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Analytical Geometry

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30 Jun 2007 19:16:13 IST

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Que.>There is a point P inside an equilateral triangle ABC of side 'd' whose distance from vertices are 3,4,5.Rotate the triangle and point P through 60 degrees about C and let A goes to A' , B goes to the position of A and P to P'.Then find :-

a)area of triangle PAP',

b)value of d.

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10904him

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Joined: 29 Dec 2006

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30 Jun 2007 19:20:29 IST

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Real tough cant solve Anybody pls help it out.........

Ramya Hegde

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30 Jun 2007 23:55:29 IST

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Ok, i've tried out the question, but i'm getting some weird answer for the area. So, i'm posting the method, plz tell me if there's any mistake.

Area is coming out to be =2(1-d²)(d²-49) +(d²-7)*d²-4

Let us take coordinates of A,B,, C as A(0,0), B(d,0) and C(d/2, 3d/2)

Let P(p,q)

now using the distance formula from the given condition,

(p-d)²+q²=16

p²+q²=9

on solving, p=(d²-7)/2d

q= (1-d²)(d²-49)/2d

Now, on rotating the triangle, the distances of P'from the vertices still remain the same. And angle between A^'C and AC is 60⁰. slope of AC=3

so using the slope formula, u get two values of m, m=0,-3

if u consider the negative slope, it'll represent the original line CB, and A^' will coincide with B, and since thats not the case, we should take m=0

so equation of A^'C is,

y=3d/2

hence A^'(-d/2,3d/2) (using distance relation between A^' and B, which is still d)

Now, let P^' be (r,s)

using distance formulae,

r²+s²=16

(r+d/2)²+(s-3d/2)²=9

(r-d/2)²+(s-3d/2)²=25

on solving, r=-8/d

and s=4/d*d²-4

Now P(d²-7/2d , (1-d²)(d²-49)/2d )

A(0,0)

P^'(-8/d, 4/d*d²-4)

on solving the area is coming out to be,

2(1-d²)(d²-49)+ (d²-7)*d²-4

Plz tell me if i'm wrong. I'm waiting for ur reply.

Thank u.

Amrita bhavaraju

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2 Jul 2007 17:39:10 IST

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Hii ramya....how is A'=(-d/2,

3d/2)????...isn't it (d/2,-

3d/2)??? plzz do help me come out of this doubt...

Bipin Dubey

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Joined: 23 Jan 2007

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22 Oct 2007 05:59:22 IST

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Well done Ramya.

Ramya Hegde

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Joined: 8 Feb 2007

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23 Oct 2007 02:48:57 IST

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Thank you!!

To amrita-
well, the triangle is rotated in such a way that A' is now in the second quadrant (bcoz B now coincides with A) and it lies on the line parallel to x-axis and passing thru C. So the y-coordinate thruout the line for all pts. remain the same and is equal to the y-coordinate of C. If u still have any doubts, do nudge me!!

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