How do you find the equation of the line perpendicular to y+5=3(x-2) that passes through the point (6, 2)?

2 Answers
Mar 6, 2018

Let,the equation of line be #y=mx+c#, where, #m# is its slope and #c# is the #Y# intercept.

Now,arranging the given equation in the above mentioned form to get its slope,

Given, #y+5=3(x-2)=y=3x-11#

So,its slope is #3#

Now,for two lines to be mutually perpendicular,their product of slope must be #-1#

So, #m*3=-1#

or. #m=-1/3#

So,our required line equation becomes, #y=-1/3 x+c#

Now,given,that the line passes through #(6,2)#,so putting the values in the equation to get #c#

So, #2=(-1/3)*6 +c#

or, #c=4#

So,the equation of the line is #y=-1/3 x+4#

or, #3y+x=12# graph{3y+x=12 [-10, 10, -5, 5]}

Mar 6, 2018

#y-2 = -1/3(x-6)#

Or #y = -1/3x +4#

Or #x+3y =12#

Or #x+3y-12=0#

Explanation:

There are several forms for the equation of a straight line:

#y = mx+c," and " ax +by +c =0" "# are well known.

Another is #y-y_1 = m(x-x_1)#

where #(x_1, y_1)# is a point and #m# is the slope.

This is exactly the form we have for the given equation:

#y+5=color(blue)(3)(x-2)" "rarr :. color(blue)(m= 3)#

If lines are perpendicular, the product of their slopes is #-1#

One is the negative reciprocal of the other, so:

#m_1 = 3 " "rArr" "m_2 =-1/3#

Using #m= -1/3# and the point #(6,2)#

#y-y_1 = m(x-x_1)" "# gives#" "y-2 = -1/3(x-6)#

Or in another form:

#y= -1/3x+2+2" "rarr y = -1/3x +4#

Or even

#3y =-x+12" "rarr" "x+3y =12#