Finding Three Lines that Form a TriangleDate: 08/30/2001 at 16:30:56 From: Joe Kowalski Subject: Triangle Area Puzzle The question is: Find an equation for each of the lines that goes through the point (3,2) forming a triangle of area 12 with the coordinate axes. There are three answers. I have been able to come up with one answer, which is y = (-2\3)x+4. The only way I've been trying so far is trial and error. My teacher said we really wouldn't be able to figure it out any other way with our current knowledge. Date: 08/31/2001 at 09:25:14 From: Doctor Rob Subject: Re: Triangle Area Puzzle Thanks for writing to Ask Dr. Math, Joe. In general, there can be as many as four such triangles, but in the particular case you were given, the number is just three. Here is a rough diagram: ^y | | A o. | `-. | `-. | `-. | `-. P(3,2) | __`o. C | __,,--''_- : `-. __,,o-'' _- : `-. __,,--'' | _- : `-. x --------o---------------o------o-------o-----------o---> D O| _- E Q B | _- F o | | (Actually, points D and F are much farther from the origin than shown in the diagram.) The three triangles are OAB, OCD, and OEF. You have found OAB by finding line AB to be y = (-2/3)*x + 4, and lengths OA = 4 and OB = 6. If you know something about similar triangles, the formula for area, how to solve quadratic equations, and how to calculate with expressions involving square roots, you can find all three lines exactly, and without trial-and-error. To find CD, drop a perpendicular from P to the x-axis, meeting it at point Q. We know length QP = 2 and OQ = 3. Then the right triangles DPQ and DCO are similar, because they share angle D and each has a right angle (<DQP and <DOC), so all three angles must be respectively equal. Call length DO a, and length OC b. Then we have two equations involving a and b: one from the area being equal to 12, and the other because corresponding sides of similar triangles are proportional: 12 = area = (1/2)*a*b, b/2 = OC/QP = DO/DQ = a/(a+3). From the first equation, b = 24/a, and, substituting that into the second equation, we get 12/a = a/(a+3), 12*(a+3) = a^2, 0 = a^2 - 12*a - 36, a = 6 + 6*sqrt(2) (ignoring the negative root as extraneous), b = 24/a, = 24/(6+6*sqrt[2]), = 4/(1+sqrt[2]). Now to rationalize the denominator, multiply both numerator and denominator by -1 + sqrt(2): b = 4*(-1+sqrt[2])/([1+sqrt(2)]*[-1+sqrt(2)]), = 4*(-1+sqrt[2])/(-1+2), = -4 + 4*sqrt(2). Now the points (0,b) and (3,2) are on the line (so is (-a,0), but we only need two), so we can find the equation of line CDP using the two-point form: (y-y1)*(x2-x1) = (y2-y1)*(x-x1), (y-b)*(3-0) = (2-b)*(x-0), 3*(y-b) = (2-b)*x, y = ([2-b]/3)*x + b, y = ([6-4*sqrt(2)]/3)*x + 4*(-1+sqrt[2]). The analysis of the other case, line EFP, is similar, using the fact that right triangles EQP and OEF are similar. We get that the equation of line EFP is y = ([6+4*sqrt(2)]/3)*x + 4*(-1-sqrt[2]). Do you think this is beyond what you know? - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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