Straight Line Question 66

Question: The equations of two equal sides of an isosceles triangle are $ 7x-y+3=0 $ and $ x+y-3=0 $ and the third side passes through the point (1, ? 10). The equation of the third side is [IIT 1984]

Options:

A) $ y=\sqrt{3}x+9 $ but not $ x^{2}-9y^{2}=0 $

B) $ 3x+y+7=0 $ but not $ 60^{o} $

C) $ 3x+y+7=0 $ or $ x-3y-31=0 $

D)Neither $ 3x+y+7 $ nor $ x-3y-31=0 $

Show Answer

Answer:

Correct Answer: C

Solution:

  • Any line through (1, ?10) is given by $ y+10=m(x-1) $ Since it makes equal angle say ? $ \alpha $ ? with the given lines $ 7x-y+3=0 $ and $ x+y-3=0 $ , therefore $ \tan \alpha =\frac{m-7}{1+7m} $ $ =\frac{m-(-1)}{1+m(-1)}\Rightarrow m=\frac{1}{3} $ or ? 3 Hence the two possible equations of third side are $ 3x+y+7=0 $ and $ x-3y-31=0 $ .