Question

$\huge{\mathcal{\underline{\pink{Question:-}}}}$

The equation of two straight lines throught (7,9) and making an angle of 60° with the line x - √3y -2√3 = 0

a] x - 7 = 0

b] $\small\rm{(y-9)\:=\:-\frac{1}{\sqrt{3}}\times\:(x-7)}$

c] x + 7 = y

d] $\small\rm{(y+9)\:=\:\frac{1}{\sqrt{3}}\times\:(x+7)}$

[Please solve with clear explanation.This question is telated to Angle between two lines topic. Answers are (a),(b). Copied answers and spammers will be reported]

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Answer 1

GIVEN :–

• Straight lines passing throught (7,9).

• And making an angle of 60° with the line x - √3y -2√3 = 0.

TO FIND :–

▪︎ Straight line = ?

SOLUTION :–

▪︎ First Let's find the Slope of line x - √3y -2√3 = 0.

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$\bf \implies \large{ \boxed{ \bf \: slope = \dfrac{ - coffieciant \: \: of \: \: x}{coffieciant \: \: of \: \: y}}}$

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$\bf \implies m_{1} = \dfrac{1}{ \sqrt{3} }$

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• We know that –

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$\bf \implies \large{ \boxed{ \bf \: \tan( \theta) = \left| \dfrac{m_{2}- m_{1}}{ 1+m_{1}m_{2}} \right| }}$

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• Here –

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$\bf \: \to \: \: \theta = {60}^{ \circ} \: \:$

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$\bf \: \to \: \: m_1 = \dfrac{ 1}{ \sqrt{3} } \: \:$

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$\bf \: \to \: \: m_2 =slope \: \: of \: \: required \: \: line$

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• So that –

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$\bf \implies \tan( {60}^{ \circ} ) = \left| \dfrac{m_{2}- \left(\dfrac{ 1}{ \sqrt{3} } \right) }{ 1+ \left(\dfrac{1}{ \sqrt{3} } \right) m_{2}} \right|$

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$\bf \implies \sqrt{3} = \left| \dfrac{m_{2} - \dfrac{ 1}{ \sqrt{3} } }{ 1 + \dfrac{m_{2}}{ \sqrt{3} } } \right|$

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$\bf \implies \sqrt{3} = \left| \dfrac{m_{2} \sqrt{3} -1}{ \sqrt{3} + m_{2}} \right|$

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$\bf \implies \sqrt{3} = \pm \left( \dfrac{m_{2} \sqrt{3} -1}{ \sqrt{3} + m_{2}} \right)$

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• Take positive sign –

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$\bf \implies \sqrt{3} = \dfrac{m_{2} \sqrt{3} - 1}{ \sqrt{3} + m_{2}}$

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$\bf \implies \sqrt{3}( \sqrt{3} + m_{2}) = m_{2} \sqrt{3} - 1$

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$\bf \implies 3 + \sqrt{3} m_{2} = m_{2} \sqrt{3} - 1$

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$\bf \implies 4 = m_{2} (0)$

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$\bf \implies { \boxed{ \bf{m_{2} = \infty }}}}$

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• Take negetive sign –

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$\bf \implies - \sqrt{3} = \dfrac{m_{2} \sqrt{3} - 1}{ \sqrt{3} + m_{2}}$

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$\bf \implies - \sqrt{3}( \sqrt{3} + m_{2}) = m_{2} \sqrt{3} - 1$

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$\bf \implies - 3 - \sqrt{3} m_{2} = m_{2} \sqrt{3} - 1$

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$\bf \implies - 2 = 2 \sqrt{3} m_{2}$

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$\bf \implies { \boxed{ \bf{m_{2} = - \dfrac{1}{\sqrt{3}}}}}$

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• Standard form of line if it's passing from (p,q) –

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$\bf \implies \large{ \boxed{ \bf y-q= m_{2}(x - p) }}$

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• First Required line –

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$\bf \implies y-9= \infty (x - 7)$

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$\bf \implies y-9= \dfrac{1}{0} (x - 7)$

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$\bf \implies \large{ \boxed { \bf x - 7 = 0}}$

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• Second Required line –

$\bf \implies \large{ \boxed{ \bf \: y-9= - \dfrac{1}{ \sqrt{3} } (x - 7)}}$

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▪︎ Hence Option (a) & (b) are correct.

Answer 2

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