Answer to Question #96884 in Analytic Geometry for Aakash

Question #96884
Find the equation of the line which passes through the point (1,√3) and makes anangle of 30◦ with the line x+√3y+√3 = 0.
1
Expert's answer
2019-10-21T10:33:51-0400

The equation of a straight line with gradient m, passing through the point "(x_1, y_1)", is 

.

"y \u2212 y_1 = m(x \u2212 x_1)."



Point  "A(1,\u221a3)" :

"y \u2212 \\sqrt{3}= m_1(x \u2212 1 )."


The equation of the second line


"x+\\sqrt{3}y+\\sqrt{3} = 0."



Or


"y = -\\frac{1}{\\sqrt{3}} (x+\\sqrt{3}) \\Rightarrow m_2=-\\frac{1}{\\sqrt{3}}"


The angle between the lines:


"\\tan(\\alpha)= |\\frac{m_2-m_1}{1-m_2m_1} |"

then


"\\tan(\\frac{ \\pi }{6} )=|\\frac{-\\frac{1}{\\sqrt{3}}-m_1}{1+\\frac{1}{\\sqrt{3}}m_1} | \\Rightarrow \\frac{1}{\\sqrt{3}}=| \\frac{-\\frac{1}{\\sqrt{3}}-m_1}{1+\\frac{1}{\\sqrt{3}}m_1} | \\Rightarrow m_1=0 ," or "m_1=-\\frac{\\sqrt{3}}{2}"



The sought equations:



"1) y=\\sqrt{3} ."

.

and



"y \u2212 \\sqrt{3}= -\\frac{\\sqrt{3}}{2} (x \u2212 1 ) \\Rightarrow"


"2) y = -\\frac{\\sqrt{3}}{2} x +\\frac{3\\sqrt{3}}{2}"



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