Write in slope-intercept form the equation
of the line passing through (1,3) and (-5,-1). 

Use the slope formula:

     y2 - y1
m = —————————
     x2 - x1

where (x1, y1) = (1, 3) and (x2, y2) = (-5, -1)

    (-1) - (3)     -4     4     2
m = ——————————— = ———— = ——— = ———
    (-5) - (1)     -6     6     3

Now substitute in the point slope formula:

   y - y1 = m(x - x1)

   y - 3 = (2/3)(x - 1)

Multiply both sides by LCD = 3

3(y - 3) = 3(2/3)(x - 1)

  3y - 9 = 2(x - 1)

  3y - 9 = 2x - 2

      3y = 2x + 7

Divide every term by 3

      y = (2/3)x + 7/3

That's the slope-intercept form y = mx + b

where m = 2/3 is the slope and (0,b) = (0, 7/3)
is the y-intercept. 

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Show that this line is perpendicular to the line y = (-3/2)x + 17.

Comparing this equation to the slope-intercept form

                                                 y = mx + b

we see that the slope m = -3/2 and we observe that -3/2 is
the "negative reciprocal" of 2/3.  (i.e., the fraction 2/3 is 
inverted and its sign changed).  Lines are perpendicular 
whenever their slopes are "negative reciprocals" of each other.

Another way of stating this is that lines are perpendicular
when the product of their slopes is -1.

When interpreting it this way we need only note that

        (2/3)(-3/2) = -1

Edwin