Find the equation of a line passing through (15,-7) perpendicular to the line 5x+6y=11
+592 5x + 6y = 11
6y = -5x + 11
y = \(-\frac{5}{6} x + \frac{11}{6}\)
Since the slope is \(-\frac{5}{6}\) , the perpendicular slope to it is \(\frac{6}{5}\)
y = mx + b --> (15, -7)
\(-7 = \frac{6}{5}(15) + b\)
\(-7 = 18 + b\)
\(b = -25\)
This means the equation of this line is:
\(y = \frac{6}{5}x - 25\)
