• # 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 $x-3y-31=0$

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 $x-3y-31=0$.