Question #202771

Find the equation of the line which passes through the point (1, √ 3) and makes an angle of 30◦ with the line x− √ 3y + √ 3 = 0


1
Expert's answer
2021-06-07T13:29:15-0400

Write the equation of the line in slope-intercept form


y=mx+by=mx+b

Line 

x3y+3=0x-\sqrt{3}y+\sqrt{3}=0

y=13x+1y=\dfrac{1}{\sqrt{3}}x+1

slope1=m1=13slope_1=m_1=\dfrac{1}{\sqrt{3}}

The nagle between two lines


tanθ=±m2m11+m1m2\tan \theta=\pm\dfrac{m_2-m_1}{1+m_1m_2}

Given θ=30°\theta=30\degree


tan30°=±m2m11+m1m2\tan 30\degree=\pm\dfrac{m_2-m_1}{1+m_1m_2}

13=±m2131+13m2\dfrac{1}{\sqrt{3}}=\pm\dfrac{m_2-\dfrac{1}{\sqrt{3}}}{1+\dfrac{1}{\sqrt{3}}m_2}

3+m2=±(3m23)\sqrt{3}+m_2=\pm(3m_2-\sqrt{3})

Let


3+m2=(3m23)\sqrt{3}+m_2=-(3m_2-\sqrt{3})

m2=0m_2=0

The equation of the line which passes through the point (1,3)(1, \sqrt{3}) is


y=3y=\sqrt{3}


Let


3+m2=(3m23)\sqrt{3}+m_2=(3m_2-\sqrt{3})


m2=3m_2=\sqrt{3}

Point (1,3)(1, \sqrt{3})

3=3(1)+b=>b=0\sqrt{3}=\sqrt{3}(1)+b=>b=0

The equation of the line which passes through the point (1,3)(1, \sqrt{3}) is


y=3xy=\sqrt{3}x

y=3y=\sqrt{3} or y=3xy=\sqrt{3}x




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