Question

Question:-
Find the angle between the lines y - √3x - 5 = 0 and √3y - x + 6 = 0.

Need the answer ASAP!​

Answer 1

Given lines

• There are two given lines :-

$\tt\longrightarrow{y - \sqrt{3}x - 5 = 0}$

$\tt\longrightarrow{\sqrt{3}y - X + 6 = 0}$

⠀

To find

• Angle between the given two lines.

⠀

Solution

• We have to equation of lines, given in the question.

⠀

★ First line

$\tt\longrightarrow{y - \sqrt{3}x - 5 = 0}$

⠀

$\mathcal{\bigg\lgroup{In\: slope\: intercept\: form}\bigg\rgroup}$

$\tt\longrightarrow{y = \sqrt{3}x + 5 }$⠀⠀.....[1]

⠀

★ Second line

$\tt\longrightarrow{\sqrt{3}y - X + 6 = 0}$

⠀

$\mathcal{\bigg\lgroup{In\: slope\: intercept\: form}\bigg\rgroup}$

$\tt\longrightarrow{y = \dfrac{1}{\sqrt{3}}x - 2\sqrt{3}}$⠀⠀.....[2]

⠀

$\: \: \: \: \: \: \: \: \: \underline{\sf{\red{According\: to\: slope\: intercept\: form}}}$

⠀

$\sf\longmapsto{Slope\: of\: first\: line\: (m_1) = \sqrt{3}}$

⠀

$\sf\longmapsto{Slope\: of\: second\: line\: (m_2) = \dfrac{1}{\sqrt{3}}}$

⠀

• The acute angle (θ) between two lines is given by

$\large{\boxed{\boxed{\bf{tan \theta = \bigg| \dfrac{m_2 - m_1}{1 + m_1 m_2} \bigg|\: \: \: }}}}$

⠀

$\: \: \: \: \: \: \: \: \: \underline{\sf{\red{Putting\: the\: values}}}$

⠀

$\rm:\implies\: \: \: \: \: \: \: \: {tan \theta = \bigg| \dfrac{\dfrac{1}{\sqrt{3}} - \sqrt{3}}{1 + \sqrt{3} \times \dfrac{1}{\sqrt{3}}}\bigg|}$

⠀

$\rm:\implies\: \: \: \: \: \: \: \: {tan \theta = \bigg| \dfrac{\dfrac{1 - 3}{\cancel{\sqrt{3}}}}{\dfrac{\sqrt{3} + \sqrt{3}}{\cancel{\sqrt{3}}}}\bigg|}$

⠀

$\rm:\implies\: \: \: \: \: \: \: \: {tan \theta = \bigg| \dfrac{-2}{2\sqrt{3}} \bigg|}$

⠀

$\rm:\implies\: \: \: \: \: \: \: \: {tan \theta = \bigg| \dfrac{-1}{\sqrt{3}} \bigg|}$

⠀

$\rm:\implies\: \: \: \: \: \: \: \: {tan \theta = \dfrac{1}{\sqrt{3}}}$

⠀

$\rm:\implies\: \: \: \: \: \: \: \: {tan \theta = tan30°}$

⠀

$\rm:\implies\: \: \: \: \: \: \: \: {\underline{\boxed{\orange{\theta = 30°}}}}$

⠀

Hence,

• Angle between the two given lines is either 30° or 180° - 30° = 150°.

━━━━━━━━━━━━━━━━━━━━━━

Answer 2

We know that :-

If slope of two lines are m₁ and m₂ then acute angle θ between two lines is given as :

$\sf tan\theta = \bigg | \dfrac{m_1 -m_2 }{1 +m_1 m_2 } \bigg |$

We also know that :-

$\tt slope (m) \: \: of \:\: line \: \: ax + by + c = 0 \: is : \\ \\ \boxed{ \sf m = \frac{ - a}{b} }$

$\tt slope (m _1) \: \: of \: line \: \: y - \sqrt{3} x - 5 = 0 : \\ \\ \rightarrow \sf m _1 = - \: \frac{ - \sqrt{3} }{1} = \sqrt{3} \\ \\ \\ \tt slope (m _2) \: \: of \: line \: \: \sqrt{3} y - x + 6 = 0 : \\ \\ \sf \rightarrow m _2 = - \frac{ - 1}{ \sqrt{3} } = \frac{ 1}{ \sqrt{3}}$

$\sf \longrightarrow tan\theta = \bigg | \dfrac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 +(\sqrt{3} \times \frac{1}{ \sqrt{3}}) } \bigg | \\ \\ \sf \longrightarrow tan\theta = \bigg | \dfrac{ \frac{3 - 1}{ \sqrt{3} } }{1 +1} \bigg |\\ \\ \sf \longrightarrow tan\theta = \bigg | \dfrac{ \frac{2}{ \sqrt{3} } }{2} \bigg |\\ \\ \sf \longrightarrow tan\theta = \bigg | \frac{1}{ \sqrt{3} } \bigg |\\ \\ \sf \longrightarrow tan\theta = \frac{1}{ \sqrt{3} } \\ \\ \therefore \theta \: = 30^{\circ}$

Angle between the lines y - √3x - 5 = 0 and √3y - x + 6 = 0 = 30°