A st. line through the point (2, 2) intersects the lines 3x+y=0 and 3x−y=0 at the points A and B. The equation to the line AB so that the triangle OAB is equilateral is:

Given equations 3x+y=0−−(1)⇒ slope =−3and 3x−y=0−−(2)⇒ slope =3Let equation of the line passing through (2, 2) isy−2=m(x−2)−−(3)∴ Δle OAB is Equilateral∴ Angle b/w (2)&(3)=600⇒ Tan 600=|m−31+3m|⇒ 3 = |m−31+3m|S.B.S⇒ 3(1+3m)2=(m−3)2⇒ 3(1+3m2+23m)=m2+3−25m⇒ 3+9m2+63=m2+3−23m⇒ 8m2+83=0⇒ 8m(m+3)=0m=0, m=−3If m=0 then from (3)y−2=0(x−2)⇒ y−2=0

A st. line through the point (2, 2) intersects the lines 3x+y=0 and 3x−y=0 at the points A and B. The equation to the line AB so that the triangle OAB is equilateral is:

Given equations 3x+y=0−−(1)⇒ slope =−3and 3x−y=0−−(2)⇒ slope =3Let equation of the line passing through (2, 2) isy−2=m(x−2)−−(3)∴ Δle OAB is Equilateral∴ Angle b/w (2)&(3)=600⇒ Tan 600=|m−31+3m|⇒ 3 = |m−31+3m|S.B.S⇒ 3(1+3m)2=(m−3)2⇒ 3(1+3m2+23m)=m2+3−25m⇒ 3+9m2+63=m2+3−23m⇒ 8m2+83=0⇒ 8m(m+3)=0m=0, m=−3If m=0 then from (3)y−2=0(x−2)⇒ y−2=0

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$A s t . l i n e t h r o u g h t h e p o int (2, 2) int e r \sec t s t h e l i n e s \sqrt{3} x + y = 0 a n d \sqrt{3} x - y = 0 a t$ the points A and B. The equation to the line AB so that the triangle OAB is equilateral is:

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$x - 2 = 0$

$x + y - 4 = 0$

$x + y + 4 = 0$

$y - 2 = 0$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n e q u a t i o n s \sqrt{3} x + y = 0 - - (1) \Rightarrow s l o p e = - \sqrt{3} \\ a n d \sqrt{3} x - y = 0 - - (2) \Rightarrow s l o p e = \sqrt{3} \\ L e t e q u a t i o n o f t h e l i n e p a s \sin g t h r o u g h (2, 2) i s \\ y - 2 = m (x - 2) - - (3) \\ ∴ Δ^{l e} O A B i s E q u i l a t e r a l \\ ∴ A n g l e b / w (2) & (3) = 60^{0} \\ \Rightarrow Tan 60^{0} = | \frac{m - \sqrt{3}}{1 + \sqrt{3} m} | \\ \Rightarrow \sqrt{3} = | \frac{m - \sqrt{3}}{1 + \sqrt{3} m} | \end{array}$

$\begin{array}{l} S . B . S \\ \Rightarrow 3 {(1 + \sqrt{3} m)}^{2} = {(m - \sqrt{3})}^{2} \\ \Rightarrow 3 (1 + 3 m^{2} + 2 \sqrt{3} m) = m^{2} + 3 - 2 \sqrt{5} m \\ \Rightarrow 3 + 9 m^{2} + 6 \sqrt{3} = m^{2} + 3 - 2 \sqrt{3} m \\ \Rightarrow 8 m^{2} + 8 \sqrt{3} = 0 \\ \Rightarrow 8 m (m + \sqrt{3}) = 0 \\ m = 0, m = - \sqrt{3} \\ I f m = 0 t h e n f r o m (3) \\ y - 2 = 0 (x - 2) \\ \Rightarrow y - 2 = 0 \end{array}$