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Find Slope, Equation, Midpoint

Date: 06/04/2003 at 23:50:08
From: Ashley
Subject: Slopes

A = (-4,6) and B = (-2,1)

   1) find the slope;

   2) find the equation of the line containing A and B;

   3) find the coordinates for the midpoint AB;

   4) find the distance between A and B.


Date: 06/05/2003 at 09:57:01
From: Doctor Dotty
Subject: Re: Slopes

Hi Ashley,

Thanks for the question.

Any (non-vertical) line can be represented by the equation:

           y - y1 = m(x - x1)

Where m is the gradient (slope) and two points it passes through are 
(x, y) and (x1, y1).

Let's think about a line passing through (2, 4) and (-3, 6) as an 
example.

We have two points, so x = 2, y = 4; x1 = -3, y1 = 6.

This gives:

            4 - 6 = m(2 - -3)

               -2 = 5m

                m = -2/5

So the gradient (slope) is -2/5.

Next we need an equation of the line. We can find this using one point 
and the gradient. So:

           y - y1 = m(x - x1)

            y - 6 = (-2/5)(x - -3)

            y - 6 = (-2/5)(x + 3)

Multiply by 5:

          5y - 30 = -2(x + 3)

          5y - 30 = -2x - 6

     5y + 2x - 24 = 0

Which is the equation of the line.

Now, the midpoint.

That is the point that lies horizontally halfway between x and x1, and 
vertically halfway between y and y1.

To remind us: x = 2, y = 4; x1 = -3, y1 = 6.

Halfway between 2 and -3 is -0.5.

Halfway between 4 and 6 is 5.

Therefore the coordinates of the midpoint are (-0.5, 5).

For easier use in the future, you can write this method as:

                    x + x1     y + y1
   Midpoint is at   ------  ,  ------
                       2          2

Now, the distance from A to B.

        .               |                      
            .           |                      
               A      6 +                      
               |    .   |                      
               |        |                      
               |        |  .                   
               |_ _ _ 4_+_ _ _.B              
                        |        .             
                        |           .          
                        |              .       
                        |                 .    
    -----------+--------+------+---------------
              -3        |      2               .
                        |                      
                        |                      

Let's look at that triangle:

               A                              
               |    .   d                      
              2|                              
               |_          .                   
               |_|_ _ _ _ _ _ .B
                      5

Using Pythagoras' theorem,

   2^2 + 5^2 = d^2             (where ^2 means squared)

      4 + 25 = d^2

    sqrt(29) = d

Which is the distance between the two points.

For easier use in the future, you can write this method as an 
equation:

   d = sqrt[ (x - x1)^2  +  (y - y1)^2 ]

Can you apply this to your question now?

Write back if I can be of any more help - on this or anything else.

- Doctor Dotty, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Linear Equations

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