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December 25, 2014

December 25, 2014

Posted by **Anonymous** on Wednesday, November 30, 2011 at 10:51am.

- geometry -
**Steve**, Wednesday, November 30, 2011 at 3:09pmIf we have to use the distance formula, then we have

LK = sqrt(9+16) = 5

so, JL also is 5.

The question is, where is J?

Note that since LK = 5, it is the hypotenuse of a right triangle, with legs parallel to the axes. Thus, the base of the triangle is parallel to the x-axis.

So, since the base extends to the left of the line x = -1 the same distance it extends to the right, namely 3,

J = (-4,1)

This problem would have been harder if the triangle had been oriented differently.

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