The equation of lines $2 x^{2} + 3 x y - 2 y^{2} + 3 x - 9 y - 9 = 0$ represents a pair of straight lines, the distance of the point of intersection from the origin, is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{3}{5}$

$\frac{9}{5}$

$\frac{3 \sqrt{10}}{5}$

$\frac{18}{5}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Given Equation $2 x^{2} + 3 x y - 2 y^{2} + 3 x - 9 y - 9 = 0$

Comparing with $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$

$\begin{array}{l} a = 2, h = \frac{3}{2}, b = - 2, g = \frac{3}{2}, f = \frac{- 9}{2}, c = - 9 \\ P = p o int o f int e r \sec t i o n = (\frac{h f - b g}{a b - h^{2}}, \frac{g h - a f}{a b - h^{2}}) \\ = (\frac{(\frac{3}{2}) (\frac{- 9}{2}) - (- 2) (\frac{3}{2})}{2 (- 2) - \frac{9}{4}}, \frac{(\frac{3}{2}) (\frac{9}{2}) - 2 (\frac{- 9}{2})}{2 (- 2) - \frac{9}{4}}) \\ = (\frac{- 27 + 12}{- 16 + 9}, \frac{9 + 36}{- 16 - 9}) \\ = (\frac{- 15}{- 25}, \frac{45}{- 25}) \\ = (\frac{3}{5}, \frac{- 9}{5}) \\ ∴ O P = \sqrt{\frac{9}{25} + \frac{81}{25}} = \sqrt{\frac{90}{25}} = \frac{3 \sqrt{10}}{5} \end{array}$