The coordinates of the three points O, A and B are (0,0),(0,4) and (6,0) respectively. If a point P moves so that the area of ΔPOA is always twice the area of ΔPOB, the locus of P is
A
x−3y=0
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B
x+3y=0
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C
3x+4y=0
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D
3x−4y=0
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Solution
The correct options are Cx−3y=0 Dx+3y=0 Let the coordinates of point P be (x,y) ∴ΔPOA=∣∣
∣∣∣∣
∣∣xy1001041∣∣
∣∣∣∣
∣∣=|−4x|=|4x|, (expanding along first column) And ΔPOB=∣∣
∣∣∣∣
∣∣xy1001601∣∣
∣∣∣∣
∣∣=|6y|, (expanding
along second column) Now given ΔPOA=2⋅ΔPOB ⇒|4x|=2⋅|6y|⇒|x|=3|y| Squaring both sides we get, x2=9y2⇒x2−9y2=0 Hence required locus of P is x2−9y2=0 i.e. x+3y=0 or x−3y=0