IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

vivazcool7

Consider an equilateral triangle ABC.Vertices B and C lie on the line x+y=2.Find the coordinate of vertices A,B,C.Given that centroid of the triangle is (

2,2).

ramyadiamond

Take coordinates as A(h,k) B(m,2-m) C(p,2-p).

h+m+p/3=

2,

so h+m+p =3

2 -{1}

similarly, k+2-m+2-p=6

k-m-p=2 -{2}

Also,

slope_AB =k+m-2/h-m

and slope_BC= -1 (given)

slope_AC= k+p-2/h-p

angle between AB and BC is 60⁰ and similarly, for BC and AC. In this way u can make conditions, and solve the problem.

Hope this helps

_{{if it has, plz rate me}}

joyfrancis

well done

vivazcool7

its gud idea but could get coordinates of A only.
What about others???

ramyadiamond

dunno if it'll work, but just a suggestion, coz i've not tried it on paper yet. Just try it.

Write the equaton of a line perp. to the given line, which should be of type x-y+c=0 satisfy coordinates of A and find const. c. now, the midpoint of B and C lie on this line, as the given triangle is equilateral, right? so, u might get another equation for m and p. Try solving this way and let me know.

vivazcool7

coordinates of A(2

2 , 2+

2)

coordinates of B and C are pretty tuf to find. plz give it a try

fwks_phoenix

@ramya

for A(h,k) B(m,2-m) C(p,2-p).

wouldn't slopeAB= (2-m-k)/(m-h) ???????

m=y2-y1/x2-x1right ?

vivazcool7

B=(

2 - 1/

2 -

3/2 ,

2 - 1/

2 +

3/2) and

C=(

2 - 1/

2 +

3/2 , 2 - 1/

2 -

3/2)

plz rate if i'm correct

ramyadiamond

fwks_phoenix,

both k+m-2/h-m and 2-k-m/m-h are the same. if u take out -1 from both numerator and denominator as common, then u'll get the same.

vivazcool7

is the answer i gave correct???A is given in the scrap above B and C
plz rate me if i'm correct

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