The straight line 2x – 3y = 1 divides the circular region x2 + y2 ≤ 6 into two parts. If S = \left{\right. \left(\right. 2 , \textrm{ }\textrm{ } \frac{3}{4} \left.\right) , \textrm{ }\textrm{ } \left(\right. \frac{5}{2} , \textrm{ }\textrm{ } \frac{3}{4} \left.\right) , \textrm{ }\textrm{ } \left(\right. \frac{1}{4} , \textrm{ }\textrm{ } \frac{- 1}{4} \left.\right) , \textrm{ }\textrm{ } \left(\right. \frac{1}{8} , \textrm{ }\textrm{ } \frac{1}{4} \left.\right) \left.\right}, then the number of point(s) in S lying inside the smaller part is

Engineering

Mathematics

Point and Circle

Question

The straight line 2x – 3y = 1 divides the circular region x² + y² ≤ 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

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Pont (x₁, y₁) lies inside the region if x²₁+ y²₁ – 6 $\leq$ 0 & 2x₁ – 3y₁ – 1 $\leq$ 0.

$P_{1} \equiv (2, \frac{3}{4}) 4 + \frac{9}{16} - 6 \geq 0$ Ture

$4 - \frac{9}{4} - 1 > 0$ True

$P_{2} \equiv (\frac{5}{2}, \frac{3}{4}) \frac{25}{4} + \frac{9}{16} - 6 \leq 0$ False

$P_{3} \equiv (\frac{1}{4}, \frac{- 1}{4}) \frac{1}{16} + \frac{1}{16} - 6 \leq 0$ True

$\frac{2}{4} + \frac{3}{4} - 1 > 0$ True

$P_{4} \equiv (\frac{1}{8}, \frac{1}{4}) \frac{1}{64} + \frac{1}{16} - 6 \leq 0$ True

$\frac{2}{8} - \frac{3}{4} - 1 > 0$ False

So P₁ & P₃ lies in the interval