Intersection PointsDate: 03/13/99 at 04:55:17 From: sara Subject: Intersection Points A line is drawn through the point A(1, 2) to cut the line 2y = 3x - 5 in P and the line x + y = 12 in Q. If AQ = 2AP, find the coordinates of P and Q. The answers given in the book are P(4, 3.5) Q(7, 5) or P(2/5, -19/10) Q(2 1/5, 9 4/5) I have no idea how they got those answers. Please help. Date: 03/13/99 at 07:34:04 From: Doctor Mitteldorf Subject: Re: Intersection Points Here is how I would tackle it if I were you. First, I would get a piece of graph paper and find the point A, plot the two lines, and draw some lines from A out in the direction of those two lines to see how the line can be situated to make AP about half the length of AQ. So you have solved the problem approximately 'by eye'. That will give you a feel for what you are being asked to do. The next thing I would do would be to try the answers that the book gives you: convince yourself that point P is really on line 2y = 3x - 5. Find the distance between the point P and A. Do the same for Q, and see that their solution works. Do the same for their second solution. ---------------- Next, I would say: suppose the slope of the line we draw from A is m. Then, we can write the equation of the line as (y - 2)/(x - 1) = m. Solve this equation simultaneously with the first equation 2y = 3x - 5 and you will have an expression for the coordinates of P in terms of m. Now subtract the coordinates (1, 2) of the point A, and use the Pythagorean theorem to find an expression for the distance between point P and point A, all in terms of your one variable, m. Now do the same thing for point Q: end by finding an expression for the distance between A and Q, written as an expression in m. Then you are finally ready to write an equation in m that says one expression is twice the other. If you are lucky, this will not be too messy, and you will be able to turn it into a quadratic equation for m. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 07/19/99 at 19:36:45 From: Tim Greene Subject: Re: Intersection Points Dr. Math - My first inclination for solving this problem was apparently the solution you described (although I did not follow through on the details) - to solve the problem by using the Pythagorean theorem (distance formula) on segments AP and AQ and the fact that the slopes of segments AP and AQ are equal. When I attacked the problem that way, the algebra quickly got ugly, so I digressed to look for a quicker and easier approach. I had first, as you suggested at the beginning of your solution description, sketched a rough graph of the problem, and by examination I had determined that there would be two solutions: one line through A satisfying the conditions of the problem will have a small slope, with points P and Q in the same direction from A; another line through A satisfying the conditions of the problem will have a much steeper slope, with points P and Q on opposite sides of A. I began by labeling the two sets of solution points as P1 = (a,1.5a-2.5) and Q1 = (b,12-b); and P2 = (c,1.5c-2.5) and Q2 = (d,12-d). My simplification of the problem came from using the distance formula not on the diagonal segments AP and AQ but on the x- and y-components of those segments. For points P1 and Q1, for example, the equations which say that the x- and y-components of AQ are twice the corresponding components of AP are (b-1) = 2(a-1) and (10-b) = 2(1.5a-4.5); this is a set of linear equations which quickly gives the solution P=(4,3.5) and Q = (7.5). Similarly, the same approach for points P2 and Q2 gives another set of linear equations which give the solution P = (0.4,-1.9) and Q = (2.2,9.8) - Tim Greene |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/