Question 829415
<pre>
We will use the point-to-line formula:

The perpendicular distance from the point (x<sub>1</sub>,y<sub>1</sub>)
to the line Ax+By+C=0 is
d = {{{abs(Ax[1]+By[1]+C)/sqrt(A^2+B^2)}}}

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 = {{{abs(Ax[1]+By[1]+C)/sqrt(A^2+B^2)}}}

Since we want d to equal 1, and (x<sub>1</sub>,y<sub>1</sub>) = (2,1)

1 = {{{abs(m(2)+(-1)(1)+(-4-7m))/sqrt(m^2+(-1)^2)}}}

1 = {{{abs(2m-1-4-7m)/sqrt(m^2+1)}}}

1 = {{{abs(-5m-5)/sqrt(m^2+1)}}}

Multiply both sides by {{{sqrt(m^2+1)}}}

{{{sqrt(m^2+1)}}}{{{""=""}}}{{{abs(-5m-5)}}}

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={{{-3/4}}}   m={{{-4/3}}}

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

Using m={{{-3/4}}}

-4-7({{{-3/4}}}) = b

-4+({{{21/4}}}) = b

Clear of fractions:

-16+21 = 4b
     5 = 4b
     {{{5/4}}} = b

Substituting in y = mx+b

   {{{y}}}{{{""=""}}}{{{expr(-3/4)x+5/4}}}

That's one solution.

Using m={{{-4/3}}}

-4-7({{{-4/3}}}) = b

-4+({{{28/3}}}) = b

Clear of fractions:

-12+28 = 3b
    16 = 3b
     {{{16/3}}} = b

Substituting in y = mx+b

   {{{y}}}{{{""=""}}}{{{expr(-4/3)x+16/3}}}

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

   {{{y}}}{{{""=""}}}{{{expr(-3/4)x+5/4}}}
   {{{y}}}{{{""=""}}}{{{expr(-4/3)x+16/3}}}

{{{drawing(300,300,-9,9,-9,9, graph(300,300,-9,9,-9,9),

line(19,-13,-13,11), line(-14,24,16,-16),
circle(2,1,0.15),circle(2,1,0.13),circle(2,1,0.11),circle(2,1,0.09),circle(2,1,0.07),circle(2,1,0.05),circle(2,1,0.03),circle(2,1,0.01),

circle(7,-4,0.15),circle(7,-4,0.13),circle(7,-4,0.11),circle(7,-4,0.09),circle(7,-4,0.07),circle(7,-4,0.05),circle(7,-4,0.03),circle(7,-4,0.01) )}}}

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</pre>