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