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

2 Answers
Nov 10, 2016

Answer:

#y=-3/2x+5#

Explanation:

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

Nov 10, 2016

Answer:

#y = -3/2x+5#

Explanation:

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#