Question 30

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Jambmaths question: 

Find the locus of a particle which moves in the first quadrant so that it is equidistant from the line x = 0 and y = 0 (where k is a constant)

Option A: 

$x+y=0$

Option B: 

$x-y=0$

Option C: 

$x-y-k=0$

Option D: 

$x+y+k=0$

Jamb Maths Solution: 

Answer Option B

The locus of the particle is the line passing through the origin (which make k = 0) and is the bisector of AO and OD (i.e line OC will bisect $\angle AOD$) which makes $\angle COD={{45}^{\circ }}$

The slope is $m=\tan \theta =\tan {{45}^{\circ }}=1$

Using one point slope equation $(x,y)=(0,0)$

$(y-{{y}_{1}})=m(x-{{x}_{1}})$

$y-0=1(x-0)$

$y=x$

$x-y=0$

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