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 Community Discussion Question:
locus
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Dec 2007 19:28:44 IST
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a line intersects the X axis at A (9,0)and Y axis at B (0,-7).a variable line perpendicular to AB cuts X axis at p and y axis at Q .if AQ and bp itersects at R then the locus of R
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8 Dec 2007 20:16:24 IST
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this took me a long time to figure out, i hope this is correct.
the let points p be (h,0) and q (0,k). since the slope of the original line is -7/9 and perpindicular line is 9/7 ,implying k/h=9/7.-(a)
the eqn of the line joining A and q is y-0=k-0/0-9(x-()
or k(9-x)=9y-let this be (1)
the eqn of the line joining B and p is y+7=0+7/h-0(x-0)
or h(7+y)=7x -let this be (2)
dividing 1 and 2 we get
k/h=9/7(y^2 + 7y)/(9x-x^2)
byt k/h=9/7( proved above)
hence the locus comes to be x^2+y^2+7y-9x=0 which is a circle with centre (-7/2,9/2).
good question... cheers:d
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