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.

☛ Check: Class 9 Maths NCERT Solutions Chapter 7

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.

☛ Related Questions:

• Show that in a right-angled triangle, the hypotenuse is the longest side.
• In Fig. 7.48, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
• In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
• AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that ∠A > ∠C and ∠B > ∠D.