Search 75,448 tutors
FIND TUTORS
Ask a question
0 0

I need help on finding out how to write the given equation in point-slope form: Through point (-3, -3), perpendicular to y-7/3x+3.

Tutors, please sign in to answer this question.

2 Answers

Hi Jason;
(-3, -3), perpendicular to y-7/3x+3.
I am assuming that the equation is...
y=(7/3)x+3
If not, please let me know.
The slope of this line is 7/3.
The slope of the line perpendicular to this is the negative inverse; -3/7.
y=(-3/7)x+b is a formula written in slope-intercept form.
You need point-slope, especially since the y-intercept, value of b, is still unknown.
y-y1=m(x-x1)
y--3=(-3/7)(x--3)
Subtracting a negative is the same as adding a positive.
y+3=(-3/7)(x+3)
Let's distribute...
y+3=(-3/7)(x)-(9/7)
Let's subtract 3 from both sides as we proceed to isolate y...
-3+y+3=(-3/7)x-(9/7)-3
y=(-3/7)x-(9/7)-3
Considering the fact that 3=21/7...
y=(-3/7)x-(9/7)-(21/7)
y=(-3/7)x-(30/7)
Another method: Be careful writing your questions
 
Passing through  ( -3, -3) , perpendicular to Y = 7/3X + 3 (L1)
 
   Equation of line is always    Y = mx + b (1)
 
      Now you have to find m. b from the given information and plug it into  (1).
 
                The m is equal to -3/7 ( being perpendicular to line (L1).
 
                so  the equation of line is
   
                Y = -3/7X + b
                 
               next have to find b, we can plug ( -3,-3) into equation and solve for b
 
              - 3 = -3/7( -3) +b          b = -3- 9/7 = -30/7 
                  
                  plugging back into the equation the answer is:
 

                  Y = -3/7 X -30/7