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 February 3rd, 2009, 12:14 PM #1 Newbie   Joined: Feb 2009 Posts: 2 Thanks: 0 Hard Question for Me Hi, been having problems with this question.. I got to be honest.. I need this to help me get through my college course, but it's a high school question.. just find it really hard.. thanks in advance: 1.) Find the equation of a straight line that passes through the point (3, -1) and is perpendicular to the line y = 5x - 2. Give your answer in the form ax + by + c = 0 Thanks, if I find anymore difficult.. I will shout back. Thanks, Dan
 February 3rd, 2009, 01:50 PM #2 Senior Member   Joined: Sep 2008 Posts: 116 Thanks: 0 Re: Hard Question for Me Point slope form of a line is given by: y - y1 = m (x - x1) You are given a point (3, -1) and just need to find the slope. You know the slope of the new line is perpindicular to the line y = 5x - 2 which has a slope of 5. A perpindicular slope is the negative recipricole of that; so -1/5. Plug in these values to the point slope equation of a line: y - (-1) = -1/5 ( x - 3) y + 1 = -(x - 3) / 5 Then, rewrite the equation in standard form, the form being asked for: 5 (y + 1) = -(x -3) 5y +5 = 3 - x 5y + x + 5 -3 = 0 5y + x +2 = 0 Wrote this all in a rush, headed out of my office for the day, though if you have a question, feel free to ask and I'll reply later tonight at my house.
 February 4th, 2009, 09:23 AM #3 Newbie   Joined: Feb 2009 Posts: 2 Thanks: 0 Re: Hard Question for Me Thank you good sir

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