Q64 / 150 · MHT CET 2023 (12 May Shift 1) · Mathematics
The co-ordinates of the points on the line $2 x-y=5$ which are the distance of 1 unit from the line $3 x+4 y=5$ are
- A$\left(\frac{30}{11}, \frac{-5}{11}\right),\left(\frac{20}{11}, \frac{15}{11}\right)$
- B$\left(\frac{-30}{11}, \frac{5}{11}\right),\left(\frac{-20}{11}, \frac{15}{11}\right)$
- Correct answerC$\left(\frac{30}{11}, \frac{5}{11}\right),\left(\frac{20}{11}, \frac{-15}{11}\right)$
- D$\left(\frac{-30}{11}, \frac{5}{11}\right),\left(\frac{-20}{11}, \frac{-15}{11}\right)$
Solution
Let $\left(x_1, y_1\right)$ be the required point
$\therefore \quad 2 x_1-y_1=5$
Also, $\left(x_1, y_1\right)$ is at the distance of 1 unit from line $3 x+4 y=5$
$$
\begin{array}{ll}
\therefore & 1=\left|\frac{3 x_1+4 y_1-5}{\sqrt{9+16}}\right| \\
\therefore & \pm 5=3 x_1+4 y_1-5 \\
\therefore & 3 x_1+4 y_1-5=5 \quad \text { or } 3 x_1+4 y_1-5=-5 \\
\therefore & 3 x_1+4 y_1=10 \\
& \text { or } \\
& 3 x_1+4 y_1=0
\end{array}
$$
Solving equations (i) and (ii), we get $x_1=\frac{30}{11}$ and $y_1=\frac{5}{11}$
Solving equation (i) and (iii), we get $x_1=\frac{20}{11}$ and $y_1=\frac{-15}{11}$
$\therefore \quad\left(\frac{30}{11}, \frac{5}{11}\right)$ and $\left(\frac{20}{11}, \frac{-15}{11}\right)$ are the required points.
Hence, the correct answer is option: 'C'.