JEE Main & Advanced Mathematics Straight Line Question Bank Critical Thinking

  • question_answer 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]

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

    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 \[x-3y-31=0\]

    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 \[x-3y-31=0\].


You need to login to perform this action.
You will be redirected in 3 sec spinner