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If the slope of a straight line passing through $\mathrm{A}(3,2)$ 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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The correct answer is:
$(7,5),(-1,-1)$
$\tan \theta=\frac{3}{4}$
$\therefore \quad \sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$
$\begin{aligned} P(x, y)= & (3+5 \cos \theta, 2+5 \sin \theta) \\ & =\left(3+5 \cdot \frac{4}{5}, 2+5 \cdot \frac{3}{5}\right)=(7,5) \\ Q(x, y)= & (3-5 \cos \theta, 2-5 \sin \theta) \\ & =\left(3-5 \cdot \frac{4}{5}, 2-5 \cdot \frac{3}{5}\right)=(-1,-1) .\end{aligned}$
$\therefore \quad \sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$
$\begin{aligned} P(x, y)= & (3+5 \cos \theta, 2+5 \sin \theta) \\ & =\left(3+5 \cdot \frac{4}{5}, 2+5 \cdot \frac{3}{5}\right)=(7,5) \\ Q(x, y)= & (3-5 \cos \theta, 2-5 \sin \theta) \\ & =\left(3-5 \cdot \frac{4}{5}, 2-5 \cdot \frac{3}{5}\right)=(-1,-1) .\end{aligned}$
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