If P(alpha, beta) is a point on the line 3x + | vedclass

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(A) Let the line be $L(x, y) = 3x - 4y - 8 = 0$. For point $Q(1, 1)$,$L(1, 1) = 3(1) - 4(1) - 8 = 3 - 4 - 8 = -9$. Since $L(1, 1) < 0$,for $P(\alpha,

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$\alpha > \frac{8}{15}, \beta < -\frac{8}{5}$

Vedclass · Answer

(A) Let the line be $L(x, y) = 3x - 4y - 8 = 0$. For point $Q(1, 1)$,$L(1, 1) = 3(1) - 4(1) - 8 = 3 - 4 - 8 = -9$. Since $L(1, 1)  0$. Substituting $P(\alpha, \beta)$ into the line equation: $3\alpha - 4\beta - 8 > 0$. Since $P$ lies on $3x + y = 0$,we have $\beta = -3\alpha$. Substituting $\beta = -3\alpha$ into the inequality: $3\alpha - 4(-3\alpha) - 8 > 0$. $3\alpha + 12\alpha - 8 > 0$ $\Rightarrow 15\alpha > 8$ $\Rightarrow \alpha > \frac{8}{15}$. Now,using $\alpha = -\frac{\beta}{3}$ in the inequality $3\alpha - 4\beta - 8 > 0$: $3(-\frac{\beta}{3}) - 4\beta - 8 > 0$ $\Rightarrow -\beta - 4\beta - 8 > 0$ $\Rightarrow -5\beta > 8$ $\Rightarrow \beta Thus,$\alpha > \frac{8}{15}$ and $\beta < -\frac{8}{5}$.

If $P(\alpha, \beta)$ is a point on the line $3x + y = 0$ such that the point $P$ and the point $Q(1, 1)$ lie on opposite sides of the line $3x - 4y - 8 = 0$,then:

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