How do you write the standard form of a line given #(-6,7)# and perpendicular to #3x+4y-12=0#?

1 Answer
Jul 25, 2017

Answer:

#4x-3y=-45#

Explanation:

Since the equation must be expressed in standard form, that is, #Ax+By=C#, we can find the perpendicular line by switching the coefficients of #x# and #y# and changing the operation sign.

But first let's rewrite the given equation in #Ax+By=C# form.

#3x+4y-12=0#

#3x+4y=12#

To find the perpendicular line, switch the coefficients of #x# and #y# and change the operation sign. By doing so we get:

#4x-3y=C#; where #C# is any constant but we are given the point #(-6,7)# so we can substitute these values for #x# and #y# to find #C#.

If we let #(-6,7)->(x,y)# then,

#4(-6)-3(7)=C#

#-24-21=C#

#-45=C#

So we found our constant. We can now substitute this value for #C# for the equation of the perpendicular line. In doing so, we get our final equation:

#4x-3y=-45larr# And this is the equation of the perpendicular line