Equation of a LineDate: 01/27/98 at 20:57:30 From: amanda Subject: Equation of a line How do I find the equation of a line through origin perpendicular to 2x + 3y = 5 ? Date: 01/28/98 at 13:09:19 From: Doctor Jaffee Subject: Re: Equation of a line There are several properties of equations of lines that you have to know in order to understand how to solve this problem. First, any time you have an equation in the form y = mx + b, where m and b are real numbers, m will be the "slope" of the line and b will be the "y-intercept" of the line. So if we take your equation 2x + 3y = 5 and add -2x to both sides as follows: -2x + 2x + 3y = -2x + 5 we get 3y = -2x + 5 then divide both sides by 3 and the result is y = (-2/3)x + 5/3 This means that the slope of the line (our value of m) is -2/3 and the y-intercept (our value of b) is 5/3. Check the Dr. Math archives, select the link to the high school section, and then the link to "equations, graphs, and translations" - the question from 6/28/96 entitled graphing y = mx + b provides some more information about this topic. Also, it would be worth your while to try the link entitled "linear equations" in the high school section. Look for the problem dated 12/1/94. Now, back to your problem. The second important property of equations of lines is that if two lines are perpendicular, their slopes are negative reciprocals of each other. This means that if one of the slopes is negative, the other will be positive, and furthermore, once you adjust the sign you must "flip" the number you get. In the previous paragraph we calculated that the slope of the given line was -2/3, so to find the slope of the perpendicular line we change the -2/3 to 2/3 and then flip it so that it ends up as 3/2, which will be the slope of the line we are looking for. Finally, you said that the line goes through the origin, so the y-intercept of the line will be 0. If we substitute 3/2 for m and 0 for b in the equation y = mx + b, the result is y = (3/2)x + 0 or y = (3/2)x which is the answer to the problem. I hope this has helped. -Doctor Jaffee, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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