VITEEE 2011 Paper

The equation of the bisectors of the angle between the lines given by $ax^2$ + 2hxy + $by^2$ = 0 is $\frac{x^2-y^2}{a-b}$ = $\frac{xy}{h}$ ... (i) And the equation of the bisectors of the angle between the lines given by $a^2x^2$ + 2h (a + b) xy + $b^2y^2$ = 0 is $\frac{x^2-y^2}{a-b}$ = $\frac{xy}{h(a+b)}$ ⇒ $\frac{x^2-y^2}{a-b}$ = $\frac{x}{h}$ ... (ii) From eqs. (i) and (ii), it is clear that both the pair of straight lines have the same bisector, hence, the given two pairs of straight lines are equally inclined.