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

sudarshan kumar, 11 years ago

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

10 years ago

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