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 points in $S$ lying inside the smaller part is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

The point $(\frac{5}{2},\frac{3}{4})$ lies outside the circle as ${\left(\frac{5}{2}\right)}^2+{\left(\frac{3}{4}\right)}^2\ge 6$.
So, this point is cancelled.

To check, we will look for the points that are on the non-origin side of $2x_1-3y_1-1=0$, that is $(x_1,y_1)$ will lie inside the smaller part if $2x_1-3y_1-1>0$

Now, for the point $(2,\frac{3}{4})$, $2x-3y-1>0$
So, $(2,\frac{3}{4})$ lies inside the smaller part.

Now, for the point $(\frac{1}{4},\frac{-3}{4})$, $2x-3y-1>0$
So, $(2,\frac{3}{4})$ lies inside the smaller part.

Next , for the point $(\frac{1}{8},\frac{1}{4})$, $2x-3y-1<0$
So, $(\frac{1}{8},\frac{1}{4})$ will not lie inside the smaller part.

Hence , 2 points of S will lie inside the smaller part.