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(C) First,find the point of intersection of the lines $x - y + 1 = 0$ and $2x - 3y + 5 = 0$. Solving these,we get $x = 2$ and $y = 3$. The point of in

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$3x - 4y + 6 = 0$ and $4x - 3y + 1 = 0$

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(C) First,find the point of intersection of the lines $x - y + 1 = 0$ and $2x - 3y + 5 = 0$. Solving these,we get $x = 2$ and $y = 3$. The point of intersection is $(2, 3)$.The equation of any line passing through $(2, 3)$ is given by $y - 3 = m(x - 2)$,which simplifies to $mx - y + (3 - 2m) = 0$.The distance of this line from the point $(3, 2)$ is given as $\frac{7}{5}$. Using the perpendicular distance formula:$\frac{|m(3) - 2 + 3 - 2m|}{\sqrt{m^2 + (-1)^2}} = \frac{7}{5}$$\frac{|m + 1|}{\sqrt{m^2 + 1}} = \frac{7}{5}$Squaring both sides: $25(m^2 + 2m + 1) = 49(m^2 + 1)$$25m^2 + 50m + 25 = 49m^2 + 49$$24m^2 - 50m + 24 = 0$$12m^2 - 25m + 12 = 0$$(3m - 4)(4m - 3) = 0$Thus,$m = \frac{4}{3}$ or $m = \frac{3}{4}$.For $m = \frac{4}{3}$,the line is $y - 3 = \frac{4}{3}(x - 2)$ $\Rightarrow 3y - 9 = 4x - 8$ $\Rightarrow 4x - 3y + 1 = 0$.For $m = \frac{3}{4}$,the line is $y - 3 = \frac{3}{4}(x - 2)$ $\Rightarrow 4y - 12 = 3x - 6$ $\Rightarrow 3x - 4y + 6 = 0$.Therefore,the equations are $3x - 4y + 6 = 0$ and $4x - 3y + 1 = 0$.

The equations of the lines passing through the point of intersection of the lines $x - y + 1 = 0$ and $2x - 3y + 5 = 0$ and whose distance from the point $(3, 2)$ is $\frac{7}{5}$ are:

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