Question 736111
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Let's use C to refer to the point of intersection of the segment AB with the line *[tex \LARGE x\ =\ 2].  By definition of bisector, point C must be the midpoint of AB, which is to say that the segments AC and CB must be equal in measure.  Since *[tex \LARGE x\ =\ 2] is a vertical line and is given as perpendicular to the line containing the segment AB, the line containing the segment AB must be a horizontal line.  Horizontal lines in *[tex \LARGE \mathbb{R}^2] have the characteristic that the *[tex \LARGE y]-coordinates of every point on the line are equal.  Since A and B are both on a horizontal line, and the *[tex \LARGE y]-coordinate of point A is 1, the *[tex \LARGE y]-coordinate of point B must be 1 as well.  Since C is not only an element of the set of ordered pairs that comprise the line that contains the segment AB, it is also an element of the set that comprises the line *[tex \LARGE x\ =\ 2].  That means that the *[tex \LARGE x]-coordinate of point C must be 2 as is the *[tex \LARGE x]-coordinate of every point on the line *[tex \LARGE x\ =\ 2].  Since A and C have identical *[tex \LARGE y]-coordinates (for the same reason that A and B have identical *[tex \LARGE y]-coordinates), the distance from A to C is quickly calculated by taking the absolute value of the difference between the *[tex \LARGE x]-coordinate of point A and the *[tex \LARGE x]-coordinate of point C, that is to say *[tex \LARGE |-4\ -\ 2|\ =\ 6].  Since AC and CB are congruent, the distance from C to B must be the same as the distance from A to C, namely 6 units.  6 units to the right of the point *[tex \LARGE (1,\,2)] is the point *[tex \LARGE (1,\,2\,+\,6)] which is to say *[tex \LARGE (1,\,8)].


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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