The distance of the point (2, 5) from the lin | vedclass

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(C) Let the required distance be $r$.The line along which the distance is measured is $3x - 4y + 8 = 0$. Its slope is $m = \frac{3}{4}$.Thus,$	an 	h

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$5$

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(C) Let the required distance be $r$.The line along which the distance is measured is $3x - 4y + 8 = 0$. Its slope is $m = \frac{3}{4}$.Thus,$	an 	heta = \frac{3}{4}$,which implies $\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$.The coordinates of any point $P$ on the line passing through $(2, 5)$ with slope $m = \frac{3}{4}$ are given by $(2 + r \cos 	heta, 5 + r \sin 	heta) = (2 + r \cdot \frac{4}{5}, 5 + r \cdot \frac{3}{5})$.Since $P$ lies on the line $3x + y + 4 = 0$,we substitute these coordinates into the equation:$3(2 + \frac{4r}{5}) + (5 + \frac{3r}{5}) + 4 = 0$$6 + \frac{12r}{5} + 5 + \frac{3r}{5} + 4 = 0$$15 + \frac{15r}{5} = 0$$15 + 3r = 0$$3r = -15$$r = -5$Since the distance $r$ must be positive,we take the magnitude: $|r| = |-5| = 5$.

The distance of the point $(2, 5)$ from the line $3x + y + 4 = 0$,measured parallel to the line $3x - 4y + 8 = 0$,is

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