posted by vaughn on .
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
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:
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)
B1: 6x+3y-1 = 0
B2: 6x+12y+1 = 0
B3: 2x-2y-1 = 0