# Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

**Solution:**

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, 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^{o}

∠P + ∠N + ∠M = 180^{o }(Angle sum property of a triangle)

∠P + ∠M = 90^{o}

Clearly, ∠M is an acute angle.

∠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.

**Video Solution:**

## Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

### NCERT Maths Solutions Class 9 - Chapter 7 Exercise 7.4 Question 6:

**Summary:**

Thus, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.