Question 1099454
<pre>
Here's the line through (-6,3) and (0,12)

{{{drawing(500,500,-20,20,-20,20,grid(1),
line(-22,-21,18,39) )}}}

We use the fact that parallel lines have the same slope.

{{{m}}}{{{""=""}}}{{{(y[2]-y[1])/(x[2]-x[1])}}}
where (x<sub>1</sub>,y<sub>1</sub>) = (-1,b)
and where (x<sub>2</sub>,y<sub>2</sub>) = (c,8)

{{{m}}}{{{""=""}}}{{{(8-b)/(c-(-1))}}}
{{{m}}}{{{""=""}}}{{{(8-b)/(c+1)}}}

That must equal the slope of the line though (-6,8) and (0,12)


{{{m}}}{{{""=""}}}{{{(y[2]-y[1])/(x[2]-x[1])}}}
where (x<sub>1</sub>,y<sub>1</sub>) = (-6,3)
and where (x<sub>2</sub>,y<sub>2</sub>) = (0,12)

{{{m}}}{{{""=""}}}{{{(12-3)/(0-(-6))}}}
{{{m}}}{{{""=""}}}{{{9/6}}}
{{{m}}}{{{""=""}}}{{{3/2}}}

So we set the slopes equal.

{{{(8-b)/(c+1)}}}{{{""=""}}}{{{3/2}}}

There are an infinite number of possible answers for that

b = -7, c = 9 the line through (-1,-7) and (9,8).
b = -4, c = 7 the line through (-1,-4) and (7,8).
b = -1, c = 5 the line through (-1,-1) and (5,8).
b = 2, c = 3 the line through (-1,2) and (3,8).
b = 5, c = 1 the line through (-1,5) and (1,8).
b = 11, c = -3 the line through (-1,11) and (-3,8).

Here are the graphs of those solutions.  There are trillions
more.  Did your teacher warn you that there are many many different
solutions to this problem?

{{{drawing(500,500,-20,20,-20,20,grid(1),
line(-22,-21,18,39),
line(-27,-43,29,41),
line(-31,-46,19,29),
line(-25,-34,25,41),
line(-27,-34,23,41),
line(23,47,-23,-22)





 )}}}

Edwin</pre>