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

Explanation

Let variable Point P(x,y) lie in 1 st quadrant.
`:." "x,ygt0`

Also, point P(x,y) is nearer to origin that to the line x=3.
Now, `OP=sqrt(x^(2)+y^(2))`
Distance of P from x-3=0 is PM=3-x.
According to the question,
`OPltPM`
`rArr" "sqrt(x^(2)+y^(2))lt(3-x)`
`rArr" "x^(2)+y^(2)ltx^(2)-6x+9`
`rArr" "y^(2)lt-6x+9`
`rArr" "y^(2)lt-6(x-(3)/(2))`
Points satisfying above inequality lie inside parabola `y^(2)=9-6x` in first quadrant.

Parabola `y^(2)=-6(x-(3)/(2))` is concave to the left, ahving axis as x-axis, vertex at `((3)/(2),0)` and the directrix x=3.
The required region is shown in the figure.