Question

plzzzz solve it fast. ​

Answer 1

Given lines:-

There are two given lines :-

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

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

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To find:-

Angle between the given two lines.

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Solution:-

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

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★ First line:-

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

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

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

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★ Second line:-

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Hence,

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

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Answer 2

Answer:

y - √3x- 5 = 0

y= √3x+5 ( y = mx + c )

Slope m1 = √3

√3y-x+6 =0

√3y = x -6

y = x/√3–6/√3

slope m2= 1/√3

tan theta =( m1-m2)/(1+m1m2)

tan theta = (√3–1/√3)/[1+(√3*1/√3)]

tan theta = [(√3*√3–1)/√3]/ 1+1)

= [(3–1/√3)]/ 2

= (2/√3)/2

= 1/√3

tan theta = 1/√3

tan30° = 1/√3

The angle between the lines is 30°.