The lines 2x + 3y = 6 and 2x + 3y = 8 cut the | vedclass

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(A) The given lines are $2x + 3y = 6$ and $2x + 3y = 8$.To find the $X$-intercepts,set $y = 0$:For $2x + 3y = 6$,$2x = 6 \Rightarrow x = 3$. So,$A = (

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$2x + 3y = 10$

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(A) The given lines are $2x + 3y = 6$ and $2x + 3y = 8$.To find the $X$-intercepts,set $y = 0$:For $2x + 3y = 6$,$2x = 6 \Rightarrow x = 3$. So,$A = (3, 0)$.For $2x + 3y = 8$,$2x = 8 \Rightarrow x = 4$. So,$B = (4, 0)$.The line $L$ passes through $(2, 2)$ and cuts the $X$-axis at $C(x_1, 0)$.Given that the abscissae of $A, B,$ and $C$ are in arithmetic progression,we have $3, 4, x_1$ in $AP$.Thus,$2(4) = 3 + x_1$ $\Rightarrow 8 = 3 + x_1$ $\Rightarrow x_1 = 5$.So,the point $C$ is $(5, 0)$.The equation of the line $L$ passing through $(2, 2)$ and $(5, 0)$ is given by:$y - 0 = \frac{2 - 0}{2 - 5}(x - 5)$$y = \frac{2}{-3}(x - 5)$$-3y = 2x - 10$$2x + 3y = 10$

The lines $2x + 3y = 6$ and $2x + 3y = 8$ cut the $X$-axis at $A$ and $B$,respectively. $A$ line $L$ drawn through the point $(2, 2)$ meets the $X$-axis at $C$ in such a way that the abscissae of $A, B,$ and $C$ are in arithmetic progression. Then,the equation of the line $L$ is

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