What is the equation of the line that is perpendicular to the line y=3/4x - 6 and passes through the point (-3,5)?

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 the line crosses the y-axis)
Perpendicular lines always have slopes that are negative reciprocals of each other (ex. -1/3 and 3, 1/2 and -2, etc.)

1) Determine the slope (m)

[tex]y=\frac{3}{4} x - 6[/tex]

Looking at the given equation, we can identify clearly that [tex]\frac{3}{4}[/tex] is in the place of m, making it the slope of the line. Because perpendicular lines always have slopes that are negative reciprocals, we know that the slope of the line we're solving for will be [tex]-\frac{4}{3}[/tex]. Plug this into [tex]y=mx+b[/tex]:

[tex]y=-\frac{4}{3}x+b[/tex]

2) Determine the y-intercept (b)

[tex]y=-\frac{4}{3}x+b[/tex]

Plug in the given point (-3,5)

[tex]5=-\frac{4}{3}(-3)+b\\5=4+b[/tex]

Subtract 4 from both sides

[tex]5-4=4+b-4\\1=b[/tex]

Therefore, the y-intercept is 1. Plug this back into [tex]y=-\frac{4}{3}x+b[/tex]:

[tex]y=-\frac{4}{3}x+1[/tex]

I hope this helps!