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(A) The equation of the line passing through $(5, -7)$ and perpendicular to $3x - 5y + 1 = 0$ is $5x + 3y + k = 0$.Since it passes through $(5, -7)$,w

Vedclass · Accepted Answer

$\frac{1}{2\sqrt{29}}$

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(A) The equation of the line passing through $(5, -7)$ and perpendicular to $3x - 5y + 1 = 0$ is $5x + 3y + k = 0$.Since it passes through $(5, -7)$,we have $5(5) + 3(-7) + k = 0$,which gives $25 - 21 + k = 0$,so $k = -4$.The line is $5x + 3y - 4 = 0$.To find $M$,we solve the system:$3x - 5y = -1$ (multiplied by $3 \implies 9x - 15y = -3$)$5x + 3y = 4$ (multiplied by $5 \implies 25x + 15y = 20$)Adding these,$34x = 17$,so $x = \frac{1}{2}$.Substituting $x = \frac{1}{2}$ into $5x + 3y = 4$,we get $\frac{5}{2} + 3y = 4$,so $3y = \frac{3}{2}$,which gives $y = \frac{1}{2}$.Thus,$M = (\frac{1}{2}, \frac{1}{2})$.The perpendicular distance from $M(\frac{1}{2}, \frac{1}{2})$ to $2x + 5y - 3 = 0$ is given by $d = \frac{|2(\frac{1}{2}) + 5(\frac{1}{2}) - 3|}{\sqrt{2^2 + 5^2}} = \frac{|1 + 2.5 - 3|}{\sqrt{4 + 25}} = \frac{|0.5|}{\sqrt{29}} = \frac{1}{2\sqrt{29}}$.

Let $M$ be the foot of the perpendicular drawn from the point $(5, -7)$ to the line $3x - 5y + 1 = 0$. Then the perpendicular distance from $M$ to the line $2x + 5y - 3 = 0$ is

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