Equation of locus

Geometry Level 2

Given three points O = ( 0 , 0 ) , O=(0, 0), A = ( 3 , 0 ) A=(3, 0) and B = ( 0 , 1 ) , B=(0, 1), find the equation of the locus of point P P that satisfies 2 O P ‾ 2 = A P ‾ 2 + B P ‾ 2 . {2\overline{ OP } }^{2}={\overline{ AP } }^{2}+{\overline{ BP } }^{2}.

y = 4 x − 9 y=4x-9 y = − 3 x + 5 y=-3x+5 y = − 5 x − 12 y=-5x-12 y = 2 x + 6 y=2x+6

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Lawahar Qasid by Lawahar Qasid

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1 solution

let locus be P(h,k) now op^2=h^2 +k^2.....by distance formula similarly AP^2=(h-3)^2+k^2 and BP^2=h^2+(k-1)^2 Putting these values in the equation 2(OP)^2=(AP)^2 +(BP)^2 , we get 3 h + k =5 replacing h & k with x & y respectively we get y = - 3 x +5

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