InterviewSolution
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InterviewSolution
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Find the equation of the straight line which passes through the originand makes angle `60^0`with the line `x+sqrt(3)y+sqrt(3)=0`. |
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Answer» Correct Answer - x=0, x`-sqrt(3)` y =0 The given line is `x+sqrt(3)y+3sqrt(3) = 0.` `"or " y=(-(1)/(sqrt(3)))x-3` Therefore, the slope of (1) is `-1//sqrt(3).` Let the slope of the required line be m. Also, the angle between these line is `60^(@)`. Therefore, `" tan " 60^(@) = |(m-(-1//sqrt(3)))/(1+m(1//sqrt(3)))|` `"or " sqrt(3) = |(sqrt(3)m+1)/(sqrt(3)-m)|` `"If "(sqrt(3)m+1)/(sqrt(3)-m) =sqrt(3)` `"or "m=(1)/(sqrt(3))` Using y=mx +c, the equation of the required line is `y=(1)/(sqrt(3))x +0` `"i.e.," x-sqrt(3) y = 0` (as the line passes through the origin, c=0) `(sqrt(3)m-1)/(sqrt(3)-m) = -sqrt(3)` `"or " sqrt(3)m+1 =-3+sqrt(3)m` Therefore, finite m does not exist. Therefore, the slope of the required line is infinity. Thus, the required line is a vertical line. This line passes through the origin. Therefore, the equation of the required line is x=0. |
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