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Find the equations of the bisectors of the angles between the lines $y=x$ and $y=7x+4$. Identify the bisectpr of the acute angle.

My Attempt: I have solved the first part of the question and got the equations of bisectors as: $$x+2y+2=0$$ $$6x-3y+2=0$$

I couldn't solve the second part of the question.

nonuser
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pi-π
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    Hint : The angle between a line and the x.axis can be determined via $\tan(\alpha)=m$, if the line has the equation $y=mx+b$ – Peter Aug 24 '17 at 12:30
  • All is done, you got the bisectors, draw it. Acute angle bisector has positive slope, cf.ajotatxe.Bisector of obtuse angle has negative slope, they are perpendicular to each other: m_1 = - (1/-m_2); m_1 = -1/2, m_2 = 2, so all is well :)). – Peter Szilas Aug 24 '17 at 14:08

5 Answers5

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The bisector of the acute angles between two lines with positive slope has positive slope.

A line with positive slope and the positive $X$ axis form an acute angle. Then, if two lines with positive slopes $m$ and $m'$ (with $m>m'$) form angles with the positive $X$ axis $\alpha$ and $\beta$ with $\alpha>\beta$, the acute angle between both lines is $\alpha-\beta$.

Now, the slope $m''$ of the bisector of this angle must meet the inequality $m'<m''<m$, so $m''>0$.

Since the bisectors that you have found have slopes of opposite signs (as it must be, since they are perpendicular), only one of them can be the bisector of the acute angle.

ajotatxe
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I like the following way.

Let $A\left(-\frac{2}{3},-\frac{2}{3}\right)$, $B(0,4)$ and $C(1,1)$.

We knew that $A$ is a common point of our lines,

$B$ is placed on $y=7x+4$ and $C$ is placed on $y=x$.

It's obvious that $\vec{AB}\cdot\vec{AC}>0$, which says $\measuredangle BAC<90^{\circ}$.

Now we can calculate $AB$, $AC$ and we can find $D\in BC$ such that $$BD:DC=AB:AC.$$

Thus, the line $AD$ is our bisector.

Finally I got $AB:AC=2:1$, $D\left(\frac{2}{3},2\right)$ and the eqution of $AD$ it's: $$y=2x+\frac{2}{3}.$$

Michael Rozenberg
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Let $T(x,y)$ be on angle bisector of given lines. Theb $$d(T,p) = d(T,q)$$so

$${|x-y|\over \sqrt{2}} = {|-7x+y-4|\over \sqrt{50}}$$

so we have $5x-5y = \pm (7x-y+4)$, thus first a.b. is $x+2y+2=0$ and the second $6x-3y+2=0$. Now calculate angle between $y=x$ and $6x-3y+2=0$:

$$ \tan \alpha = {2-1\over 1+2}={1\over 3}< {\sqrt{3}\over 3} \Longrightarrow \alpha < 30^{\circ} $$ so $6x-3y+2=0$ is a.b. of acute angle.

nonuser
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The lines intersect at the point: $$A(-\frac23,-\frac23)$$ The lines have $y$-intercepts at the points: $$O(0,0), B(0,4)$$ Let $C(0,t)$ be the foot of the bisector of the triangle $ABO$ from the vertex $A$. Then the property of bisector is: $$\frac{AB}{BC}=\frac{AO}{OC} \Rightarrow \frac{\sqrt{\left(4+\frac23 \right)^2+\left(\frac23 \right)^2}}{4-t}=\frac{\sqrt{\left(\frac23 \right)^2+\left(\frac23 \right)^2}}{t} \Rightarrow$$ $$t=\frac23.$$ The bisector line passing through $A(-\frac23, -\frac23)$ and $C(0,\frac23)$ is: $$\frac{y+\frac23}{\frac23+\frac23}=\frac{x+\frac23}{0+\frac23} \Rightarrow y=2x+\frac23.$$

farruhota
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Per this question, if you rearrange the two equations into the form $ax+by+c=0$ and $px+qy+r=0$, then the acute angle bisector of the two lines is given by the equation $${ax+by+c\over\sqrt{a^2+b^2}}=-\operatorname{signum}(ap+bq){px+qy+r\over\sqrt{p^2+q^2}}.$$

In this case, we have the equations $x-y=0$ and $7x-y+4=0$, from which $1\cdot7+(-1)\cdot(-1)=8\gt0$. The acute bisector is therefore $${x-y\over\sqrt2}=-{7x-y+4\over5\sqrt2}$$ which can be rearranged into $$\frac{\sqrt2}5(6x-3y+2)=0,$$ i.e., it’s the second of the equations that you computed.

amd
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