How do you prove geometrically that the perpendicular segment fomr a point to a line is the shortest segment..

Question

from the point to the line?

I mean like... prove it! No compasses no constructions, steps/reasons, cuz i cant think of one

and Paekm:

you're telling me what i already know, but it isn'te going to help me prove it

Can anyone prove the Pythagorean theorem then?...

Answer 1

Proof By Contradiction

set the length of the perpendicular line to A
assume another line from the point to the line is the shortest distance. (Length B) (it will be the hypotenuse of a right triangle)

Disprove your assumption because every hypotenuse is longer than the leg intersecting the line at a right angle i.e.
B = sqrt(A^2 + C^2) C is the separation of the intersection points on the straight line.

Thus for every C, B > A because sqrt(A^2 + C^2) > A
Therefore your assumption that there is a B shorter than A is false and A is the shortest distance.

Answer 2

You can draw a right triangle. Since the hyp > leg, you found contradiction. So, you have proved that the shortest line segment joining a point to a line is the perpendicular line segment.

Answer 3

Let P be the point that is not on the line, and let X be the other endpoint of the perpendicular segment.

Assume that the perpendicular segment PX is not the shortest distance from P to the line. Then there must be some other point (call it Y) on the line such that PY < PX. But since PY is now the hypotenuse of right triangle PXY, we know that
(PY)^2 = (XY)^2 + (PX)^2.

and

(PY)^2 - (PX)^2 = (XY)^2.

As all segment lengths are positive, the square of PY must be less than the square of PX, in which case

(PY)^2 - (PX)^2 < 0.

This leads to a contradiction, since this same difference,
(PY)^2 - (PX)^2, was already stated to be a positive amount, (XY)^2. Therefore, the original assumption is absurd; The perpendicular segment PX must be the shortest distance from point P to the line.

Answer 4

Let AB be a straight line and P be a point outside it.
Let PQ be the perpendicular from P on AB.One more point R is taken on AB.PR is joined.
To prove that PQ Proof: In triangle PQR, m Therefore m In triangle PQR,
m m Therefore PR>PQ [As in a triangle the side opposite a greater angle is greater than the side opposite a smaller angle]
Similarly we may take as many points on line AB but all shall be greater than PQ.
Therefore,it is proved that the perpendicular segment from a point to a line is always the shortest segment.

Answer 5

Think of a triangle on the line. the hypotenuse is the longest and the opposite is the shortest which is also the perpendicular line. if you want to prove pythogoreas theorem, there are many ways to do that. i suggest you check the internet

Answer 6

Let AB be a line and P be a point not on the line. Let PM be the perpendicular from to AB. Now, draw a line PN from P to any other point on AB. We must prove that PM < PN

After the drawings finish, we get a right triangle PMN with
From triangle inequalities, we know that in a triangle, the side opposite larger angle is longer. Since PN is opposite to
PM < PN which is what we had to prove.

So, The shortest line segment connecting a line to a distict point not on the line is perpendicular to the line.

Answer 7

Use a compass to draw a circle with the radius as the length of that perpendicular.


*edit: well it's the same as proving. You proved that there are no other lines that are within the circle drawn, meaning no lines are shorter than the perpendicular


*edit: c^2 = a^2 + b^2 - 2ab*cosC
where a, b and c are sides of any triangle and C is the angle between a and b
so we actually always have cosC = 0 <---> C = 90 deg for the lengths to be the largest

Answer 8

Since the segment is PERPENDICULAR to the line, they already form a RIGHT ANGLE.

With this formation, we can create a RIGHT TRIANGLE. The shortest of all the sides is the OPPOSITE. The longest is the HYPOTENUSE.

From your problem, the SEGMENT becomes the OPPOSITE side of a RIGHT TRIANGLE. Which is the SHORTEST side of a RIGHT TRIANGLE.

Answer 9

Draw two lines from the point: one prependicular to the line (line a) and one is not prependicular (line b). It forms a triangle with one of its angle is prependicular. It is obvious that in that triangle, line a is always shorter than line b (phytagoras role). It is valid for any line b drawn from the point to the line will be longer than line a or will be the same length as a only if line b is in conjunction with line a.

Hope this will explains.

Answer 10

Wow... this is really sad. So far I seem to be the only one that actually understands your question. Unfortunately, I dont have an answer for you

Some of the simplest, obvious truths in nature have such of a difficult proof to them. It isnt so obvious, is it, how one might prove it.

What I find particularly annoying is this phenomenon I am observing right now taking place in the school systems. Teachers that fail to teach some things that they take for granted... so obvious to them that they dont even teach it. Its sickening when someone who proclaims their own intelligence, who can do calculus behind their backs, cant explain a basic algebraic concept.