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(B) Given the slope $m = 	an 	heta = \frac{3}{4}$.Since $	an 	heta = \frac{3}{4}$,we have $\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{

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$(7, 5), (-1, -1)$

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(B) Given the slope $m = 	an 	heta = \frac{3}{4}$.Since $	an 	heta = \frac{3}{4}$,we have $\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$.The coordinates of points at a distance $r = 5$ from $A(x_1, y_1) = (3, 2)$ are given by $(x_1 \pm r \cos 	heta, y_1 \pm r \sin 	heta)$.For the first point $P$:$x = 3 + 5 	imes \frac{4}{5} = 3 + 4 = 7$$y = 2 + 5 	imes \frac{3}{5} = 2 + 3 = 5$So,$P = (7, 5)$.For the second point $Q$:$x = 3 - 5 	imes \frac{4}{5} = 3 - 4 = -1$$y = 2 - 5 	imes \frac{3}{5} = 2 - 3 = -1$So,$Q = (-1, -1)$.Thus,the required points are $(7, 5)$ and $(-1, -1)$.

If the slope of a straight line passing through $A(3, 2)$ is $3/4$,then the coordinates of the two points on the same line that are $5$ units away from $A$ are

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