If the point (1, a) lies between the lines x | vedclass

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(A) The given lines are $L_1: x + y - 1 = 0$ and $L_2: 2x + 2y - 3 = 0$. For the point $(1, a)$ to lie between these two lines,the expressions $L_1(1,

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$(0, 1/2)$

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(A) The given lines are $L_1: x + y - 1 = 0$ and $L_2: 2x + 2y - 3 = 0$. For the point $(1, a)$ to lie between these two lines,the expressions $L_1(1, a)$ and $L_2(1, a)$ must have opposite signs. Let $f(x, y) = x + y - 1$ and $g(x, y) = 2x + 2y - 3$. At $(1, a)$,$f(1, a) = 1 + a - 1 = a$. At $(1, a)$,$g(1, a) = 2(1) + 2(a) - 3 = 2a - 1$. Since the point lies between the lines,$f(1, a) \cdot g(1, a) Therefore,$a(2a - 1) This inequality holds when $a$ is between the roots of $a(2a - 1) = 0$,which are $a = 0$ and $a = 1/2$. Thus,$a \in (0, 1/2)$.

If the point $(1, a)$ lies between the lines $x + y = 1$ and $2(x + y) = 3$,then in which interval does $a$ lie?

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