# Bisector of an acute angle Q2 (Help)

• June 27th 2009, 12:13 AM
mj.alawami
Bisector of an acute angle Q2 (Help)
[FONT="Arial"]Find the equation of the bisector of the acute angle whose equation is
$4x-5y-10$ and $x-4y+6$

Attempt

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

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

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

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

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

Is my answer correct?
Thank you.
• June 27th 2009, 04:13 AM
earboth
Quote:

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

Attempt

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

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

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

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

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

Is my answer correct?
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:

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

I've attached the graph of my solution.
• June 27th 2009, 04:16 AM
mj.alawami
Quote:

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:

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

I've attached the graph of my solution.

So is my answer wrong?
• June 27th 2009, 04:35 AM
earboth
Quote:

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

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

...

I take this equation:

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

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

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

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

$
-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}

$

$
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.
• June 27th 2009, 04:39 AM
Soroban
Hello, mj.alawami!

Your game plan is correct,
but you have some incorrect minus-signs.

Quote:

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

Attempt

$L_1 \:=\:\frac{4x-5y-10}{{\color{red}-}\sqrt{41}}\qquad L_2\:=\:\frac{x'-4y'+6}{{\color{red}-}\sqrt{17}}$ . ??
There is no minus-sign in the denominator.

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

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

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

Is my answer correct? .
. . . Sorry, no

But you were close . . .

Try it again without those extra minus-signs.

• June 27th 2009, 07:28 AM
pankaj
Here is a general method which I have been using.
Let the two lines be
$a_{1}x+b_{1}y+c_{1}=0$
$a_{2}x+b_{2}y+c_{2}=0$

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

The eqution of the bisectors are

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

i.e.
$
\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

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

i.e.
$
\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

$
\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) $
\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, $c_{1}>0,c_{2}>0$

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

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