First we have to transform 2x-3y=11 into common form y=ax+b, so
2x-3y=11 /-2x both side
-3y=11-2x /:(-3) both side
y=[tex] \frac{2}{3}x- \frac{11}{3} [/tex]
Parallel condition is [tex] a_{1} = a_{2} [/tex]
Our [tex] a_{1}= \frac{2}{3} [/tex], so we are sure that new equation looks
y=[tex] \frac{2}{3}x+b[/tex]
To find b, we can substitute x and y from point (6,-5).
-5=[tex] \frac{2}{3}*6+b [/tex]
-5=4+b /-4 both side
-9=b
So, the result is
y=[tex] \frac{2}{3}-9 [/tex]
