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

Explanation

Point `P(x, y)` lies in the first quadrant.
So `x, y gt 0`

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 the line `x-3=0` is PM =`3-x`.
According to the questions, OP `lt` PM
`rArr" "sqrt(x^(2)+y^(2)) lt (3-x)`
`rArr" "x^(2) + y^(2) lt x^(2) - 6x +9`
`rArr" "y^(2) lt -6x+9`
`rArr" " y^(2) lt -6 (x- (3)/(2))`
Points satisfying the above inequality lie inside the parabola `y^(2) =9-6x` in the first quadrant.
Parabola `y^(2) = -6(x-(3)/(2))` is concave to the left, having axis as the x-axis, vertex at `((3)/(2), 0)` and directrix x=3.
The required region as shown in the following figure.