document.write( "Question 6664: Dear Sir/Madam,
\n" ); document.write( "I am confronted with the following problem:\r
\n" ); document.write( "\n" ); document.write( "\"Let parallelogram ABCD have vertices on the xy-coordinate plane A(0,0), B(b,0), and C(a,c). Find an equation for the distances between the points B and D. Assume a, b, and c are all positive and assume a > b.\"\r
\n" ); document.write( "\n" ); document.write( "I visualize this parallelogram (well, at least partially) and have drawn myself a sketch of it, but how would you go about solving this question because I can't seem to figure it out?!
\n" ); document.write( "Thanks in advance.
\n" ); document.write( "Regards,
\n" ); document.write( "-Mike \n" ); document.write( "

Algebra.Com's Answer #3677 by prince_abubu(198) $\"\"$ $\"About$

You can put this solution on YOUR website!
It's tough to show you how this works if I can't show you a drawing. For sure, though, you know that point B lies further to the positive direction along the x-axis from A, so segment AB forms the very base of your parallelogram. They said that point C is at (a,c). If a > b, then point C is further to the right (positive direction) than point B, and it's c units above the x-axis. If you connect a line from AB and then from BC, you should get an obtuse angle opening \"northwest\".\r
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\n" ); document.write( "\n" ); document.write( "The trick here is getting the coordinates of point D, the missing point that would complete the parallelogram. Since we now know that CD must be parallel to AB, the y-coordinate of point D must be c. So far, the coordinate point of point D is (?,c). We don't know yet what D's x-coordinate is.\r
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\n" ); document.write( "\n" ); document.write( "Examine the diagonal segment BC (the right edge of the parallelogram). Its slope is $\"+%28c-0%29%2F%28a+-+b%29+\"$ which actually is $\"+c%2F%28a+-+b%29+\"$ . We'll have to examine the (a - b) because that's the change in x from point B to point C. The change in x from point A to point D MUST also be the same change in x as the one from point B to point C. That way, BC and AD are parallel. So, from point A, which is coodinate point (0,0), we need to shift-copy the x-coodinate by (a - b) (to the right) and bring that \"copy\" up vertically by c so that it's horizontally aligned with point C. This makes point D have the coordinate point (a-b,c).\r
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\n" ); document.write( "\n" ); document.write( "Now, we have to plug in the x and y values from point C and D to the distance formula:\r
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\n" ); document.write( "\n" ); document.write( " $\"+sqrt%28+%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D-y%5B2%5D%29%5E2%29%29+\"$ <----- formula
\n" ); document.write( " $\"+sqrt%28+%28%28b+-+a%29+-+b%29%5E2+%2B+%28c+-+0%29%5E2%29%29+\"$ <---- plugged values in.
\n" ); document.write( " $\"+sqrt%28+%28-a%29%5E2+%2B+%28c%29%5E2%29%29+\"$ <---- further simplification.
\n" ); document.write( " $\"+sqrt%28+a%5E2+%2B+c%5E2+%29\"$ <---- This ultimately should be the expression for the distance between points C and D. \n" ); document.write( "

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