Let a neq 0, b neq 0, c be three real numbers | vedclass

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(B) Given $L(p, q) = \frac{ap + bq + c}{\sqrt{a^2 + b^2}}$.Substituting the given values:$L\left(\frac{2}{3}, \frac{1}{3}\right) = \frac{a(\frac{2}{3}

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$(1, 1)$

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(B) Given $L(p, q) = \frac{ap + bq + c}{\sqrt{a^2 + b^2}}$.Substituting the given values:$L\left(\frac{2}{3}, \frac{1}{3}\right) = \frac{a(\frac{2}{3}) + b(\frac{1}{3}) + c}{\sqrt{a^2 + b^2}} = \frac{2a + b + 3c}{3\sqrt{a^2 + b^2}}$$L\left(\frac{1}{3}, \frac{2}{3}\right) = \frac{a(\frac{1}{3}) + b(\frac{2}{3}) + c}{\sqrt{a^2 + b^2}} = \frac{a + 2b + 3c}{3\sqrt{a^2 + b^2}}$$L(2, 2) = \frac{a(2) + b(2) + c}{\sqrt{a^2 + b^2}} = \frac{2a + 2b + c}{\sqrt{a^2 + b^2}} = \frac{6a + 6b + 3c}{3\sqrt{a^2 + b^2}}$Summing these values:$\frac{2a + b + 3c + a + 2b + 3c + 6a + 6b + 3c}{3\sqrt{a^2 + b^2}} = 0$$\frac{9a + 9b + 9c}{3\sqrt{a^2 + b^2}} = 0$$9(a + b + c) = 0 \implies a + b + c = 0$This implies that the line $ax + by + c = 0$ satisfies the condition $a(1) + b(1) + c = 0$ for the point $(1, 1)$.Thus,the line always passes through $(1, 1)$.

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