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Plot the region in the first quadrant in...

Plot the region in the first quadrant in which the points are nearer to the origin thatn to the line `x=3`.

Text Solution

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Point P(x,y) lies in the first quadrant. Therefore, `x,ygt0`

Also point P(x,y) is nearer to the origin than to the line x=3
`"Now "OP=sqrt(x^(2)+y^(2))`
Distance of P from line x-3=0 is PM =3-x
According to the question, `OPltPM`
`rArr" "sqrt(x^(2)+y^(2))lt(3-x)`
`"or "x^(2)+y^(2)ltx^(2)-6x+9`
`"or "y^(2)lt-6x+9`
`"or "y^(2)lt-6(x-(3)/(2))`
Points satisfying above inequality lie inside parabola `y^(2)=9-6x` in the first quadrant.
`"Parabola "y^(2)=-6(x-(3)/(2))` is concave to the left, having axis as
x-axis and vertex at `((3)/(2),0)` and directrix x=3
The required region as shown in the figure.

`"Area of the region "=overset(3)underset(0)int(9-y^(2))/(6)dy=(1)/(6)(9y-(y^(3))/(3))_(0)^(3)`
`=(1)/(6)(27-9)=3` sq. units
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