Question 1184939
{{{y=mx+b}}}
{{{m=(y2-y1)/(x2-x1)}}}

your given 1 point on the line where {{{x1=8}}} and {{{y1=(-6)}}}
{{{m=-5/2}}}
{{{(-5/2)=(y2-y1)/(x2-x1)}}}
{{{highlight((-5/2)=(y-8)/(x+6))}}}
or
{{{-6=(-5/2)8+b}}}
{{{-6=(-20)+b}}}
{{{14=b}}}
{{{highlight(y=(-5/2)x+14)}}}

if you need to solve:
{{{-5=(y2-y1)}}} and {{{2=(x2-x1)}}}
{{{-5=(y2-8)}}} and {{{2=(x2-(-6))}}}
{{{-5=(y-8)}}} and {{{2=(x+6))}}} solve for the variables
{{{8-5=(y)}}} and {{{2-6=(x))}}} 
{{{3=y}}} and {{{-4=x)}}}