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Question
Two sides of an isosceles triangle are given by the equation 7xāy+3=0 and x+yā3=0. lf its third side passes through the point (1,ā10), then its equations are
Two sides of an isosceles triangle are given by the equation 7xāy+3=0 and x+yā3=0. lf its third side passes through the point (1,ā10), then its equations are
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Solution
The correct option is D x−3y−31=0 or 3x+y+7=0
The equation of any line passing through (1,−10) is y+10=m(x−1).
Since it makes equal angles, say θ, with the given lines, thereforetanθ=m−71+7m=−m−(−1)1+m(−1)
⇒m=13 or −3
Thus the equations of third side are y+10=13(x−1) or y+10=−3(x−1)
i.e. x−3y−31=0 or, 3x+y+7=0
The equation of any line passing through (1,−10) is y+10=m(x−1).
Since it makes equal angles, say θ, with the given lines, therefore
tanθ=m−71+7m=−m−(−1)1+m(−1)
⇒m=13 or −3
Thus the equations of third side are y+10=13(x−1) or y+10=−3(x−1)
i.e. x−3y−31=0 or, 3x+y+7=0
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