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:
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
To find the bisectors of the interior angles of a triangle, we need to find the equations of the bisectors, which are lines that divide an angle into two equal angles.
Let's start by finding the vertices of the triangle corresponding to the given lines. To do that, we need to rewrite each line equation in the slope-intercept form (y = mx + b).
Line 1: 7x + y - 7 = 0
Rewrite it as: y = -7x + 7
Line 2: x + y + 1 = 0
Rewrite it as: y = -x - 1
Line 3: x + 7y - 4 = 0
Rewrite it as: y = -(1/7)x + 4/7
Now, we can find the intersection points of these lines by solving the system of equations formed by them.
To find the intersection of lines 1 and 2:
Substitute the value of y from line 2 into line 1:
-7x + 7 = -x - 1
Simplify and solve for x:
-6x = -8
x = 8/6
x = 4/3
Substitute the value of x back into line 2 equation:
y = -(4/3) - 1
y = -7/3
So, the intersection point of lines 1 and 2 is (4/3, -7/3).
Similarly, we can find the intersection points of lines 2 and 3:
Substitute the value of y from line 3 into line 2:
-(1/7)x + 4/7 = -x - 1
Multiply through by 7 to clear the fraction:
-x + 4 = -7x - 7
Simplify and solve for x:
6x = 11
x = 11/6
Substitute the value of x back into line 3 equation:
y = -(1/7)(11/6) + 4/7
y = -5/6
So, the intersection point of lines 2 and 3 is (11/6, -5/6).
Next, we can find the intersection points of lines 1 and 3:
Substitute the value of y from line 3 into line 1:
7x + (-(1/7)x + 4/7) - 7 = 0
6x = 6
x = 1
Substitute the value of x back into line 3 equation:
y = -(1/7)(1) + 4/7
y = 1/7
So, the intersection point of lines 1 and 3 is (1, 1/7).
Now that we have the vertices of the triangle, we can find the equations of the bisectors of the interior angles.
To find the bisector of the angle formed at vertex (4/3, -7/3):
First, find the slope of the line connecting this vertex to the origin.
Slope = -7/3 divided by 4/3 = -7/4
Next, find the equation of the line using the point-slope form:
y - y1 = m(x - x1), where (x1, y1) is the vertex point.
Using the values, we have:
y - (-7/3) = (-7/4)(x - 4/3)
Simplifying, we get:
4y + 28 = -7x + 7
Rearranging the equation, we get:
7x + 4y - 21 = 0
Similarly, you can find the other two bisectors for the remaining angles formed by the given vertices.
The bisector of the angle formed at vertex (11/6, -5/6) is given by:
5x + 11y + 1 = 0
The bisector of the angle formed at vertex (1, 1/7) is given by:
105x - 56y - 34 = 0
These are the equations of the bisectors of the interior angles of the given triangle.