Question 774723
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"Diagonal Segment" is really non-standard terminology, however I suspect you mean a segment that is not orthogonal to any of the axes in your coordinate system.  Endpoints of segments orthogonal to the axes in *[tex \LARGE \mathbb{R}^2] either have identical *[tex \LARGE x]-coordinates in the case of a segment perpendicular to the *[tex \LARGE x]-axis and parallel to (or coincident with) the *[tex \LARGE y]-axis, or they have identical *[tex \LARGE y]-coordinates in the case of a segment perpendicular to the *[tex \LARGE y]-axis and parallel to the *[tex \LARGE x]-axis.  In your case, the segment is non-orthogonal because *[tex \LARGE x_1\ \not =\ x_2\ \ ] AND *[tex \LARGE \ \ y_1\ \not =\ y_2]


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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