UPSC NDA 2 2017 Mathematics Solved Paper

Concept:The distance from a point to a line, measured parallel to another line, is found by intersecting the given line with a line through the point that has the same slope as the parallel line.Then compute the distance between the given point and the intersection point.Explanation:The given line is $2x + 3y = 6$.The direction of measurement is parallel to $4x + y = 4$.The slope of $4x + y = 4$ is $m = -4$.A line through the point $(1, 3)$ with slope $-4$ is $(y - 3) = -4(x - 1)$.This simplifies to $4x + y = 7$.Now find the intersection of this line with $2x + 3y = 6$.From $4x + y = 7$: $y = 7 - 4x$.Substitute into $2x + 3y = 6$: $2x + 3(7 - 4x) = 6$.This gives $2x + 21 - 12x = 6$.Simplify: $-10x = -15$, so $x = \frac{3}{2}$.Then $y = 7 - 4 \cdot \frac{3}{2} = 7 - 6 = 1$.The intersection point is $\left(\frac{3}{2}, 1\right)$.Now compute the distance between $(1, 3)$ and $\left(\frac{3}{2}, 1\right)$:$\sqrt{\left(\frac{3}{2} - 1\right)^2 + (1 - 3)^2} = \sqrt{\left(\frac{1}{2}\right)^2 + (-2)^2}$.This equals $\sqrt{\frac{1}{4} + 4} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2}$.Answer:$\frac{\sqrt{17}}{2}$ units, which corresponds to option D.