SOLUTION: find the equation of the line through (7,-4) passing at a distance 1 from the point (2,1)

We will use the point-to-line formula:

The perpendicular distance from the point (x1,y1)
to the line Ax+By+C=0 is
d = 

Suppose the required line has the equation

    y = mx+b

Since it passes through (7,-4)

   -4 = m(7)+b
   -4 = 7m + b
-4-7m = b 

So y = mx+b becomes

   y = mx-4-7m

Let's get it in the form Ax+By+C=0 so we can use
the point-to-line formula above.

       mx-4-7m = y

     mx-y-4-7m = 0

where A = m, B=-1, C=-4-7m

Using the point-to-line formula

d = 

Since we want d to equal 1, and (x1,y1) = (2,1)

1 = 

1 = 

1 = 

Multiply both sides by 



Square both sides

m²+1 = (-5m-5)²

m²+1 = (-5m-5)(-5m-5)

m²+1 = 25m²+25m+25m+25

m²+1 = 25m²+50m+25

   0 = 24m²+50m+24

Divide through by 2

   0 = 12m²+25m+12

   0 = (4m+3)(3m+4)

Use the zero-factor property:

      4m+3=0  3m+4=0
        4m=-3   3m=-4       
         m=   m=

So we have two solutions.  From above we have
-4-7m = b

Using m=

-4-7() = b

-4+() = b

Clear of fractions:

-16+21 = 4b
     5 = 4b
      = b

Substituting in y = mx+b

   

That's one solution.

Using m=

-4-7() = b

-4+() = b

Clear of fractions:

-12+28 = 3b
    16 = 3b
      = b

Substituting in y = mx+b

   

Now let's draw the graphs and see if it looks right.  We plot the points
(2,1) and (7,-4) and the lines

   
   



Those lines both pass through (7,-4) and they look like they are both
1 unit from the point (2,1).  So that must be correct

Edwin