Question 826947
Two parallel lines
one through (a,1) and (-10,9) and another through (-3,8) and (a+2,1)
Since parallel lines have the same slope we can
use the general equation for slope
{{{(y2-y1)/(x2-x1)}}}
For (a,1) and (-10,9) , x1=a x2=-10 y1=1 y2=9
{{{(9-1)/(-10-a)}}}= {{{(8)/(-10-a)}}}
For (-3,8) and (a+2,1) , x1=-3 x2=a+2 y1=8 y2=1
{{{(1-8)/((a+2)-(-3))}}} = {{{(7)/((a+2)+3)}}}={{{(7)/((a+5))}}}
Now set the two slope equations to be equal
{{{(7)/((a+5))}}}={{{(8)/(-10-a)}}}
do cross products
{{{7(-10-a)=8(a+5)}}}
-70 -7a = 8a + 40
add 70 to each side
 -7a = 8a + 110
add -8a to each side
 -15a = 110
divide each side by -15
a = -110/15
a = -22/3
Checking 
For (a,1) and (-10,9) , x1=a x2=-10 y1=1 y2=9
{{{(9-1)/(-10-a)}}}?= {{{(8)/(-10-a)}}}
with a = -22/3
{{{(9-1)/(-10-(-22/3))}}}?= {{{(8)/(-10-(-22/3))}}}
?= {{{(8)/(-10+22/3)}}}
{{{(8)/(-30/3+22/3)}}}
{{{(8)/(-8/3)}}} = -3
So we have the slope for the first set of points = -3
Let's look at the second set.
For (-3,8) and (a+2,1) , x1=-3 x2=a+2 y1=8 y2=1
{{{(1-8)/((a+2)-(-3))}}} = {{{(7)/((a+2)+3)}}}={{{(7)/((a+5))}}}
with a = -22/3
{{{(1-8)/(((-22/3)+2)-(-3))}}}
{{{(7)/(((-22/3)+6/3)+3)}}}
{{{(7)/(((-22/3)+6/3)+9/3)}}}
{{{7/(-7/3)}}}= -3
So we see that a = -22/3 produces two lines with the same -3 slope.