How do you write an equation of a line passing through (2, 4), perpendicular to #y + 4x= 5#?

1 Answer
Shwetank Mauria
Jul 14, 2016

#x-4y+14=0#

Explanation:

Let us write the equation #y+4x=5# in slope intercept form as #y=-4x+5#. Hence it's slope is #-4#.

Now, the product of slopes of two lines perpendicular to each other is #-1#, hence as one line has slope #-4#, line perpendicular to it will have a slope of #(-1)/(-4)=1/4#.

As equation of a line passing through #(x_1,y_1)# and slope of #m# is

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

Equation of a line passing through #(2,4)# and having a slope of #1/4# is

#(y-4)=1/4(x-2)# or

#4y-16=x-2# or

#x-4y+16-2=0# or

#x-4y+14=0#

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