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Find the slope-intercept equation of the line perpendicular to the line x - 3y = 4 and containing the point (6,-9). Explain/show work.
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2 Answers

L1 : X - 3Y = 4 
 
       Y = X/3 + 4/3 
 
 
L2  l  L1
 
        m L2 = -3 
 
 L2 :   Y = -3X +b
 
      Passes through ( 6 , -9) , then:
 
       - 9 = -3( 6 ) +b
 
         b = -9 + 18 = 9
 
    L 2 :   Y = -3X + 9 

Find the slope-intercept equation of the line perpendicular to the line x - 3y = 4 and containing the point (6,-9).

 
First, let's rewrite the equation in standard slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept:
 
x-3y = 4
-3y = 4 - x               (Subtract x from both sides)
y = (1/3)x -4           (Divide both sides by -3)
 
The slope of this line is m = 1/3.  If the slope of a line is m, the slope of the line perpendicular to it is -1/m.  So the slope of a line that's perpendicular to y = (1/3)x - 4 is mp = -1/(1/3) = -3.
 
OK, we're half way there.  We want to find the formula of line perpendicular to y = (1/3)x - 4.  We know it will have a slope mp = -3 and the line passes through the point (6,-9)
y = (-3)x + b
-9 = (-3)(6) + b        (Plug in -9 for y, 6 for x)
-9 = -18 + b
9 = b                        (Solve for b)
 
So our equation is:
 
y = -3x + 9