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*Alan Hale - ajh4216  Attendance #16                              *
*MATH 311- 503                                  &nbs p;                 *
*2-13-99                                   &nbs p;                      *
*Week 3                                     ;                       *
*Problem # 1.2.2                                  &n bsp;               *
*http://www.geocities.com/collegepark/lounge/8250/prob1_2_2.html  *
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Problem Solution to 1.2.2

Find an equation of the form ax + by = c for these lines in R²

a) The line through the points (4,7) and (2,-1)
b) The line with parametric equation  x = t(1,1) + (4,-1)
c) The line through the origin parallel to (5,1)
d) The line with parametric equation x = t(0,1) + (-2,-1)

Start by finding the slope "m" to get equation of the line in form y = mx + b

m = (y₂ - y₁)/(x₂ - x₁) = (-1-7)/(2-4) = 4

Use formula y - y₁ = m(x - x₁)  using (x₁,y₁) = (4,7)

y - 7 = 4(x - 4)
y = 7 + 4x - 16 = 4x - 9

Rearrange to get the form ax + by = c
 
 

Solution to b)

-Using method 2 in the textbook.
x = t(1,1) + (4,-1)      where x = tu + x₀
    u = (1,1)  x₀ = (4,-1)

Since u  = (1,1), a vector perpendicular to it is:
    a = (1,-1)
a*x₀ = (1,-1)*(4,-1) = 4 + 1 = 5

So the equation of the line is:

Solution to c)
Find the line through (0,0) and (5,1)

determine slope of line:
    m = (y₂ - y₁)/(x₂ - x₁) = 1/5

equation of the line:  y - y₁ = m(x - x₁)
 y - 0 = 1/5(x - 0)

Solution to d)

Find the line with parametric equation  x = t(0,1) + (-2, -1)

x = t(0,1) + (-2,-1)  where x = tu + x₀

u = (9 ,1)  x₀ = (-2,-1)

We need a that is perpendicular to u.
a = (-1,0 ) satisfies this.                [using a = (1,0 ) will also work.  Try it and see.]

Use this a in the equation  a*x = a*x₀
(-1,0)*(x,y)  =  (-1,0)*(-2,-1)
which yields:    -x = 2

So the equation of the line is x = -2