What is the equation of the line perpendicular to #y=-1/16x # that passes through # (3,4) #?

1 Answer
Feb 16, 2016

Answer:

Equation of desired line is #y=16x-44#

Explanation:

The equation of line #y=−(1/16)x# is in slope-intercept form #y=mx+c#, where #m# is slope and #c# is intercept on #y# axis.
Hence its slope is #−(1/16)#.

As product of slopes of two perpendicular lines is #-1#, slope of line perpendicular to #y=−(1/16)x# is #16# and slope-intercept form of the equation of line perpendicular will be #y=16x+c#.

As this line passes through (3,4), putting these as #(x, y)# in #y=16x+c#, we get #4=16*3+c# or #c=4-48=-44#.

Hence equation of desired line is #y=16x-44#