Pair of straight lines

Find the equation of the bisector of the obtuse angle between the lines $x – 2y + 4 = 0$ and $4x – 3y + 2 = 0$.

I don’t even know how to proceed here. I know how to find the angle between two lines, but not sure whether that would help in this case.

Solutions Collecting From Web of "Pair of straight lines"

Given that for $ax+by+c=0$ the vector $n=(a,b)$ is normal to the line, then take the normalized vectors of the two lines:
the vectors $n_1+n_2$ and $n_1-n_2$ give the direction normal to the two bisectors.


The two bisectors are
where $P_0=(x_0,y_0)$ is the intersection point of the lines, and I have not used the $5$ in the denominator of the vectors because it is inessential.

If you know the equations for two straight lines, then you can find the point of intersection $ (8/5, 14/5) $ and also we know slope of both the lines $ (1/2), (4/3) $.

From slopes we can find the angle made by both the lines with the positive X-axis $(\tan \theta)$

So we can find the acute angle between the two lines, subtract twice of this angle from $360$ and divide that by 2 to get the obtuse angle made between the two lines. Divide this obtuse angle by 2 and add the angle corresponding to slope $(4/3)$.

Now we have the angle made by the obtuse angle bisector with the positive X-axis, get slope $ ( \tan \theta ) $ and we already calculated the point on this bisector $ (8/5, 14/5) $, so use $y=mx+c$ to find $c$ and you have the equation needed.

Suppose $P(X,Y)$ is a point on the bisector. If $PA$ and $PB$ are perpendiculars from $P$ to the two lines, then $PA = PB$.

$$\frac{X-2Y+4}{\sqrt{1^2+2^2}} = \frac{4X-3Y+2}{\sqrt{4^2+3^2}}$$

The two bisectors are: