If the equation of a line parallel to 3x - 2y | vedclass

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(A) The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Giv

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$5(1 \pm \sqrt{13})$

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(A) The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Given lines are $3x - 2y + 5 = 0$ and $3x - 2y + C = 0$.Here,$a = 3$,$b = -2$,$c_1 = 5$,$c_2 = C$,and $d = 5$.Substituting these values into the formula:$5 = \frac{|C - 5|}{\sqrt{3^2 + (-2)^2}}$$5 = \frac{|C - 5|}{\sqrt{9 + 4}}$$5 = \frac{|C - 5|}{\sqrt{13}}$$|C - 5| = 5\sqrt{13}$$C - 5 = \pm 5\sqrt{13}$$C = 5 \pm 5\sqrt{13} = 5(1 \pm \sqrt{13})$.

If the equation of a line parallel to $3x - 2y + 5 = 0$ and at a distance of $5$ units from it is $3x - 2y + C = 0$,then $C$ is equal to

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