UPSC NDA Math Model Paper 5

Here, we have to find the points which are on the y-axis such that the perpendicular distance between the points and the line $\frac{x}{3}-\frac{y}{4}=1$ is 3 units. The given equation of line can be re-written as: $4x - 3y - 12 = 0$ Let P = (0, y) Here a = 4, b = - 3 and d = 3 Now substitute $x_1 = 0$ and $y_1 = y$ in the equation $4x - 3y - 12 = 0$ $\Rightarrow \left|4\dot{x}_1 - 3\dot{y}_1 - 12\right| = \left|0 - 3y - 12\right|$ $\Rightarrow \sqrt{a^2+b^2} = \sqrt{4^2+(-3)^2} = 5$ As we know that, the perpendicular distance d from $P(x_1, y_1)$ to the line $ax+by+c=0$ is given by $d = \left|\frac{a x_1 + b y_1 + c}{\sqrt{a^2+b^2}}\right|$ $\Rightarrow d = \left|\frac{4\dot{x}_1 - 3\dot{y}_1 - 12}{\sqrt{4^2+(-3)^2}}\right| = 3$ $\Rightarrow \left| -3y - 12 \right| = 15$ ⇒ y = 1 or - 9 So, the points are: (0, 1) and (0, - 9) Hence, option D is the correct answer.