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

We observe that the points $\{(2,\frac{3}{4}),(\frac{5}{2},\frac{3}{4}),(\frac{1}{4},-\frac{1}{4}),(\frac{1}{8},\frac{1}{4})\}$


lie on the opposite side of origin with respect to the given line. But 5/2, 3/4 lied outside the circle. Hence there are exactly two points of the set lie in the smaller part.