Where on the line x - y = 8 does the perpendicular from (2, 1) intersect.

From the given
line equation: x-y=8, express it in slope-intercept form (y=mx+b).

x-y
=8

  +y   +y

---------------

x     =
 8+y

   -8   -8 

----------------

x -8
=y            Or y = 1x-8 where slope  `m_1 = 1.`

Note that perpendicular
lines follows` m_2 = -1/m_1`

Then `m_2=- 1/(-1)=-1`

Determine the other line equation using `m_2=-1` and that will pass through point
(2,1).

Plug-in the values in y=mx+b

                     
       1 = (-1)(2)+b

                              1= -2 +b

                              +2  +2

                             
------------------

                              3 =b

Line
equation:  y=-1x+3 or x+y =3  based from `m_2=-1` and b =3

the two lines are
x-y=8 and x+y =3.

Applying elimination method to solve for x. 

         x-y=8           Add the to equations.

    +  x+y
=3

      -------------

        2x   = 11        Cancels
out y's since -y+y = 0.

       `(2x)/2=11/2`           Divide both sides by 2
to isolate x.

        `x = 11/2`

To solve for y, subtract
the equations as:

        x -y = 8          Or            x -y =8

-     ( x +y =3)  ------->  +  (-x -y = -3 ).    Subtraction rules of
signs. 

   -----------------            

           -2y =
5

            `(-2y)/-2=5/2 `      Divide both sides by -2 to isolate
y.

                `y = - 5/2`

Intersection point:  `
(11/2, - 5/2)`

This is the same as (5.5, -2.5)