Q1 / 109 · NTA Abhyas (JEE Main) 2020 · Mathematics · Straight Lines

Let the lengths of the altitudes from the vertices $A (- 1,1), B (5,2), C (3, - 1)$ of $Δ A B C$ are $p_{1}, p_{2}, p_{3}$ units respectively, then the value of $\frac{{(\frac{1}{p_{1}})}^{2} + {(\frac{1}{p_{3}})}^{2}}{{(\frac{1}{p_{2}})}^{2}}$ is equal to

Correct answer: 2.5

Solution

\frac{{(\frac{1}{p_{1}})}^{2} + {(\frac{1}{p_{3}})}^{2}}{{(\frac{1}{p_{2}})}^{2}} = \frac{{(\frac{2 Δ}{p_{1}})}^{2} + {(\frac{2 Δ}{p_{3}})}^{2}}{{(\frac{2 Δ}{p_{2}})}^{2}} = \frac{{(B C)}^{2} + {(A B)}^{2}}{{(A C)}^{2}}

{where

Δ

represents the area of triangle

A B C

}

= \frac{4 + 9 + 36 + 1}{16 + 4} = \frac{50}{20} = 2.5

Hence, the correct answer is 2.5.