How do you find the line that is perpendicular to #y=2/3x+1# and passes through (0, 5)?

Perpendicular lines have opposite reciprocal slope, #m#, in an equation. Meaning, if the slope of a line is a positive fraction, its reciprocal slope would be negative.

The given equation is a linear equation, #y=mx+b#

Where #m# is the slope, #(rise)/(run#, and #b# is the y intercept

To find a line that is perpendicular to the given equation, use the opposite reciprocal slope and point and solve using the point-slope formula

#y-y_1=m(x-x_1)#

#m=-3/2#

#y_1=5#

#x_1=0#

#y-5=-3/2(x-0)#

Distribute the #-3/2# throughout the set of parenthesis

#y-5=-3/2x+0#

Add #5# on both sides of the equation

#y=-3/2x+5#

And now you can see the slope is the opposite reciprocal of the original equation

If lines are perpendicular, one slope is the negative reciprocal of the other.
(Their product is -1)

If #m_1 = 2/3," " m_2 = -3/2#

So we have the slope of the line perpendicular to the one given.

The point #(0,5)# is the y-intercept because the x-value is 0.
This is also known as 'c' #rarr c = 5#

The equation of a line is #y = mx +c#.
We have #m and c#

#y = -3/2x+5#