Question

o Prove that the perpendicular is the shortest line segment joining an external point to a line, i.e prove the following theorem. THEOREM. Let / be a line. let P be an external point for the line and let F be the foot of the perpendicular from P to l. If R is any point on that is different from F, then PR > PF

Answer 1

To prove this theorem stating that 'The perpendicular is the shortest line segment joining an external point to a line', a simple graphical approach has been used.

The problem can be solved in two methods such as 'Pythagoras theorem' and 'Distance calculation between points'. The latter one has been used here to find lengths of PF and PR where P, F and R are an external point, foot of the perpendicular and a random point respectively corresponding to the line 'l' that is taken on X-axis here. The points on graph are identified and the distances between them are calculated.

The calculation of lengths PF and PR proved that ''PR > PF'' and hence the theorem is proved.

gaven datar Question : - Let "l be a dne. → up' be an external point for the line and & be the foot of the perpendicular from Ptol. • Here, we have to prove that of 'R' is any point on 'l that as different from f', then PRYPE. • In other words to perpendicular is the shortest line. gegnent joming an external point to a lime”. • we can prove this in 2 ways &..., using Pythagoras theorem or using distance calculations. • we will do this úgma distance calculations on a graph. Scale: On X-oxis 1 unit=1cm. On Yorkis T Unit=lcom P(3,4) Aly 3 į į § ¢ io 7 - F(30sem IRCG i 2 4 5 K al - • In the above graph, * is a time on X-asers. →P (3,4) is an external point, > F(3,0) is the foot of the perpendicular and → R (60) is another point on l? ..The distance between two points AG, 9,) and Brea, g) en the length of alme rs given as To ABC 522-X,)+ cya yu • Using this formula, let us find PF and PR

3, 4, X2, Z • hength of the perpendicular PEP (34) & E(3,0) I PE = √(22-x,]²+ (% 5,2 here x=3, 9, =4 12=3 , 42=0 . PF = √(3-3)²+ (0-4)2 .PF = √(0)²+ (-432 = 50+ 16 = 576 PF = 4cm • Gemilarly, dength of dine PR, P (3,4) & R (6,0) PR = {(2, -x)+ (% -9,22 Here *,=3, 9,=4 PR = √(6-3)² + (0-42 X2=6, 4, 20 PRE 5039*+(-4)2 89+16 - 25 PR = 5 cm → clearly 574 and hence PR >PE • Therefore, for a line P, & being an extemal point and of being the foot of the perpendicular from "p to d', if R as omy point and different from t, then PR > PE n..., the perpendicular as the shortest time segment joining an external point to a line. Hence proved