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the sides of a parallelogram are on the lines $$x-3y+20=0,\\ x+y+6=0,\\ x-3y-10=0 \text{ and} \\ x+y+2=0.$$ Find its area. solve using the fourth standard equation of the straight line.

Dominic Michaelis
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hannah
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    What is the fourth standard equation of the straight line? – superAnnoyingUser Feb 17 '13 at 13:26
  • xcos(theta) + ysin(theta) = p – hannah Feb 17 '13 at 13:29
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    What are the first, second, and third standard equation of the straight line? – Julien Feb 17 '13 at 13:41
  • Maybe there is a smarter way. But this works. Find the intersections of your lines, then use http://www.jtaylor1142001.net/calcjat/Solutions/VCrossProduct/VCPAParallelogram.htm – Julien Feb 17 '13 at 13:43
  • julien, im not quite sure what are the first, second, and third standard equation of the straight line. thanks for the link you sent, but our teacher told us to solve using the fourth standard equation. – hannah Feb 17 '13 at 13:47

2 Answers2

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A less mysterious answer than rlgordonma's would be to solve for the four intersections (four sets of 2x2 linear systems of equations; 3 will do in fact), and use Heron's formula for the area of one of the resulting triangles, twice that is your answer.

vonbrand
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    Fair enough, but there is a solution still less mysterious! Find the vertices, and then use scalar product to compute the area - it will lead to much less computation. – Jakub Konieczny Feb 17 '13 at 16:48
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First you should solve the eqns of the adjacent side and Not the eqns which are parallel.then you would get the value of coordinates then find the area of the triangle Separated by the diagonal then 2* area of the triangle

user445971
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