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**Question:**

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.

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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:

$$

n_1=\frac{(a_1,b_1)}{\sqrt{a_1^2+b_1^2}}\\

n_2=\frac{(a_2,b_2)}{\sqrt{a_2^2+b_2^2}}

$$

the vectors $n_1+n_2$ and $n_1-n_2$ give the direction normal to the two bisectors.

**Edit**

$$

n_1=\frac{(1,-2)}{\sqrt{5}}=\sqrt{5}\frac{(1,-2)}{5}\\

n_2=\frac{(4,-3)}{5}

$$

then

$$

n_1+n_2=\frac{(\sqrt{5}+4,-2\sqrt{5}-3)}{5}\\

n_1-n_2=\frac{(\sqrt{5}-4,-2\sqrt{5}+3)}{5}\\

$$

The two bisectors are

$$

(\sqrt{5}+4)(x-x_0)+(-2\sqrt{5}-3)(y-y_0)=0\\

(\sqrt{5}-4)(x-x_0)+(-2\sqrt{5}+3)(y-y_0)=0

$$

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:

$(4-\sqrt5)x-(3-2\sqrt5)y+(2-4\sqrt5)=0$

$(4+\sqrt5)x-(3+2\sqrt5)y+(2+4\sqrt5)=0$

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