What is the equation of the line perpendicular to y=5/4x-10 and passes through the point (11,6)?

Linear equations are typically organized in slope-intercept form: [tex]y=mx+b[/tex] where m is the slope and b is the y-intercept (the value of y when x is equal to 0)
Perpendicular lines always have slopes that are negative reciprocals (ex. 3 and -1/3, 5/6 and -6/5, etc.)

1) Determine the slope (m)

[tex]y=\frac{5}{4} x-10[/tex]

From the given equation, we can identify clearly that the slope of this line is

[tex]\frac{5}{4}[/tex]. The negative reciprocal of [tex]\frac{5}{4}[/tex] is [tex]-\frac{4}{5}[/tex], so therefore, the slope of the line we're currently solving for is [tex]-\frac{4}{5}[/tex]. Plug this into [tex]y=mx+b[/tex]:

[tex]y=-\frac{4}{5}x+b[/tex]

2) Determine the y-intercept (b)

[tex]y=-\frac{4}{5}x+b[/tex]

Plug in the given point (11,6) and solve for b

[tex]6=-\frac{4}{5}(11)+b\\6=-\frac{44}{5}+b[/tex]

Add [tex]-\frac{44}{5}[/tex] to both sides to isolate b

[tex]6+\frac{44}{5}=-\frac{44}{5}+b+-\frac{44}{5}\\\frac{74}{5} =b[/tex]

Therefore, the y-intercept is [tex]\frac{74}{5}[/tex]. Plug this back into [tex]y=-\frac{4}{5}x+b[/tex]:

[tex]y=-\frac{4}{5}x+\frac{74}{5}[/tex]

I hope this helps!