Two points, A and B, are on the Cartesian plane at (-1,4) and (9,6).

A. What point on the x-axis is equidistant from each of these two points?

B. What point on the y-axis is equidistant from these two points?

To obtain the X point on the x-axis, we realize that it has to be between -1 and 9. Therefore, making a right-angle triangle of sides x, 4 and L (where L is the equidistant point from X to point A and X to point B), we obtain L²=X²+16 using the Pitagorian theorem. Then we make another right-angle triangle but with sides (10-X), 6 and L. Again using the Pitagorian theorem, we obtain L²=(10-X)²+36=X²-20*X+136. So, L²=L²=X²+16=X²-20*X+136, this gives X=120/20=6.
The X point in the x-axis is (5,0), based from the origin.
Now, for the Y point on the y-axis, imagine an isoceles triangle of sides M, M and the distance between points AB (where the distande AB=√104). In an isoceles triangle, two angles are equal an the the third is the subtraction of 180 minus those two equal angles. The angle of aperture between the distance AB and the y-axis is ArcTan[(9+1)/(6-4)]=ArcTan[5]=78.690067526 degrees, so the angle between M and M should be 180-2*78.690067526=22.619864948 degrees. Using the law of cosines, we formulate an equation 104=√(M²+M²-2*M*M*Cos(22.619864948)), this equation results in M=√(52/(1-cos(22.619864948)))=26
The Y point in the y-axis is (0,30), based from the origin.

Posted by Antonio on 2003-09-03 09:34:09