UPSC NDA I 2018 Mathematics Solved Paper

Concept:The perpendicular distance from the origin $(0,0)$ to a line $Ax+By+C=0$ is given by $p=\frac{|C|}{\sqrt{A^2+B^2}}$.Explanation:Convert each line to standard form, apply the formula, and compare with the given relation.Statement (1):Line: $ax+by=c$ can be written as $ax+by-c=0$.Distance: $p=\frac{|-c|}{\sqrt{a^2+b^2}}=\frac{c}{\sqrt{a^2+b^2}}$.Square: $p^2=\frac{c^2}{a^2+b^2}$. Matches the given relation. So, correct.Statement (2):Line: $\frac{x}{a}+\frac{y}{b}=1$ becomes $bx+ay=ab$ or $bx+ay-ab=0$.Distance: $p=\frac{|-ab|}{\sqrt{b^2+a^2}}=\frac{ab}{\sqrt{a^2+b^2}}$.Square: $p^2=\frac{a^2b^2}{a^2+b^2}$. Reciprocal: $\frac{1}{p^2}=\frac{a^2+b^2}{a^2b^2}=\frac{1}{a^2}+\frac{1}{b^2}$. Matches the given relation. So, correct.Statement (3):Line: $y=mx+c$ can be written as $mx-y+c=0$.Distance: $p=\frac{|c|}{\sqrt{m^2+(-1)^2}}=\frac{c}{\sqrt{m^2+1}}$.Square: $p^2=\frac{c^2}{m^2+1}$. Reciprocal: $\frac{1}{p^2}=\frac{m^2+1}{c^2}$. The given relation is $\frac{1}{p^2}=\frac{1+m^2+c^2}{c^2}$, which is different. So, incorrect.Thus, only statements 1 and 2 are correct.Answer:Option C (1 and 2 only).