An equilateral triangle ABC has its centroid at the the origin and the base BC lies along #x+y+1=0#. Gradient of the other two lines are?

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Ans- #2+sqrt3, 2-sqrt3#

1 Answer
CW
Oct 10, 2017

#2+-sqrt3#

Explanation:

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Given that #BC# lies on the line #x+y+1=0#,
#=># slope of #BC=-1#
#=> alpha=45^@#
given that #DeltaABC# is an equilateral triangle,
angle between #BC and CA =60^@#,
#=> theta=180-45-60=75^@#
#=># slope of #AC=tantheta=tan75=color(red)(2+sqrt3)#
Similarly,
slope of #AB=tan(60-45)=tan15=color(red)(2-sqrt3)#

Or you can use the following formula to find the slopes of the other two lines:
formula for the angle #beta# between two lines with slopes #m_1 and m_2# is given by :
#tanbeta=|m_1-m_2|/|1+m_1*m_2|#
let #m_1# be the slope of #BC=-1#
#=> tan60=sqrt3=|-1-m_2|/|1-m_2|#
#=>m_2=2+sqrt3, 2-sqrt3#

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