Please help with this geometry question
1 answer:
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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Answer:
5 units
Step-by-step explanation:
The distance from a point (m, n ) to the line Ax + By + C = 0 is given by
d =
For the point (- 1, 3 ) , with m = - 1 and n = 3
3x - 4y = 10 ( subtract 10 from both sides )
3x - 4y - 10 = 0
with A = 3 , B = - 4 , C = - 10
d =
=
=
=
= 5