Question

The slope of a straight line through A (3, 2) is $\frac{3}{4}$. Find the coordinates of the points on the line that are 5 units away from A.

Answer 1

$\mathrm{Here},\left(x_1,y_1\right)=\left(3,2\right),\mathrm{tan\theta }=\frac{3}{4}$ and r = 5 units

$$\begin{gathered} \mathrm{Now},\mathrm{sin\theta }=\frac{3}{\sqrt{3^2+4^2}}\mathrm{and}\mathrm{cos\theta }=\frac{4}{\sqrt{3^2+4^2}} \\ \Rightarrow \mathrm{sin\theta }=\frac{3}{5}\mathrm{and}\mathrm{cos\theta }=\frac{4}{5} \end{gathered}$$

So, the equation of the line is $\frac{x-3}{\mathrm{cos\theta }}=\frac{y-2}{\mathrm{sin\theta }}$
$$\begin{gathered} \Rightarrow \frac{x-3}{{\displaystyle \frac{4}{5}}}=\frac{y-2}{{\displaystyle \frac{3}{5}}} \\ \Rightarrow \frac{x-3}{4{\displaystyle }}=\frac{y-2}{3{\displaystyle }} \\ \Rightarrow 3x-9=4y-8 \\ \Rightarrow 3x-4y=1 \end{gathered}$$


Hence, the points on the line at a distance of 5 units from A (3, 2) are
$\left(x_1\pm r\mathrm{cos\theta },y_1\pm r\mathrm{sin\theta }\right)$

$$\begin{gathered} \left(3\pm 5\times \frac{4}{5},2\pm 5\times \frac{3}{5}\right)\mathrm{or}\left(3\pm 4,2\pm 3\right) \\ \mathrm{or}\left(7,5\right)\mathrm{and}\left(-1,-1\right) \end{gathered}$$