Question

If $\theta$ is an angle between the lines given by the equation $6x^2+5xy-4y^2+7x+13y-3=0$, then equation of the line passing through the point of intersection of these lines and making an angle $\theta$ with the positive x - axis is

• A. $2x+11y+13=0$
• B. $11x-2y+13=0$
• C. $2x-11y+2=0$
• D. $11x+2y-11=0$

Answer 1

The correct option is B

$11x-2y+13=0$

Writing the given equation as a quadratic in x we have $6x^2+(5y+7)x-(4y^2-13y+3)=0$
$\Rightarrow x=\frac{-(5y+7)\pm \sqrt{(5y+7{)}^2+24(4y^2-13y+3)}}{12}=\frac{-(5y+7)\pm \sqrt{121y^2-242y+121}}{12}=\frac{-(5y+7)\pm 11(y-1)}{12}=\frac{6y-18}{12},\frac{-16y+4}{12}\Rightarrow 2x-y+3=0and2x+4y-1=0$
which are the two lines represented by the given equation and the point of intersection is (-1, 1), obtained by solving these equations.
Also $tan\theta =\frac{2\sqrt{h^2-ab}}{a+b}$ where a = 6, b = -4, $h=\frac{5}{2}$
$=\frac{2\sqrt{{\left(\frac{5}{2}\right)}^2-6(-4)}}{6-4}=\sqrt{\frac{121}{4}}=\frac{11}{2}$
So the equation of the required line is
$y-1=\frac{11}{2}(x+1)\Rightarrow 11x-2y+13=0$