Posted by **vaughn** on Monday, January 7, 2013 at 1:46pm.

Find the bisectors of the interior angles of the triangle whose sides are the line 7x+y-7=0, x+y+1=0 and x+7y-4=0.

Please I really need help. I do not know how to do this problem

- geometry -
**Steve**, Monday, January 7, 2013 at 2:14pm
It really helps to draw a diagram, so you know which three angles you're dealing with. Label the three lines L1,L2,L3 in order.

First, get the slope of each line:

L1: -7

L2: -1

L3: -1/7

Now convert those slopes to angles, measured counter-clockwise from the x- axis:

L1: θ = 98.13°

L2: θ = 135.00°

L3: θ = 171.87°

Now, the angle bisector is the average of the two angles made by the lines:

B1: (L1+L2)/2 = 116.57° tan = -2

B2: (L2+L3)/2 = 153.43° tan = -1/2

B3: (L1+L3)/2 = -135.00° tan = 1

Now you need the intersections of the lines:

P1(L1=L2): (4/3, -7/3)

P2(L2=L3): (-11/6, 5/6)

P3(L1=L3): (15/16, 7/16)

Now we have a point-slope for each angle bisector:

B1: (y+7/3) = -2(x-4/3)

B2: (y-5/6) = -1/2 (x+11/6)

B3: (y-7/16) = 1(x-15/16)

or,

B1: 6x+3y-1 = 0

B2: 6x+12y+1 = 0

B3: 2x-2y-1 = 0

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