If the straight line L equiv 3x+4y-k=0 cuts t | vedclass

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(C) The point of intersection $R$ of the line segment $PQ$ dividing it in the ratio $4:1$ is given by the section formula:$R = \left(\frac{4(1)+1(2)}{

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$5x-5y-3=0$

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(C) The point of intersection $R$ of the line segment $PQ$ dividing it in the ratio $4:1$ is given by the section formula:$R = \left(\frac{4(1)+1(2)}{4+1}, \frac{4(1)+1(-1)}{4+1}\right) = \left(\frac{6}{5}, \frac{3}{5}\right)$Since $R$ lies on the line $L \equiv 3x+4y-k=0$,we have:$3\left(\frac{6}{5}\right) + 4\left(\frac{3}{5}\right) - k = 0$ $\Rightarrow \frac{18+12}{5} = k$ $\Rightarrow k=6$Thus,$L \equiv 3x+4y-6=0$.The equation of the line $PQ$ passing through $(2,-1)$ and $(1,1)$ is:$y - 1 = \frac{1-(-1)}{1-2}(x-1)$ $\Rightarrow y-1 = -2(x-1)$ $\Rightarrow 2x+y-3=0$The family of lines passing through the intersection of $L=0$ and $PQ=0$ is given by:$(3x+4y-6) + \lambda(2x+y-3) = 0$$(3+2\lambda)x + (4+\lambda)y - (6+3\lambda) = 0$Since this line is parallel to $y=x$ (slope $m=1$),its slope must be $1$:$-\frac{3+2\lambda}{4+\lambda} = 1$ $\Rightarrow -3-2\lambda = 4+\lambda$ $\Rightarrow 3\lambda = -7$ $\Rightarrow \lambda = -\frac{7}{3}$Substituting $\lambda = -\frac{7}{3}$ into the equation:$(3+2(-\frac{7}{3}))x + (4-\frac{7}{3})y - (6+3(-\frac{7}{3})) = 0$$(3-\frac{14}{3})x + (\frac{12-7}{3})y - (6-7) = 0$$-\frac{5}{3}x + \frac{5}{3}y + 1 = 0$ $\Rightarrow -5x+5y+3=0$ $\Rightarrow 5x-5y-3=0$

If the straight line $L \equiv 3x+4y-k=0$ cuts the line segment joining the points $P(2,-1)$ and $Q(1,1)$ in the ratio $4:1$,then the equation of the line parallel to the line $y=x$ and concurrent with the lines $PQ$ and $L=0$ is

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