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Trignometry
18 Nov 2006 03:09:35 IST
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Manasi
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Joined: 1 Dec 2006
Posts: 2108
2 Dec 2006 12:23:45 IST
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it can be done using concept of family of straight lines
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31 Dec 2006 22:42:24 IST
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vectors suggest no way...........it will be very lengthy by that method........easiest way is to find the equation using family of lines and then to find the parameter the best way is to see that all medians divide the triangle into two triangles of equal areas so let the first eqn be
a1 x + b1 y + c1 =
(a2 x + b2 y + c2)
then this eqn and
a2 x + b2 y + c2 = 0 .... (2)
a3 x + b3 y + c3 = 0 .... (3)
should form a triangle whose area is half the area of the tiangle formed by
a1 x + b1 y + c1 = 0 .... (1)
a2 x + b2 y + c2 = 0 .... (2)
a3 x + b3 y + c3 = 0 .... (3)
which
gives the requied value of
thats simple by using determinants method............similarly two other equations can be found out...................easy 2 step approach
a1 x + b1 y + c1 =
(a2 x + b2 y + c2)then this eqn and
a2 x + b2 y + c2 = 0 .... (2)
a3 x + b3 y + c3 = 0 .... (3)
should form a triangle whose area is half the area of the tiangle formed by
a1 x + b1 y + c1 = 0 .... (1)
a2 x + b2 y + c2 = 0 .... (2)
a3 x + b3 y + c3 = 0 .... (3)
which
gives the requied value of
thats simple by using determinants method............similarly two other equations can be found out...................easy 2 step approach
7 Jan 2007 00:04:21 IST
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Let triangle be ABC
Let D1(h1,k1) be the point lying on BC(a2x+b2y+c2=0) . The median bisecting this side let be (y-k1)=m1(x-h1);
this median and the other 2 sides are concurrant hence
a1x + b1y + c1=0(if c1 is not zero)
this is similar to
(a1/c1)x + (b1/c1)y +1=0 ----------1----------
(a3/c3)x + (b3/c3)y +1=0 ----------2----------
m1(x-h1) - (y-k1) =0 ----------3---------
1, 2, 3 are concurrant(they obviously meet at A)
hence the determinant of coefficients is zero(as there should be only one solution to this linear system of 3 equations)
therefore we get a relation in a1, b1, c1, a2, b2, c2, h1, k1, m1
m1= { (k1(a2.b1-a1.b2)+(a2.c1-a1.c2) } / {h1(a2.b1-a1.b2) + (c2.b1-c1.b2) }
so now by generalising h1.k1 to x,y we get the equation to the median cutting BC; also since the expression is symmetric; we can cyclically change (a1, a2,a3) ; (b1,b2,b3) and (c1,c2,c3) to get m2 and m3 and hence the equations of the other 2 medians
Thank you
Kaushik, Bangalore
Let D1(h1,k1) be the point lying on BC(a2x+b2y+c2=0) . The median bisecting this side let be (y-k1)=m1(x-h1);
this median and the other 2 sides are concurrant hence
a1x + b1y + c1=0(if c1 is not zero)
this is similar to
(a1/c1)x + (b1/c1)y +1=0 ----------1----------
(a3/c3)x + (b3/c3)y +1=0 ----------2----------
m1(x-h1) - (y-k1) =0 ----------3---------
1, 2, 3 are concurrant(they obviously meet at A)
hence the determinant of coefficients is zero(as there should be only one solution to this linear system of 3 equations)
therefore we get a relation in a1, b1, c1, a2, b2, c2, h1, k1, m1
m1= { (k1(a2.b1-a1.b2)+(a2.c1-a1.c2) } / {h1(a2.b1-a1.b2) + (c2.b1-c1.b2) }
so now by generalising h1.k1 to x,y we get the equation to the median cutting BC; also since the expression is symmetric; we can cyclically change (a1, a2,a3) ; (b1,b2,b3) and (c1,c2,c3) to get m2 and m3 and hence the equations of the other 2 medians
Thank you
Kaushik, Bangalore
23 Jan 2007 21:10:21 IST
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this problem consists of four small steps:
1)finding the equation of AB
2)finding M(the intersection point of AB and OP)
3)finding the mid point of PM
4)finding the equation of P'Q'
every step is a short step.
in the first step, let A be (x,y) and C be centre of OP. AC=CP completes first step
infourth step, slope and a point are known.
answer comes out to be x(x1+g)+y(y1+f)=PA/2
1)finding the equation of AB
2)finding M(the intersection point of AB and OP)
3)finding the mid point of PM
4)finding the equation of P'Q'
every step is a short step.
in the first step, let A be (x,y) and C be centre of OP. AC=CP completes first step
infourth step, slope and a point are known.
answer comes out to be x(x1+g)+y(y1+f)=PA/2
23 Jan 2007 21:18:47 IST
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this problem consists of four small steps:
1)finding the equation of AB
2)finding M(the intersection point of AB and OP)
3)finding the mid point of PM
4)finding the equation of P'Q'
every step is a short step.
in the first step, let A be (x,y) and C be centre of OP. AC=CP completes first step
infourth step, slope and a point are known.
answer comes out to be x(x1+g)+y(y1+f)=PA/2
1)finding the equation of AB
2)finding M(the intersection point of AB and OP)
3)finding the mid point of PM
4)finding the equation of P'Q'
every step is a short step.
in the first step, let A be (x,y) and C be centre of OP. AC=CP completes first step
infourth step, slope and a point are known.
answer comes out to be x(x1+g)+y(y1+f)=PA/2
23 Jan 2007 21:19:07 IST
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Let triangle be ABC
Let D1(h1,k1) be the point lying on BC(a2x+b2y+c2=0) . The median bisecting this side let be (y-k1)=m1(x-h1);
this median and the other 2 sides are concurrant hence
a1x + b1y + c1=0(if c1 is not zero)
this is similar to
(a1/c1)x + (b1/c1)y +1=0 ----------1----------
(a3/c3)x + (b3/c3)y +1=0 ----------2----------
m1(x-h1) - (y-k1) =0 ----------3---------
1, 2, 3 are concurrant(they obviously meet at A)
hence the determinant of coefficients is zero(as there should be only one solution to this linear system of 3 equations)
therefore we get a relation in a1, b1, c1, a2, b2, c2, h1, k1, m1
m1= { (k1(a2.b1-a1.b2)+(a2.c1-a1.c2) } / {h1(a2.b1-a1.b2) + (c2.b1-c1.b2) }
so now by generalising h1.k1 to x,y we get the equation to the median cutting BC; also since the expression is symmetric; we can cyclically change (a1, a2,a3) ; (b1,b2,b3) and (c1,c2,c3) to get m2 and m3 and hence the equations of the other 2 medians
Thank you
Kaushik, Bangalore
Let D1(h1,k1) be the point lying on BC(a2x+b2y+c2=0) . The median bisecting this side let be (y-k1)=m1(x-h1);
this median and the other 2 sides are concurrant hence
a1x + b1y + c1=0(if c1 is not zero)
this is similar to
(a1/c1)x + (b1/c1)y +1=0 ----------1----------
(a3/c3)x + (b3/c3)y +1=0 ----------2----------
m1(x-h1) - (y-k1) =0 ----------3---------
1, 2, 3 are concurrant(they obviously meet at A)
hence the determinant of coefficients is zero(as there should be only one solution to this linear system of 3 equations)
therefore we get a relation in a1, b1, c1, a2, b2, c2, h1, k1, m1
m1= { (k1(a2.b1-a1.b2)+(a2.c1-a1.c2) } / {h1(a2.b1-a1.b2) + (c2.b1-c1.b2) }
so now by generalising h1.k1 to x,y we get the equation to the median cutting BC; also since the expression is symmetric; we can cyclically change (a1, a2,a3) ; (b1,b2,b3) and (c1,c2,c3) to get m2 and m3 and hence the equations of the other 2 medians
Thank you
Kaushik, Bangalore
22 Feb 2007 14:20:40 IST
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eqn of median thro B is given by (1)+ k(2)=0
similarly the eqns of medians thro A and C can be given....
now, all the medians have a common soln (the centroid)
hence we have the determinant of coefficient matrix to be 0 (a homogenous system of eqns in x,y)
so, we can solve for k1, k2, k3... and hence the problem..
(Here, (1) => a1x+b1y+c1=0....)
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