How do you find a standard form equation for the line that is perpendicular to #6x-2y+5=0# and has the same y-intercept as #x-y+8=0#?

2 Answers
Jun 15, 2017

Answer:

When given an equation of a line of the form:

#ax-by+c=0#

The family of perpendicular lines are of the form:

#bx+ay+k=0#

Where k is an unknown constant.

Explanation:

Given: #6x-2y+5=0#

The family of perpendicular lines are:

#2x+6y+k=0" [1]"#

We must find the y intercept of the line #x-y+8=0#:

#0-y+8=0#

#y = 8#

Substitute the point #(0,8)# into equation [1] an then solve for k:

#2(0)+6(8)+k=0#

#k = -48#

Substitute this into equation [1]:

#2x+6y-48=0" [2]"#

Equation [2] is the answer.

Jun 15, 2017

Answer:

#y=-1/3x+8#

This is the same as: #1/3x+y-8=0" "->" " x+3y-24=0#

Explanation:

First consider #x-y+8=0" "..................Equation(1)#

This can be written as: #y=x+8#

So the y-intercept for this equation is 8.

Thus the equation we are looking for passes through the point:
#(x,y)->(0,8)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider the given equation: #6x-2y+5=0" ".....Equation(2)#

Add #2y# to both sides

#2y=6x+5#

Divide both sides by 2

#y=3x+5/2" ".......Equation(2_a)#

Thus the gradient (slope) for #Equation(2_a)# is 3 from the #3x#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Compare #Equation(2_a)# to the standardised form of#" "y=mx+c#
where #m# is the gradient (slope).

A line that is perpendicular to it will have the gradient of:
#(-1)xx1/m = -1/m" "->" "(-1)xx1/3=-1/3#

Thus our standardised form for the new line is:

#y=-1/3x+c#

From #Equation(1)# we have determined that this line passes through the point #(x,y)->(0,8)#

So #y=-1/3x+c" "->" "8=-1/3(0)+c#

Thus #c=8#

So the equation of the perpendicular line is:
#y=-1/3x+8#

Tony B