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

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(A) Given lines are $2x + 3y = 6$ and $2x + 3y = 8$. These lines meet the $X$-axis at $A(3, 0)$ and $B(4, 0)$ respectively. The line $L$ passes throug

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$2x + 3y = 10$

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(A) Given lines are $2x + 3y = 6$ and $2x + 3y = 8$. These lines meet the $X$-axis at $A(3, 0)$ and $B(4, 0)$ respectively. The line $L$ passes through $(2, 2)$ and cuts the $X$-axis at $C(x_1, 0)$,such that the $x$-coordinates of $A, B,$ and $C$ are in arithmetic progression,i.e.,$3, 4, x_1$. Since they are in arithmetic progression,$2 	imes 4 = 3 + x_1$,which gives $x_1 = 8 - 3 = 5$. So,the coordinates of $C$ are $(5, 0)$. $A$ line $L$ passing through the points $(2, 2)$ and $(5, 0)$ is given by: $y - 0 = \frac{0 - 2}{5 - 2}(x - 5)$ $y = \frac{-2}{3}(x - 5)$ $3y = -2x + 10$ $2x + 3y = 10$

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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