Question

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

Answer 1

GIVEN :-

• Equation of lines are -
• $\sf \: \sqrt{3} x - y + 5 = 0 \: \: \: \: \: \: \: \: \: ....(i)$
• $\sf \: x - \sqrt{3} y - 6 = 0 \: \: \: \: \: \: \: ....(ii)$

SOLUTION :-

$\sf \: First \: \: change \: \: the \: \: eq. \: \: of \: \: line \: \: in \: \: the \: \: form$

$\sf \: of \: \: y = \: mx + c$

From equation (i)

$\sf \: \sqrt{3} x - y + 5 = 0$

$\sf \: \implies \sqrt{3} x + 5 = y$

$\sf \: \implies y = \sqrt{3} x + 5$

$\sf \: Here \: \: slope \: \: (m_{1} )$

$\sf slope \: \: (m_{1} ) = \sqrt{3}$

From equation (ii)

$\sf \: x - \sqrt{3} y - 6 = 0$

$\sf \: \implies x - 6 = - \sqrt{3} y$

$\sf \: \implies \frac{x - 6}{ - \sqrt{3} } = y$

$\sf \: \implies y = - \frac{x }{ \sqrt{3} } + \frac{6}{ \sqrt{3} }$

$\sf \: Here \: \: slope \: \: (m_{2} )$

$\sf slope \: \: (m_{2} ) = - \frac{1}{ \sqrt{3} }$

$\sf \: Use \: \: the \: \: formula \: \: of \: \: slope -$

$\sf \: slope \: \green{ \{\tan( \theta) \} }= \pink{ \bigg | \frac{ m_{1} - m_{2}}{1 +m_{1}m_{2} } \bigg | }$

$\sf \: \tan( \theta) = \bigg | \frac{ \sqrt{3} - (\frac{ - 1}{ \sqrt{3} }) }{1 + \sqrt{3}. (\frac{ - 1}{ \sqrt{3} } ) } \bigg |$

$\sf \: \tan( \theta) = \bigg | \frac{ \sqrt{3} - (\frac{ - 1}{ \sqrt{3} }) }{1 - 1 } \bigg |$

$\tan( \theta) = \infin$

$\theta = 90 \degree$

Hence lines are perpendicular to each other.