,4)

13 m=5,    p(0,0)                      14  m=-2,   p(0,0)  

15  m=0,     p(7,4)                      16    vertical line,  p(4,5) 

 

In Problems 17-22, a particle moves with constant velocity and has the given positons yat the given times t. Find the velocity and the equation of motion.

 

17   y=0 at t=0,    y=2 at t=1

18   y=3 at t=0,    y=1 at t=2

19   y=4 at t=1,    y=2 at t=5

20   y=1 at t=2,    y=3 at t=3

21   y=4 at t=0,    y=4 at t=1

22   y=1 at t=3,    y= -2 at t=6

23   A particle moves with constant velocity 3, and at time t=2 is at the point y=8.

     Find the equation for its motion.

24   A particle moves with constant velocity ___, and at time t=0 is at y=1.

     Find the equation for its motion.

 

In Problems 25-30, find the slope of the line with the given equation, and draw the line.

25   3x-2y+5=0               26     x+y-1=0

27   2x-y=0                  28     6x+2y=0

29   3x+4y=6                 30    -2x+ 4y= -1

31   Show that the line that crosses the x-axis at a ≠0 and the y-axis at b ≠0 has the

     equation (x/a)+(y/b) -1 =0

32   What is the equation of the line through the origin with slope m?

33   Find the points at which the line ax+by+c=0 crosses the x-and y-axes.

     (Assume that a≠0 and b≠0)

34   Let C denote Celsius temperature and FFahrenheit temperature. Thus, C=0 and

     F=32 at the freezing point of water, while C=100 and F=212 at the boiling point of

     water. Use the two-point formula to find the linear equation relating Cand F.