Posted: 2/21/2006 6:01:33 PM EDT
Folks, I am having a helluva time with this question. I don't think I am doing it right. Can anyone help?
Find the equation, in standard form, of the line perpendicular to 3x – 6y = 9 and passing through (2, 1). This is what I have so far, but it can't be right. 3x – 6y = 9  6y = 3x + 9 6y/6 = 3x/6 + 9/6 y = (1/2)x + 3 y = 1x + 3 How do I turn (2, 1) into a linear equation like 3x  6y = 9 to show the slopes as coefficient? Thanks folks! 

Well the day's just not complete without seeing some guy get cornholed by a kangaroo while wearing a cream pie on his head... Aimless
Tai Gan Jo Ju You wouldn't want to ban guns if one ever saved your daughter's life. 
3x – 6y = 9 is equivalent to 6y = 3x  9 equivalent to y = (1/2)x  3/2 so m = 1/2 this is now in the form of y=mx + b where m is the slope, and b is the y intercept. a line with a Perpendicular slope to m would have a slope of 1/m (m <> 0) let n = 1/m = 2 IF m is 0 then the perpendicular is parallel to the y axis. Since the new line also must pass through the point (2, 1) we know that from y = nx + b we get 1 = 2n + b but since n = 2 we get 1 = 2*(2) + b solving for b we get 1 = 4 + b or b = 4 + (1) or b = 5 therefore our new equation is: y = 2x  5 which is perpendicular to: y = (1/2)x  3/2 or 3x – 6y = 9 and our new equation passes through point (2, 1) 


What's the use?



use your equation y = (1/2)x + 3. (1/2) equals your slope (m). since perpendicular lines have the opposite slope, the slope you have to use is 2. So then you use the equation (yy1) = m(xx1). you plug in the points that you are given (2, 1) for y1 and x1, and plug in 2 for m. and solve for y.
(y  y1) = m(x  x1) (y  (1) ) = (2)(x  (2) ) y + 1 = 2x + 4 y = 2x + 3 should be your answer someone check this to make sure I'm right 


second equation is wrong, check ur negatives.



how do you figure? the double negatives make positive 


I'd help you, but I was never any good at Spanish.



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