NDA Math Mock Test 03 – Question 104

Consider the following statements:
1. The length \(p\) of the perpendicular from the origin to the line \(ax + by = c\) satisfies \(p^2 = \frac{c^2}{a^2 + b^2}\).
2. The length \(p\) of the perpendicular from the origin to the line \(\frac{x}{a} + \frac{y}{b} = 1\) satisfies \(\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}\).
3. The length \(p\) of the perpendicular from the origin to the line \(y = mx + c\) satisfies \(\frac{1}{p^2} = \frac{1 + m^2 + c^2}{c^2}\).
Which of the above is/are correct?

• 1, 2 and 3
• 1 only
• 1 and 2 only
• 2 only

Checking each statement:
1. \(p = \frac{|-c|}{\sqrt{a^2 + b^2}} \implies p^2 = \frac{c^2}{a^2 + b^2}\) (Correct).
2. For \(\frac{x}{a} + \frac{y}{b} = 1\), \(p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \implies \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}\) (Correct).
3. For \(mx – y + c = 0\), \(p = \frac{|c|}{\sqrt{1 + m^2}} \implies \frac{1}{p^2} = \frac{1 + m^2}{c^2} \ne \frac{1 + m^2 + c^2}{c^2}\) (Incorrect).
Thus, statements 1 and 2 are correct.
Hence, the correct option is (c).