Question

The equation of the pair of bisectors of the angles between two straight line is. $12x^2-7xy-12y^2=0$. If the equation of one line is $2y-x=0$ then the equation of the other line is:

• A. $-41x+38y=0$
• B. $11x+2y=0$
• C. $38x+41y=0$
• D. $11x-2y=0$

Answer 1

The correct option is A $-41x+38y=0$

Let the pair of intersecting lines be represented by:

$a{x}^2+2hxy+b{y}^2=0$

Then the pair of bisectors of the angles between them is :

$h(x^2-y^2)=(a-b)xy$

we are given: pair of bisectors:

$12(x^2-y^2)=7xy$

So, h = 12 and (a - b) = 7

So, pair of intersecting lines:

$(7+b)x^2+24xy+b{y}^2=0$

one of the lines is $2y-x=0$

So, $(2y-x)[\frac{b}{2}y-(7+b\left)x\right]=(7+b)x^2+24xy+by2$

equating coefficients: $\frac{-b}{2}-2(7+b)=24$

$\Rightarrow b=\frac{-76}{5}$

So, $a=b+7=\frac{-41}{5}$

So, the other line :

$\frac{b}{2}y-(7+b)x=0$

$-\frac{38}{5}y+\frac{41}{5}x=0$

$38y-41x=0$