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

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

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$\left( 0, \frac{1}{2} \right)$

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(D) The given lines are $L_1: x + y - 1 = 0$ and $L_2: 2x + 2y - 3 = 0$.For a point $(1, a)$ to lie between these two parallel lines,the expressions $L_1(1, a)$ and $L_2(1, a)$ must have opposite signs.Let $f(x, y) = x + y - 1$ and $g(x, y) = 2x + 2y - 3$.Substituting $(1, a)$ into the equations:$f(1, a) = 1 + a - 1 = a$$g(1, a) = 2(1) + 2a - 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,$a \in \left( 0, \frac{1}{2} \right)$.

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

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