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Question

# A line a drawn through A (4, −1) parallel to the line 3x − 4y + 1 = 0. Find the coordinates of the two points on this line which are at a distance of 5 units from A.

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Solution

## The slope of the line 3x − 4y + 1 = 0 or $y=\frac{3}{4}x-\frac{1}{4}$ is $\frac{3}{4}$ So, the slope of the required line is also $\frac{3}{4}$ as it is parallel to the given line. $\therefore \mathrm{tan}\mathrm{\theta }=\frac{3}{4}⇒\mathrm{sin\theta }=\frac{3}{5}\mathrm{and}\mathrm{cos\theta }=\frac{4}{5}$ Here, $\left({x}_{1},{y}_{1}\right)=A\left(4,-1\right)$ So, the equation of the line passing through A (4, −1) and having slope $\frac{3}{4}$ is $\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-4}{\frac{4}{5}}=\frac{y+1}{\frac{3}{5}}\phantom{\rule{0ex}{0ex}}⇒3x-12=4y+4\phantom{\rule{0ex}{0ex}}⇒3x-4y-16=0$ Here, AP = r = 5 Thus, the coordinates of P are given by $x={x}_{1}±r\mathrm{cos\theta },y={y}_{1}±r\mathrm{sin\theta }\phantom{\rule{0ex}{0ex}}⇒x=4±5\left(\frac{4}{5}\right),y=-1±5\left(\frac{3}{5}\right)$ $⇒x=4±4,y=-1±3\phantom{\rule{0ex}{0ex}}⇒x=8,y=2\mathrm{and}x=0,y=-4$ Hence, the coordinates of the two points at a distance of 5 units from A are (8, 2) and (0, −4).

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