0
$\begingroup$

The general form of the equation of the angle bisector of two lines is:

$\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}= \pm\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$

Now, we have a point $(\alpha, \beta)$ lying in one of the angles between these two.

I have been instructed to do as follows:

  1. Find the sign of the expression of $L_1$ and $L_2$.
  2. If they are opposite then, choose the negative sign from the general form otherwise choose the positive

    but I want to know this method's proof for better understanding. Or, if there's an alternate/ better method to deal with it, please let me know (with proof).

Edit: $L_1 \equiv a_1x+b_1y+ c_1$

$L_2\equiv a_2x+b_2y+ c_2$

$\endgroup$
  • $\begingroup$ What are $L_1$ and $L_2$? $\endgroup$ – Emilio Novati Sep 22 '17 at 9:35
  • $\begingroup$ @EmilioNovati Expressions of Lines $\endgroup$ – Archer Sep 22 '17 at 9:42
  • 2
    $\begingroup$ Closely related to math.stackexchange.com/q/2403530/265466. $\endgroup$ – amd Sep 23 '17 at 0:42
  • 1
    $\begingroup$ "I have been instructed to do as follows: " Rather than telling us what someone else told you to do, you should show more of what you did do. $\endgroup$ – Simply Beautiful Art Sep 24 '17 at 18:25
  • 1
    $\begingroup$ Thanks! $\ddot\smile$ $\endgroup$ – Simply Beautiful Art Sep 24 '17 at 18:52
2
$\begingroup$

The formula $$ d_1(x,y) = \frac{\lvert a_1x+b_1y+c_1 \rvert}{\sqrt{a_1^2+b_1^2}} $$ gives you the distance from a point $(x,y)$ to the line $L_1$ whose equation is $a_1x+b_1y+c_1 = 0.$ See Distance Between A Point And A Line for a proof of this. For the line $L_2$ given by $a_2x+b_2y+c_2 = 0,$ the distance of a point to the line is given by $$ d_2(x,y) = \frac{\lvert a_2x+b_2y+c_2 \rvert}{\sqrt{a_2^2+b_2^2}}. $$

A point $(x,y)$ on an angle bisector between two lines is equidistant from the two lines, that is, it satisfies the condition $d_1(x,y) = d_2(x,y).$ Writing out the formulas for $d_1$ and $d_2$ in full, $$ \frac{\lvert a_1x+b_1y+c_1 \rvert}{\sqrt{a_1^2+b_1^2}} = \frac{\lvert a_2x+b_2y+c_2 \rvert}{\sqrt{a_2^2+b_2^2}}. $$

Now observe that $\lvert a_1x+b_1y+c_1 \rvert$ will be either $a_1x+b_1y+c_1$ or $-(a_1x+b_1y+c_1),$ whichever of those two expressions is positive. In fact, $a_1x+b_1y+c_1$ will be positive for all points on one side of the line and negative for all points on the other side.

Now if the lines $L_1$ and $L_2$ intersect, they divide the plane into four regions. Label each these regions as $+L_1$ or $-L_1$ depending on whether $a_1x+b_1y+c_1$ is (respectively) positive or negative in that region. Label each region as $+L_2$ or $-L_2$ depending on whether $a_2x+b_2y+c_2$ is (respectively) positive or negative in that region.

One of the angle bisectors of $L_1$ and $L_2$ will go through the regions labeled $+L_1,+L_2$ or $-L_1,-L_2.$ That is, on that line the signs of $a_1x+b_1y+c_1$ and $a_2x+b_2y+c_2$ are either both positive or both negative. Points on this line therefore satisfy the formula $$ \frac{a_1x+b_1y+c_1 }{\sqrt{a_1^2+b_1^2}} = \frac{a_2x+b_2y+c_2 }{\sqrt{a_2^2+b_2^2}}. $$ (For points in the region $-L_1,-L_2,$ this formula gives negative values on both sides, but their absolute values are equal.)

The other angle bisector goes through $+L_1,-L_2$ and $-L_1,+L_2$ and has the formula $$ \frac{a_1x+b_1y+c_1 }{\sqrt{a_1^2+b_1^2}} = - \frac{a_2x+b_2y+c_2 }{\sqrt{a_2^2+b_2^2}}. $$

$\endgroup$
0
$\begingroup$

I'd use other method (for lines $y+ax+b=0$).

$L: y+ax+b=0$ creates with ox axis the angle $\alpha$, then $$\sin\alpha=\frac{a}{\sqrt{1+a^2}}\\ \cos\alpha=\frac{1}{\sqrt{1+a^2}}$$

If two lines $L_1, L_2$ are not paralel ($a_1\neq a_2$), we can compute their intersection point $(x_0, y_0)$

The angle between $L_1$ and $L_2$ is $\beta=\frac{\alpha_1+\alpha_2}{2}$

Thus:

  • if $\alpha_1+\alpha_2=\pi(1+2k)$ (it can be obtained only, if $a_1=-a_2$), then $\beta=\frac{\pi}{2}+k\pi$ and the bisector line is in form $$L_3:y+b_3=0$$
  • in other cases ($a_1\neq-a_2$) we can compute $\tan \beta$: $$\tan\beta = \frac{\sin(\alpha_1+\alpha_2)}{1-\cos(\alpha_1+\alpha_2)}=\frac{\sin(\alpha_1)\cos(\alpha_2)+\sin(\alpha_2)\cos(\alpha_1)}{1+\sin(\alpha_1)\sin(\alpha_2)-\cos(\alpha_1)\cos(\alpha_2)}\\ =\frac{a_1+a_2}{\sqrt{(1+a_1^2)(1+a_2^2)}+a_1a_2-1}$$ Thus the bisector is in the form $$L_3:y-\frac{a_1+a_2}{\sqrt{(1+a_1^2)(1+a_2^2)}+a_1a_2-1}x+b_3=0$$

Now all we need is to insert our intersection point into adequate form of $L_3$ to compute $b_3$.

Edit: In case when one of the lines is in form $L=ax+b$:

We have $\sin\alpha=0$, $\cos\alpha=1$ and $\alpha=0$

Lines are paralel, if both are in this form.

If not, computation of $\tan\beta$ gives us different value (for $L_1:a_1x+b$): $$\tan\beta=\frac{a_2}{\sqrt{1+a_2^2}-1}$$

$\endgroup$
  • $\begingroup$ How did you get "$\sin \alpha= 1/\sqrt{1+a^2}$" $\endgroup$ – Archer Sep 22 '17 at 10:45
  • $\begingroup$ if You'll take a function y(x)=-(ax+b) (see, that $y(x)+ax+b=0$), you can take the right triangle $(0,y(0)),(1,y(1)),(1,y(0))$ and easily compute $\sin\alpha$. $\endgroup$ – Jaroslaw Matlak Sep 22 '17 at 11:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.