How do you write an equation for the line that passes through (4,7) and is perpendicular to the line y=2/3x-1?

Question

Genesis Allen · Accepted Answer

Please see the explanation for the process. #y = -3/2x + 19# The slope of the line: #y = 2/3x - 1# is #2/3#. The slope of any line perpendicular is the negative reciprocal: #-3/2# Use the point-slope form of the equation of a line to force the line through point #(4, 7)#: #y - 7 = -3/2(x - 4)# #y - 7 = -3/2x +12# #y = -3/2x +19#

Hazel Anglin · Answer

undefined To write an equation for the line that passes through (4,7) and is perpendicular to the line $ y = \frac{2}{3}x - 1 $, use the fact that perpendicular lines have slopes that are negative reciprocals of each other.

The given line has a slope of $ \frac{2}{3} $, so the perpendicular line will have a slope of $ -\frac{3}{2} $.

Using the point-slope form of a line equation $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the point and $ m $ is the slope, substitute $ x_1 = 4 $, $ y_1 = 7 $, and $ m = -\frac{3}{2} $:

$ y - 7 = -\frac{3}{2}(x - 4) $

Expanding and simplifying:

$ y - 7 = -\frac{3}{2}x + 6 $

$ y = -\frac{3}{2}x + 6 + 7 $

$ y = -\frac{3}{2}x + 13 $

So, the equation for the line that passes through (4,7) and is perpendicular to $ y = \frac{2}{3}x - 1 $ is $ y = -\frac{3}{2}x + 13 $.