Problems from Angles between straight lines

Development:

To find the above mentioned straight line bisector, let's call it $b$, it is necessary to bear in mind 2 things:

• We know that the intersection of the straight lines cross at the point $r$ and $s$ (since it is the bisector).
• The straight line will form the same angle with $r$ as with $s$.

We firstly look for the intersection of $r$ and $s$: $$\{\begin{array}{c}x+y-1=0\\ y=2x+3\end{array}$$ $$x+(2x+3)-1=0\to 3x+2=0\to x=-{\displaystyle \frac{2}{3}}$$ $$y=2\cdot (-{\displaystyle \frac{2}{3}})+3=-{\displaystyle \frac{4}{3}}+3={\displaystyle \frac{5}{3}}$$ Therefore the straight line b that we are looking for will cross point $P=(-2/3,5/3)$.

We look now for the director vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ of the straight lines $r$ and $s$ respectively.

$\overrightarrow{u}=(1,-1),\overrightarrow{v}=(1,2)$

Let's see the angle formed by the above mentioned vectors with horizontal axis $OX$:

$tg({\alpha }_1)={\displaystyle \frac{-1}{1}}=-1\Rightarrow {\alpha }_1=-{45}^{\circ }$

$tg({\alpha }_2)={\displaystyle \frac{2}{1}}=2\Rightarrow {\alpha }_2=63,{435}^{\circ }$

Therefore we are faced with a situation of the type:

in the figure we consider ${\alpha }_1$ and ${\alpha }_2$ in absolute value.

Let's see 4 ways of solving the problem. We will start by the least rigorous.

Now we might take the angles, add them, divide the result by two and add it to $-{45}^{\circ }$ in order to know which angle has straight line $b$ with reference to the horizontal axis $OX$.

Nevertheless, this solution would be rather inelegant and vague since we would lose the decimal on the way.

Let's do it anyway:

${\alpha }_1=-{45}^{\circ }$

${\alpha }_2=63,{435}^{\circ }$

$${\displaystyle \frac{|{\alpha }_1-{\alpha }_2|}{2}}={\displaystyle \frac{|-45-63,435|}{2}}={\displaystyle \frac{|-108,435|}{2}}=54,{2175}^{\circ }=a$$ (that is the angle between $b$ and $r$ or $b$ and $s$).

Now we do ${\alpha }_1+a=9,2175..{.}^{\circ }$ which will be the angle that forms the straight line b that we are looking for (formed with the horizontal).

Therefore the slope of the straight line $b$ is: $$m=tg(9,{2175}^{\circ })=0,16227812...$$ And this way we have a director vector of the straight line $b$, which will be $\overrightarrow{w}=(1,m)=(1,0.16228)$

And using the equation vectorial we have,

$b:(x,y)=(-2/3,5/3)+k\cdot (1,0.16228)$

The previous procedure can be applied in a similar way but with exact results using the following trigonometrical formula:

$$tg(A)=-tg(-A)\Rightarrow tg(-{\alpha }_1)=-tg({\alpha }_1)=-(-1)=1$$

$$tg(A+B)={\displaystyle \frac{tg(A)+tg(B)}{1-tg(A)tg(B)}}\Rightarrow$$ $$\Rightarrow tg(|{\alpha }_1-{\alpha }_2|)=tg({\alpha }_2+(-{\alpha }_1))={\displaystyle \frac{1+2}{1-2}}=-3$$

$$tg(A)=u\Rightarrow \cos (A)={\displaystyle \frac{1}{\sqrt{1+u^2}}}\Rightarrow$$ $$\Rightarrow \cos (|{\alpha }_1-{\alpha }_2|)={\displaystyle \frac{1}{\sqrt{1+t{g}^2(|{\alpha }_1-{\alpha }_2|)}}}={\displaystyle \frac{1}{\sqrt{1+(-3{)}^2}}}={\displaystyle \frac{-1}{\sqrt{10}}}$$

$$tg({\displaystyle \frac{A}{2}})=\pm \sqrt{{\displaystyle \frac{1-\cos (A)}{1+\cos (A)}}}$$ with the corresponding sign $\pm$ depending on the quadrant where the angle ${\displaystyle \frac{A}{2}}$ is) $$\Rightarrow tg({\displaystyle \frac{|{\alpha }_1-{\alpha }_2|}{2}})=\sqrt{{\displaystyle \frac{1-\cos (|{\alpha }_1-{\alpha }_2|)}{1+\cos (|{\alpha }_1-{\alpha }_2|)}}}=\sqrt{{\displaystyle \frac{1+{\displaystyle \frac{1}{\sqrt{10}}}}{1-{\displaystyle \frac{1}{\sqrt{10}}}}}}=$$ $$=\sqrt{{\displaystyle \frac{{\displaystyle \frac{\sqrt{10}+1}{\sqrt{10}}}}{{\displaystyle \frac{\sqrt{10}-1}{\sqrt{10}}}}}}=\sqrt{{\displaystyle \frac{\sqrt{10}+1}{\sqrt{10}-1}}}=\sqrt{{\displaystyle \frac{10+2\sqrt{10}+1}{9}}}={\displaystyle \frac{\sqrt{11+2\sqrt{10}}}{3}}$$

In fact here we have the slope of a straight line from about ${54}^{\circ }$. In fact, we want want the slope to be a straight line of $(54-45{)}^{\circ }$.

$$tg(A-B)={\displaystyle \frac{tg(A)-tg(B)}{1+tg(A)tg(B)}}\Rightarrow tg({\displaystyle \frac{|{\alpha }_1-{\alpha }_2|}{2}}-45)={\displaystyle \frac{{\displaystyle \frac{\sqrt{11+2\sqrt{10}}}{3}}-1}{1+{\displaystyle \frac{\sqrt{11+2\sqrt{10}}}{3}}}}=$$ $$={\displaystyle \frac{\sqrt{11+2\sqrt{10}}-3}{\sqrt{11+2\sqrt{10}}+3}}\simeq 0,16227766$$

Therefore we can write the equation of straight line $b$ in an exact way like: $$(x,y)=({\displaystyle \frac{-2}{3}},{\displaystyle \frac{5}{3}})+k\cdot ({\displaystyle \frac{\sqrt{11+2\sqrt{10}}-3}{\sqrt{11+2\sqrt{10}}+3}})$$

Another rigorous method used to find solutions would be the following one:

We suppose that the straight line that we are looking for has the equation $Ax+By+C=0$.

As it crosses the point $(-2/3,5/3)$ we get:

$$-{\displaystyle \frac{2}{3}}A+{\displaystyle \frac{5}{3}}B+C=0$$

Now we know that

$$\cos (\widehat{r,s})=|\cos \widehat{(\overrightarrow{u},\overrightarrow{v})}|={\displaystyle \frac{|\overrightarrow{u}\cdot \overrightarrow{v}|}{|\overrightarrow{u}||\overrightarrow{v}|}}={\displaystyle \frac{|u_1\cdot v_1+u_2\cdot v_2|}{\sqrt{u_1^2+u_2^2}\sqrt{v_1^2+v_2^2}}}=$$ $$={\displaystyle \frac{|1\cdot 1+2\cdot (-1)|}{\sqrt{1^2+2^2}\sqrt{1^2+(-1{)}^2}}}={\displaystyle \frac{|-1|}{\sqrt{5}\sqrt{2}}}={\displaystyle \frac{1}{\sqrt{10}}}$$ or if we prefer $={\displaystyle \frac{\sqrt{10}}{10}}$.

and that, for trigonometry,

$$\cos ({\displaystyle \frac{\widehat{r,s}}{2}})=\sqrt{{\displaystyle \frac{1+\cos (\widehat{r,s})}{2}}}=\sqrt{{\displaystyle \frac{1+{\displaystyle \frac{1}{\sqrt{10}}}}{2}}}\sqrt{{\displaystyle \frac{{\displaystyle \frac{\sqrt{10}+1}{\sqrt{10}}}}{2}}}=\sqrt{{\displaystyle \frac{\sqrt{10}+1}{2\sqrt{10}}}}$$

We can impose the condition of equal angles on the bisector, knowing that the vector director of $b$ is $\hat{w}=(-B,A)$, (note: $|x|=\sqrt{x^2}$).

$$\cos (\widehat{r,s})={\displaystyle \frac{|u_1\cdot v_1+u_2\cdot v_2|}{\sqrt{u_1^2+u_2^2}\sqrt{v_1^2+v_2^2}}}$$ $$\sqrt{{\displaystyle \frac{\sqrt{10}+1}{2\sqrt{10}}}}={\displaystyle \frac{\sqrt{(-B+2A{)}^2}}{\sqrt{5}\sqrt{A^2+B^2}}}$$ $$\sqrt{{\displaystyle \frac{\sqrt{10}+1}{2\sqrt{10}}}}={\displaystyle \frac{\sqrt{(-B-A{)}^2}}{\sqrt{2}\sqrt{A^2+B^2}}}$$

And if we join the 3 equations and resolve:

$$-{\displaystyle \frac{2}{3}}A+{\displaystyle \frac{5}{3}}B+C=0$$ $$\sqrt{{\displaystyle \frac{\sqrt{10}+1}{2\sqrt{10}}}}={\displaystyle \frac{\sqrt{(-B+2A{)}^2}}{\sqrt{5}\sqrt{A^2+B^2}}}$$ $$\sqrt{{\displaystyle \frac{\sqrt{10}+1}{2\sqrt{10}}}}={\displaystyle \frac{\sqrt{(-B-A{)}^2}}{\sqrt{2}\sqrt{A^2+B^2}}}$$ $$\iff$$ $$C={\displaystyle \frac{2}{3}}A-{\displaystyle \frac{5}{3}}B$$ $${\displaystyle \frac{\sqrt{10}+1}{2\sqrt{10}}}={\displaystyle \frac{(-B+2A{)}^2}{5(A^2+B^2)}}$$ $${\displaystyle \frac{\sqrt{10}+1}{2\sqrt{10}}}={\displaystyle \frac{(-B-A{)}^2}{2(A^2+B^2)}}$$

We impose $B=-1$ (To check that we get the same result since the vector director would be like this $(1,m)$) and we look for a solution that fulfill all 3 equations: $$C={\displaystyle \frac{2}{3}}A-{\displaystyle \frac{5}{3}}(-1)$$ $${\displaystyle \frac{\sqrt{10}+1}{2\sqrt{10}}}={\displaystyle \frac{(1+2A{)}^2}{5(A^2+1)}}$$ $${\displaystyle \frac{\sqrt{10}+1}{2\sqrt{10}}}={\displaystyle \frac{(1-A{)}^2}{2(A^2+1)}}$$ We divide: $$1={\displaystyle \frac{2(1+2A{)}^2}{5(1-A{)}^2}}$$ $$5-10A+5A^2=2+8A+8A^2$$ $$3A^2+18A-3=0$$ $$A={\displaystyle \frac{-18\pm \sqrt{{18}^2+4\cdot 3\cdot 3}}{6}}=-3\pm {\displaystyle \frac{\sqrt{360}}{6}}$$

If we asset the solutions we have: $$A_1=-3+{\displaystyle \frac{\sqrt{360}}{6}}\simeq 0,16227766$$ $$A_2=-3-{\displaystyle \frac{\sqrt{360}}{6}}\simeq -6,16227766$$

Obviously we take the positive solution since as it is $-B=1$, we have $A=m=tg(\widehat{b,OX})$ that is from the first quadrant. Then $C={\displaystyle \frac{2}{3}}(-3+{\displaystyle \frac{\sqrt{360}}{6}})+{\displaystyle \frac{5}{3}}$ and without arrangements the straight line's equation stays like: $$(-3+{\displaystyle \frac{\sqrt{360}}{6}})x-y+({\displaystyle \frac{2}{3}}(-3+{\displaystyle \frac{\sqrt{360}}{6}})+{\displaystyle \frac{5}{3}})=0$$

Finally, another very elegant geometric procedure, in which it is not necessary to use trigonometry, would be emulating the geometric construction of the bisector.

We take director vectors of the straight lines $r$ and $s$: $$\overrightarrow{u}=(1,-1),\overrightarrow{v}=(1,2)$$ We make them unitary, that is to say, of module 1: $$\overrightarrow{u}=({\displaystyle \frac{1}{\sqrt{2}}},-{\displaystyle \frac{1}{\sqrt{2}}}),\overrightarrow{v}=({\displaystyle \frac{1}{\sqrt{5}}},{\displaystyle \frac{2}{\sqrt{5}}})$$ We apply them to the intersection point $P$ of straight lines $r$ and $s$, obtaining in this way equidistant points to the apex of the angle to bisect on the previous straight lines:

$$a=P+\overrightarrow{u}=(-{\displaystyle \frac{2}{3}},{\displaystyle \frac{5}{3}})+({\displaystyle \frac{1}{\sqrt{2}}},-{\displaystyle \frac{1}{\sqrt{2}}})=({\displaystyle \frac{-2\sqrt{2}+3}{3\sqrt{2}}},{\displaystyle \frac{5\sqrt{2}-3}{3\sqrt{2}}})$$ $$b=P+\overrightarrow{v}=(-{\displaystyle \frac{2}{3}},{\displaystyle \frac{5}{3}})+({\displaystyle \frac{1}{\sqrt{5}}},{\displaystyle \frac{2}{\sqrt{5}}})=({\displaystyle \frac{-2\sqrt{5}+3}{3\sqrt{5}}},{\displaystyle \frac{5\sqrt{5}+6}{3\sqrt{5}}})$$

If now we find the midpoint $M$ of the segment $ab$, we will have another point of the bisector $b$ ( with $P$) and, therefore, we will already be able to construct it: $$M=({\displaystyle \frac{a_1+b_1}{2}},{\displaystyle \frac{a_2+b_2}{2}})$$ $$M=({\displaystyle \frac{-2\sqrt{2}+3}{6\sqrt{2}}}+{\displaystyle \frac{-2\sqrt{5}+3}{6\sqrt{2}}},{\displaystyle \frac{5\sqrt{2}-3}{6\sqrt{2}}}+{\displaystyle \frac{5\sqrt{2}+6}{6\sqrt{5}}})=$$ $$=({\displaystyle \frac{3\sqrt{2}+3\sqrt{5}-4\sqrt{10}}{6\sqrt{10}}},{\displaystyle \frac{6\sqrt{2}-3\sqrt{5}+10\sqrt{10}}{6\sqrt{10}}})$$ Therefore, we already have two points of the straight line $b$, bisector of $r$ and $s$, so we can construct it with point $P$ and the vector $\overrightarrow{PM}$: $$\overrightarrow{PM}=({\displaystyle \frac{3\sqrt{2}+3\sqrt{5}-4\sqrt{10}}{6\sqrt{10}}},{\displaystyle \frac{6\sqrt{2}-3\sqrt{5}+10\sqrt{10}}{6\sqrt{10}}})-(-{\displaystyle \frac{2}{3}},{\displaystyle \frac{5}{3}})=$$ $$=({\displaystyle \frac{3\sqrt{2}+3\sqrt{5}}{6\sqrt{10}}},{\displaystyle \frac{6\sqrt{2}-3\sqrt{5}}{6\sqrt{10}}})$$

And in this way straight line $b$ is: $$b:(x,y)=P+k\cdot \overrightarrow{PM}=(-{\displaystyle \frac{2}{3}},{\displaystyle \frac{5}{3}})+k\cdot ({\displaystyle \frac{3\sqrt{2}+3\sqrt{5}}{6\sqrt{10}}},{\displaystyle \frac{6\sqrt{2}-3\sqrt{5}}{6\sqrt{10}}})$$

Solution:

Any of the following solutions is valid:

$b:(x,y)=(-2/3,5/3)+k\cdot (1,0.16228)$

$(x,y)=({\displaystyle \frac{-2}{3}},{\displaystyle \frac{5}{3}})+k\cdot ({\displaystyle \frac{\sqrt{11+2\sqrt{10}}-3}{\sqrt{11+2\sqrt{10}}+3}})$

$(x,y)=P+k\cdot \overrightarrow{PM}=(-{\displaystyle \frac{2}{3}},{\displaystyle \frac{5}{3}})+k\cdot ({\displaystyle \frac{3\sqrt{2}+3\sqrt{5}}{6\sqrt{10}}},{\displaystyle \frac{6\sqrt{2}-3\sqrt{5}}{6\sqrt{10}}})$

$(-3+{\displaystyle \frac{\sqrt{360}}{6}})x-y+({\displaystyle \frac{2}{3}}(-3+{\displaystyle \frac{\sqrt{360}}{6}})+{\displaystyle \frac{5}{3}})=0$

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