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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Sep 2007 17:25:01 IST
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don't solve the sum for me but give me an idea about how to solve it.....
the sum is like this....
A is the point (-1,0) and B is the point (1,1) . Find a point on the line 4x+5y=4 which is equidistant from A and B......
please help me fast guys....!!!
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6 Sep 2007 17:31:52 IST
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use the distance formula dear
find the locus and solve both the equations (the eq u get and 4x+5y=4 )
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~William Shakespeare
You were born an original. Don't die a copy.
~John Mason |
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6 Sep 2007 17:33:23 IST
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can we apply mid point formula??
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6 Sep 2007 17:34:21 IST
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no..............midpoint is only a special case when the three points lie on the same line(they r collinear)
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God has given you one face, and you
make yourself another.
~William Shakespeare
You were born an original. Don't die a copy.
~John Mason |
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6 Sep 2007 17:35:18 IST
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A (-1,0) B (1,1) Let P(x,y) be a point equidistant form A and B So PA = PB => PA2 = PB 2 Use the distance formula d = sqrt[ (x1 - x2)2 + (y1 - y2)2] Find PA PB equate we get an equation then do simultaneous equating of the euation we got and 4x+5y=4 .
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In the process of learnin..............blunders do happen !!! |
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6 Sep 2007 17:38:15 IST
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you get the answer to be (1/4 , 0)
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In the process of learnin..............blunders do happen !!! |
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6 Sep 2007 17:42:59 IST
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see the answer given in the book is (-1/4,1)
and that is the answer i am getting
i don't think that u have to do simultaneous equation........
we use distance formula and get an equation..
the concept is that the equation we get is equal to 4x+5y-4=0
so we can find out x and y from it....!
cheers.......!!!!
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6 Sep 2007 18:02:46 IST
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I method:Verify options which satisfy the given line and equidstant
to given points.
II method:Find perpendicular bisector of line segment joining given
points and solve with given equation.
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***T.Venkat*** |
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7 Sep 2007 19:18:39 IST
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ya ya i did some mistake ........
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