The line L equiv 6x + 3y + k = 0 divides the | vedclass

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(D) Let the points be $A(3, 5)$ and $B(4, 6)$. The line $L \equiv 6x + 3y + k = 0$ divides $AB$ in the ratio $m:n = -5: 4$.Using the section formula,t

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$g + 1$

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(D) Let the points be $A(3, 5)$ and $B(4, 6)$. The line $L \equiv 6x + 3y + k = 0$ divides $AB$ in the ratio $m:n = -5: 4$.Using the section formula,the point of intersection $Q$ is given by $\left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) = \left( \frac{-5(4) + 4(3)}{-5 + 4}, \frac{-5(6) + 4(5)}{-5 + 4} \right) = \left( \frac{-20 + 12}{-1}, \frac{-30 + 20}{-1} \right) = (8, 10)$.Since $Q(8, 10)$ lies on $L = 0$,we have $6(8) + 3(10) + k = 0 \implies 48 + 30 + k = 0 \implies k = -78$.So,the line $L$ is $6x + 3y - 78 = 0$,which simplifies to $2x + y - 26 = 0$.The point $P(g, h)$ is the intersection of $2x + y = 26$ and $x - y = -1$.Adding the two equations: $(2x + y) + (x - y) = 26 - 1 \implies 3x = 25 \implies g = \frac{25}{3}$.Substituting $g$ into $x - y = -1$: $h = g + 1 = \frac{25}{3} + 1 = \frac{28}{3}$.Checking the options: $g + 1 = \frac{25}{3} + 1 = \frac{28}{3} = h$. Thus,$h = g + 1$.

The line $L \equiv 6x + 3y + k = 0$ divides the line segment joining the points $(3, 5)$ and $(4, 6)$ in the ratio $-5: 4$. If the point of intersection of the lines $L = 0$ and $x - y + 1 = 0$ is $P(g, h)$,then $h =$

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