Distance from a point to a line?

**homeylova223** · March 20th 2011, 09:19 AM

Find the equation of the lines that bisect the acute and obtuse angles formed by the lines with the given equations.

y=2/3x+1
y=-3x-2

Know I used the formula d= Ax+By+C/square root A^2+B^2
To get square root for these functions

So I got (2/3x-1y+1)/(-1.20) and (-3x-1y-2)/(3.162)

Then when I cross multiply these my answer come up wrong. Can anyone please help me?

**skeeter** · March 20th 2011, 12:34 PM

$y = \dfrac{2}{3} x + 1$

$3y = 2x + 3$

$2x - 3y + 3 = 0$

$\dfrac{2x - 3y + 3}{\sqrt{13}}$

$y = -3x - 2$

$3x + y + 2 = 0$

$\dfrac{3x+y+2}{\sqrt{10}}$

$\dfrac{3x+y+2}{\sqrt{10}} - \dfrac{2x - 3y + 3}{\sqrt{13}} = 0$

$\left(\dfrac{3}{\sqrt{10}} - \dfrac{2}{\sqrt{13}}\right)x + \left(\dfrac{1}{\sqrt{10}} + \dfrac{3}{\sqrt{13}}\right)y + \left(\dfrac{2}{\sqrt{10}} - \dfrac{3}{\sqrt{13}}\right) = 0$

the above line bisects the obtuse angle ... the bisector of the acute angle will be perpendicular to the above line.

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