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In the cartesian coordinate plane,

Let $f_k(x,y) = a_kx + b_ky + c_k$.

Let $f_1(x,y)=0$ and $f_2(x,y)=0$ be two lines,

We know their (2) angle bisectors are given by :

$\frac{f_1(x,y)}{\sqrt{a_1^2+b_1^2}} = \pm \frac{f_2(x,y)}{\sqrt{a_2^2+b_2^2}}$

Now, the question is which one is acute and which is obtuse angle bisector, Say one bisector is $B_1$ and one is $B_2$, one way to do this would be to find angle between the bisector and one line, is the modulus of the tangent of that angle is less than $1$, that is acute and otherwise, obtuse.

Another way which was given in my book was;

Let sign of $a_1a_2+b_1b_2$ be $S$, ie. $S$ is $\frac{a_1a_2+b_1b_2}{|a_1a_2+b_1b_2|}$, then the obtuse angle bisector is given by :

$\frac{f_1(x,y)}{\sqrt{a_1^2+b_1^2}} = S \frac{f_2(x,y)}{\sqrt{a_2^2+b_2^2}} \ \ \ (^@)$

Now, this is something I was just not able to prove, like I tried to do to use the tangent less than $1$ condition but it just did not work at all, at least for me it became really ugly really fast (as I had to square it to take care of modulus and signs).

But, I do think there could be a solution using VECTORS!!!

The reason I believe so is that if $\vec c = a_1 \hat i + b_1 \hat j$ and $\vec d = a_2 \hat i + b_2 \hat j$,

Then, $\vec c \cdot \vec d = a_1a_2 + b_1b_2$.

But I haven't really learnt vectors much, so I am not really sure how they would relate to a straight line.

So, in a nutshell, I just want the vector proof of $(^@)$

Aditya_math
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  • @Unknown , Oh thanks, I did not know that; but still I don't think this is duplicate coz I want the vectors solution only – Aditya_math Jul 16 '21 at 11:45
  • @MathLover , that might work, i will probably need to read it a few times but before that, can you please explain what the binomial thing means here, like $\binom ab$ – Aditya_math Jul 16 '21 at 12:36
  • ${a \choose b}$ - you mean in one of the answers? in the context of vector? That would be $\hat i$ component as $a$ and $\hat j$ component as $b$. – Math Lover Jul 16 '21 at 12:40
  • @MathLover ,yup, that's it... ok thank you so much; I have just one last question, is the vector $a \hat i + b \hat j$ same as line $ax+by+c=0$, or essentially, could you please tell how vectors relate to lines in coordinate plane (you can post as an answer if you like) – Aditya_math Jul 16 '21 at 13:31
  • No it is not the same. In fact $a \hat i + b \hat j$ is normal to line $ax + by + c = 0$. – Math Lover Jul 16 '21 at 15:21

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