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so the ODE is indeed exact and there is a solution of the form . We have
With , we have
so
It has been proven that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
<h3>How to prove a Line Segment?</h3>We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.
Let us take a line l and from point P as shown in the attached file, that is, not on line l, draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In ΔPNM, ∠N = 90°
∠P + ∠N + ∠M = 180° (Angle sum property of a triangle)
∠P + ∠M = 90°
Clearly, ∠M is an acute angle.
Thus; ∠M < ∠N
PN < PM (The side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
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