The line 2x - 3y = 1 divides the circular reg | vedclass

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(B) First,we check which points lie inside the circle $x^2 + y^2 \leq 6$. For $(2, 3/4)$,$2^2 + (3/4)^2 = 4 + 9/16 = 4.5625 < 6$ (Inside). For $(5/2,

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$2$

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(B) First,we check which points lie inside the circle $x^2 + y^2 \leq 6$. For $(2, 3/4)$,$2^2 + (3/4)^2 = 4 + 9/16 = 4.5625 For $(5/2, 3/4)$,$(5/2)^2 + (3/4)^2 = 25/4 + 9/16 = 6.25 + 0.5625 = 6.8125 > 6$ (Outside). For $(1/4, -1/4)$,$(1/4)^2 + (-1/4)^2 = 1/16 + 1/16 = 0.125 For $(1/8, 1/4)$,$(1/8)^2 + (1/4)^2 = 1/64 + 1/64 = 0.03125 The line $2x - 3y - 1 = 0$ divides the circle. The origin $(0,0)$ satisfies $2(0) - 3(0) - 1 = -1 The smaller part is the region where $2x - 3y - 1 > 0$ (since the line does not pass through the origin). Checking the points inside the circle: $1$. $(2, 3/4): 2(2) - 3(3/4) - 1 = 4 - 2.25 - 1 = 0.75 > 0$ (Inside smaller part). $2$. $(1/4, -1/4): 2(1/4) - 3(-1/4) - 1 = 0.5 + 0.75 - 1 = 0.25 > 0$ (Inside smaller part). $3$. $(1/8, 1/4): 2(1/8) - 3(1/4) - 1 = 0.25 - 0.75 - 1 = -1.5 Thus,there are $2$ points in the smaller part.

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