# Thread: Angle bisector of a pair of straight lines

1. ## Angle bisector of a pair of straight lines

hello everyone

This is a question on coordinate geometry.
One bisector of the angle between the lines given by a(x-1)^2 + 2h(x-1)y + by^2 = 0 is 2x + y – 2 = 0. We need to find the other bisector.

Any help would be appreciated. Thanks in advance.

2. Hello!

We have $a(x-1)^2 + 2h(x-1)y + by^2 = 0$.

Dividing throughout by $(x - 1)^2$, and solving the quadratic in $\frac{y}{x - 1}$, we get:
$\frac{y}{x - 1} = \frac{-h \pm \sqrt{h^2 - ab}}{b}$, which gives the equations of the pair of straight lines.

The joint equation of bisectors is given by:
$\frac{(x - 1)^2 - y^2}{a - b} = \frac{(x - 1)y}{h}$
(here x in the standard equation found in books is replaced by $(x - 1)$ since the equation of our pair of straight lines is in $(x - 1)$)

On expanding, we get:

$hx^2 + xy(b - a) - hy^2 - 2hx + y(a - b) + h = 0$ --- (1)

One of the lines represented by the above equation is $2x + y - 2 = 0$.

We know that the bisectors of the angles between the lines are perpendicular to each other (Make sketches to convince yourself. Moreover in the joint equation, Coefficient of $x^2$ + Coefficient of $y^2$ = 0 which tells you this fact immediately).

Thus, equation of the other bisector is $x - 2y + k$, where k is the constant to be determined.

(We get the above equation by applying condition of perpendicularity: $m_{1}m_{2} = -1$, where $m_{1}$ and $m_{2}$ are the respective slopes. In our case $m_{1} = -2$ which gives $m_{2} = \frac{1}{2}$, thus giving the equation of the other bisector)

Multiplying these two equations together,

$(2x + y – 2)(x - 2y + k) = 0$
$\Rightarrow 2x^2 - 3xy - 2y^2 + x(2k - 2) + y(4 + k) - 2k = 0$

Comparing with (1),
1) h = 2
2) $2k - 2 = -2h$ $\Rightarrow k = -1$

So, the other bisector is:
$\boxed{x - 2y - 1 = 0}$

3. thanks fardeen, i couldnt have asked for more.

4. Originally Posted by watsmath
hello everyone

This is a question on coordinate geometry.
One bisector of the angle between the lines given by a(x-1)^2 + 2h(x-1)y + by^2 = 0 is 2x + y – 2 = 0. We need to find the other bisector.

Any help would be appreciated. Thanks in advance.
Alternatively,

we can find the point of intersection of the angle bisector 2x+y-2 with the lines
and find the equation of the perpendicular to the bisector that goes through
that same point of intersection.

Hence, y=2-2x is substituted

$a(x-1)^2+2h(x-1)(2-2x)+b(2-2x)^2=0$

$2-2x=2(1-x)=-2(x-1)$

$(2-2x)^2=2(1-x)2(1-x)=-2(x-1)(-2)(x-1)=4(x-1)^2$

then

$a(x-1)^2-4h(x-1)+4(x-1)^2=0$

$(x-1)\left[a(x-1)-4h+4(x-1)\right]=0$

x=1

$2x+y-2=0\ \Rightarrow\ 2+y-2=0,\ y=0$

The slope of $2x+y-2=0$ is -2, as $y=-2x+2$

Any perpendicular line has slope $\frac{1}{2}$

The perpendicular bisector contains the point (1,0)

hence it's equation is

$y-0=\frac{1}{2}(x-1)$

$2y=x-1$

$x-2y-1=0$

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