The points on the straight line $3 x-4 y+1=0$ which are at a distance of 5 units from the point $(3,2)$ are

1. $\left(-2,-\frac{7}{4}\right),\left(-3, \frac{-5}{2}\right)$
2. $\left(4, \frac{11}{4}\right),(-1,-1)$
3. $\left(1, \frac{1}{2}\right),\left(2, \frac{5}{4}\right)$
4. $(7,5),(-1,-1)$

Solution

Let the point $A(x, y)$ on the line $3 x-4 y+1$ is at distance 5 units from the point $(3,2)$.
$\frac{x_1-3}{\cos \theta}=\frac{y_1-2}{\sin \theta}= \pm 5$
Since, $\left(x_1, y_1\right)$ lies on the line $3 x-4 y-1=0$ $ \begin{array}{ll} \therefore & 3(3 \pm 5 \cos \theta)-4(2 \pm 5 \sin \theta)-1=0 \\ \Rightarrow & 9 \pm 15 \cos \theta-8+20 \sin \theta-1=0 \\ \Rightarrow & \pm 15 \cos \theta-20 \sin \theta=0 \\ \Rightarrow & 3 \cos \theta= \pm 4 \sin \theta \Rightarrow \tan \theta= \pm 3 / 4 \\ & \cos \theta= \pm 4 / 5 \Rightarrow \sin \theta= \pm 3 / 5 \end{array} $ From Eqs. (i) and (ii), we get $ \begin{aligned} & x_1=3 \pm\left(\frac{4}{5}\right) 5=7,-1 \\ & y_1=2 \pm 5\left(\frac{3}{5}\right)=5,-1 \end{aligned} $ $\therefore \quad$ Coordinates are $(7,5)$ and $(-1,-1)$

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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