# Thread: Bisector of an acute angle Q2 (Help)

1. ## Bisector of an acute angle Q2 (Help)

[FONT="Arial"]Find the equation of the bisector of the acute angle whose equation is
$\displaystyle 4x-5y-10$ and $\displaystyle x-4y+6$

Attempt

L1=$\displaystyle \frac{4x'-5y'-10}{-\sqrt{41}}$

L2=$\displaystyle \frac{x'-4y'+6}{-\sqrt{17}}$

Since, the bisect line is in the middle of the lines then we use the formula $\displaystyle \boxed{ -L2=L1 }$

$\displaystyle -[-\frac{x-4y+6}{\sqrt{17}}]=-[\frac{4x-5y-10}{\sqrt{41}}]$

=$\displaystyle \boxed{(\sqrt{41}-4\sqrt{17})x+(-4\sqrt{41}-5\sqrt{17})y+(6\sqrt{41}-10\sqrt{17})=0}$

Thank you.

2. Originally Posted by mj.alawami
[FONT="Arial"]Find the equation of the bisector of the acute angle whose equation is
$\displaystyle 4x-5y-10 \bold{\color{red}= 0}$ and $\displaystyle x-4y+6\bold{\color{red}= 0}$ <<<<<< I assume that you mean these equations

Attempt

L1=$\displaystyle \frac{4x'-5y'-10}{-\sqrt{41}}$

L2=$\displaystyle \frac{x'-4y'+6}{-\sqrt{17}}$

Since, the bisect line is in the middle of the lines then we use the formula $\displaystyle \boxed{ -L2=L1 }$

$\displaystyle -[-\frac{x-4y+6}{\sqrt{17}}]=-[\frac{4x-5y-10}{\sqrt{41}}]$

=$\displaystyle \boxed{(\sqrt{41}-4\sqrt{17})x+(-4\sqrt{41}-5\sqrt{17})y+(6\sqrt{41}-10\sqrt{17})=0}$

Thank you.
I've drawn the three lines and found out that your bisector doesn't pass through the point of intersection.

I've got as the bisector the line with the following equation:

$\displaystyle \left(\dfrac{4 \cdot \sqrt{41}}{41} + \dfrac{\sqrt{17}}{17} \right)·x - \left(\dfrac{5 \cdot \sqrt{41}}{41} + \dfrac{4 \cdot \sqrt{17}}{17}\right)·y + \left(\dfrac{6\cdot \sqrt{17}}{17} - \dfrac{10 \cdot \sqrt{41}}{41} \right) = 0$

I've attached the graph of my solution.

3. Originally Posted by earboth
I've drawn the three lines and found out that your bisector doesn't pass through the point of intersection.

I've got as the bisector the line with the following equation:

$\displaystyle \left(\dfrac{4 \cdot \sqrt{41}}{41} + \dfrac{\sqrt{17}}{17} \right)·x - \left(\dfrac{5 \cdot \sqrt{41}}{41} + \dfrac{4 \cdot \sqrt{17}}{17}\right)·y + \left(\dfrac{6\cdot \sqrt{17}}{17} - \dfrac{10 \cdot \sqrt{41}}{41} \right) = 0$

I've attached the graph of my solution.

4. Originally Posted by mj.alawami
[FONT="Arial"]...

$\displaystyle -[-\frac{x-4y+6}{\sqrt{17}}]=-[\frac{4x-5y-10}{\sqrt{41}}]$

...
I take this equation:

$\displaystyle -\left[-\frac{x-4y+6}{\sqrt{17}}\right]=-\left[\frac{4x-5y-10}{\sqrt{41}}\right]$

$\displaystyle \left[-\frac{x-4y+6}{\sqrt{17}}\right]=\left[\frac{4x-5y-10}{\sqrt{41}}\right]$

$\displaystyle -(x-4y+6) \cdot \sqrt{41}=(4x-5y-10) \cdot \sqrt{17}$

$\displaystyle (-x+4y-6) \cdot \sqrt{41}=(4x-5y-10) \cdot \sqrt{17}$

$\displaystyle -x\cdot \sqrt{41}+4\cdot \sqrt{41}y-6\cdot \sqrt{41}=4\cdot \sqrt{17} x-5\cdot \sqrt{17} y-10\cdot \sqrt{17}$

$\displaystyle 0=(4\cdot \sqrt{17} + \sqrt{41})x-(5\cdot \sqrt{17}+4\cdot \sqrt{41})y+(6\cdot \sqrt{41}-10\cdot \sqrt{17})$

By comparison you'll see that you have made a tiny sign mistake in the first bracket.

5. Hello, mj.alawami!

but you have some incorrect minus-signs.

Find the equation of the bisector of the acute angle whose equation is:
. . $\displaystyle 4x-5y-10 \:=\:0$ and $\displaystyle x-4y+6 \:=\:0$

Attempt

$\displaystyle L_1 \:=\:\frac{4x-5y-10}{{\color{red}-}\sqrt{41}}\qquad L_2\:=\:\frac{x'-4y'+6}{{\color{red}-}\sqrt{17}}$ . ??

Since, the bisect line is in the middle of the lines,
then we use the formula: $\displaystyle {\color{red}-}L2\,=\,L1$ . ??
The distances are of opposite signs?

$\displaystyle -\left[-\frac{x-4y+6}{\sqrt{17}}\right]\;=\;-\left[\frac{4x-5y-10}{\sqrt{41}}\right]$

=$\displaystyle \boxed{(\sqrt{41}-4\sqrt{17})x+(-4\sqrt{41}-5\sqrt{17})y+(6\sqrt{41}-10\sqrt{17})=0}$

. . . Sorry, no

But you were close . . .

Try it again without those extra minus-signs.

6. Here is a general method which I have been using.
Let the two lines be
$\displaystyle a_{1}x+b_{1}y+c_{1}=0$
$\displaystyle a_{2}x+b_{2}y+c_{2}=0$

Remember to write the two equations in such a way that $\displaystyle c_{1}>0$ and $\displaystyle c_{2}>0$.

The eqution of the bisectors are

$\displaystyle \frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}} =\pm \frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}$

If $\displaystyle a_{1}a_{2}+b_{1}b_{2}>0$,then use of $\displaystyle +$ sign in above equation will give OBTUSE angle bisector

i.e.
$\displaystyle \frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}} =+ \frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}$

is eqution of OBTUSE angle bisector and

$\displaystyle \frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}} =- \frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}$
gives equation of ACUTE angle bisector
.................................................. .................................................. ..
If $\displaystyle a_{1}a_{2}+b_{1}b_{2}<0$,then use of $\displaystyle -$ sign in above equation will give OBTUSE angle bisector

i.e.
$\displaystyle \frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}} =- \frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}$

is eqution of OBTUSE angle bisector and

$\displaystyle \frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}} =+ \frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}$
is equation of ACUTE angle bisector
.................................................. .............................................

General rule
(i)$\displaystyle \frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}} =\pm \frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}$
where,$\displaystyle c_{1}>0,c_{2}>0$

(ii)Find sign of $\displaystyle a_{1}a_{2}+b_{1}b_{2}.$

(iii)The use of this sign always gives the OBTUSE angle bisector and the use of other sign gives ACUTE angle bisector