point-slope form of: Through point (-3,-3), perpendicular to y-7/3x+3

Question

I need help understanding how to find to find the solution.

Answer 1

y = (-7/3)x + 3  (Let's call this "Line 1")

 

This equation is already in the "standard form" so we know the slope of the line is (-7/3) and the y intercept if 3.

 

The line for which you are searching is perpendicular to Line 1.  The only way this can be true is if its slope is (3/7).

 

Line 2: y = (3x/7) + b

 

We know that the point (-3, -3) lies on Line 2.  Therefore we can substitute -3 for y and x in the previous equation:

 

Line 2: y = -3 = {[(3)(-3)]/7} + b

 

-3 = -9/7 + b

 

b = -3 + (9/7) = (-21/7) + (9/7) = (-12/7)

 

Line 2: y = (3x/7) - (12/7)

Answer 2

 

Hi Jason;

y-7/3x+3

I am assuming this is...

y=(7/3)x+3

and that 7/3 is positive.

 

7/3 is the slope.

-3/7, the negative inverse of 7/3, is the slope of the line perpendicular to the original equation...

y=(-3/7)x+b

-3=(-3/7)(-3)+b

-3=(9/7)+b

Please note that a negative number multiplied by a negative number has a positive result.

Let's subtract 9/7 from both sides to isolate x...

-(9/7)-3=(9/7)+b-(9/7)

-(9/7)-3=b

To combine -9/7 and 3, we need these to be in the same format.  Let's convert 3 into 21/7.

-(9/7)-(21/7)=b

-30/7=b

 

y=(-3/7)x-(30/7)