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prove that a point can be found which is same distance from four point (am1,a\m1), (am2,a\m2),(am3,a\m3),(a/m1m2m3,am1m2m3) sl loney ch 2 prob. 27 page 15

prove that a point can be found which is same distance from four point (am1,a\m1), (am2,a\m2),(am3,a\m3),(a/m1m2m3,am1m2m3) sl loney ch 2 prob. 27 page 15


 

Grade:11

3 Answers

RAJORSHI PAUL
36 Points
11 years ago

If you observe carefully you will find that the given points are not collinear. We will start by considering the first three points. We know that there always exist a circle that pass through any three non collinear points. So the centre of ths circle is equidistant from the three points. Now we will prove that the fourth point also lies on this circle. 

Let us represent the first three points parametrically as (am,a/m). Let the centre of this circle be (α,β). Let the radius of the circle be r. So 

 (am-α)2 + (a/m-β)2 = r2

So this is biquadratic equation. So it must have four roots. m1,m2,m3 are already roots of this eqn. Let the fourth root be m4.

If you expand the eqn you will find that the product of the roots is 1. ie m1m2m3m4 = 1

So we get m4 = 1/m1m2m3

 ie the fouth point is (am4,a/m4) ie (a/m1m2m3,am1m2m3)

Please like the solution if you understand it.

RAJORSHI PAUL

1st year 

Dept of Electrical Engineering

IIT Kharagpur

ankitesh gupta
63 Points
11 years ago

CONSIDER THAT THESE POINTS LIE ON THE CIRCLE THEN THE POINT WHICH WILL BE EQUIDISTANT FROM ALL THESE POINTS WOULD BE THE CENTER OF THE CIRCLE AND THE DISTANCE WOULD BE THE RADIUS CONSIDER THE CENTER TO BE (a,b) THEN BY USNG DISTANCE FORMULA EQUATE THE DISTANCES BETWEEN (a,b) AND ALL THOSE FOUR POINTS AND GET YOUR ANSWER 

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ankitesh gupta
63 Points
11 years ago

 

 

 

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ APPROVE THE ANSWER IF IT HELPED YOU BY CLICKING THE YES BUTTON...............Smile

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