find the equation of line passing through straight line (1, 2) and making on angle 60° with line root 3x-y+2=0
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Step-by-step explanation:
Given Line: y=(tan(60°))x+c ►y=√3 x + c ►1 = 2√3 + c ►c =1 - 2√3 ► y= √3 x + 1 - 2√3. Let (a,b) be the required point on the given line that is 3 units away from (1,2). Hence, b = √3 a + 1 - 2√3 ————-—(1), (b-2)²+(a-1)²=9 ————————-(2). Solve (1) & (2) simultaneously to obtain: a=(1/4)[7 +√3 -√(32-2√3)], b = (1/4)[7 -√3 -√(96-6√3)] Or a=(1/4)[7+√3+√(32-2√3)], b=(1/4)[7-√3+√(96-6√3)] Or a~=0.85, b~=1 and a~=3.52, b~=3.63
Hope my calculations are correct.
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