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(B) The given lines are $L_1: 2x - 3y + 4 = 0$ and $L_2: 6x - 9y + 7 = 0$. Note that $L_2$ can be written as $3(2x - 3y) + 7 = 0$,which shows $L_1$ an

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$9:4$ externally

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(B) The given lines are $L_1: 2x - 3y + 4 = 0$ and $L_2: 6x - 9y + 7 = 0$. Note that $L_2$ can be written as $3(2x - 3y) + 7 = 0$,which shows $L_1$ and $L_2$ are parallel. Let the line $L$ be perpendicular to $L_1$ and $L_2$. The equation of any line perpendicular to $2x - 3y + k = 0$ is $3x + 2y + C = 0$. Since $P(1, 1)$ lies on $L$,we have $3(1) + 2(1) + C = 0$,which gives $C = -5$. So,$L: 3x + 2y - 5 = 0$. Point $A$ is the intersection of $L_1$ and $L$: $2x - 3y = -4$ and $3x + 2y = 5$. Solving this,we get $A = (7/13, 22/13)$. Point $B$ is the intersection of $L_2$ and $L$: $6x - 9y = -7$ and $3x + 2y = 5$. Solving this,we get $B = (31/39, 9/39) = (31/39, 3/13)$. Let $P$ divide $AB$ in ratio $k:1$. Using section formula for $x$-coordinate: $1 = \frac{k(31/39) + 1(7/13)}{k+1} \implies k+1 = \frac{31k + 21}{39} \implies 39k + 39 = 31k + 21 \implies 8k = -18 \implies k = -18/8 = -9/4$. The negative sign indicates external division in the ratio $9:4$.

Let the line $L$ drawn perpendicular to the lines $2x - 3y + 4 = 0$ and $6x - 9y + 7 = 0$ meet them at $A$ and $B$ respectively. If $P(1, 1)$ is a point on $L$,then the ratio in which $P$ divides $AB$ is

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