Question

The straight line $2x-3y=1$ divides the circular region $x^2+y^2\le 6$ into two parts. If $S=\{(2,\frac{3}{4}),(\frac{5}{2},\frac{3}{4}),(\frac{1}{4},\frac{-1}{4}),(\frac{1}{8},\frac{1}{4})\},$ then the number of point(s) in $S,$ lying inside the smaller part is

Answer 1

$L:2x-3y-1,S:x^2+y^2-6$

The shorter region lies in the non-origin side of the line hence, $L_1>0$

If $L_1>0,S_1<0,$ then the point lies in smaller part.

$(2,\frac{3}{4}),(\frac{1}{4}.\frac{-1}{4})$ satisfy both the conditions.
$\therefore$ there are two points in the smaller region.