Find the equation of angular Bisector of the angle formed by the lines 12x+5y=10 and 3x+4y=5 Containing (1,1).

Answer 1

Those two equations intersect at:

12x + 16y = 20 (second eqn times 4)
12x + 5y = 10
11 y = 10
y = 10/11

3x + 40/11 = 55/11
x = 5/11

...at (5/11, 10/11)

Oops -- think I figured out what you are asking...
The angle between those two lines can be found by using some inverse tangents. arctan(rise/run) = standard-position angle of a line.

The first line is at a standard-position angle of arctan(-12/5) = -67.38 or 112.62 degrees while the second is at a standard-position angle of arctan(-4/3) = -53.13 degrees.

Split the difference: (112.62 - 53.13)/2 = 29.745 degrees
tan(29.745) = .57 will be the slope of the angle bisector.

Now use point-slope form based on the intersection:
y = .57 (x - 5/11) + 10/11

Answer 2

Find the equation of angular Bisector of the angle formed by the lines 12x+5y=10 and 3x+4y=5 Containing (1,1).

First find the intersection of the lines.

12x + 5y = 10
3x + 4y = 5

Subtract the first equation from four times the second.

11y = 10
y = 10/11

Plug the value for y back into one of the equations. Let's use the second equation.

3x + 4y = 5
3x + 4(10/11) = 5
3x = 5 - 40/11 = (55 - 40)/11 = 15/11
x = 5/11

The intersection of the two lines is the point
P(x,y) = P(5/11, 10/11)

Now find the equation of the line thru P and (1,1).

Calculate the slope m.

m = ∆y/∆x = (1 - 10/11) / (1 - 5/11) = (1/11) / (6/11) = 1/6

Now write the point slope equation of the line and then rearrange terms.

y - 1 = (1/6)(x - 1)

6y - 6 = x - 1
x - 6y = -5

This line passes thru the intersection of the two given lines but does not bisect the angles formed by them. There are two lines that bisect the two angles formed by the two given lines. One bisects the acute angle and the other bisects the obtuse angle. Neither bisector passes thru the point (1,1).
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Let's calculate the slope of the lines that bisect the angles formed by the two given lines.

First calculate the slope of the two given lines.

12x + 5y = 10
3x + 4y = 5

m = -12/5
m' = -3/4

Express the slope in terms of the tangent of the angle.

tanα = m = -12/5
α = arctan(-12/5)

tanβ = m = -3/4
β = arctan(-3/4)

The bisector will have a slope that is the average of the two given lines.

γ = (α + β)/2

tan(α + β) = (tanα + tanβ) / (1 - tanα*tanβ)
tan(α + β) = (-12/5 - 3/4) / [1 - (-12/5)(-3/4)]
tan(α + β) = (- 63/20) / (1 - 9/5) = (63/20) / (4/5) = 63/16

Now we need to calculate cos(α + β) so we can use a half-angle formula for tanγ.

sin²(α + β) + cos²(α + β) = 63²/h² + (-16)²/h² = 1
63² + (-16)² = h²
h² = 3969 + 256 = 4225
h = 65

cos(α + β) = -16/65

So the slope of the first line that bisects the intersection of the given lines is:

tanγ = tan[(α + β)/2] = -√[(1 - cos(α + β)) / (1 + cos(α + β))]

tanγ = -√[(1 + 16/65) / (1 - 16/65)] = -√(81/49) = -9/7

The point slope equation of the bisector is:

y - 10/11 = (-9/7)(x - 5/11) = (-9/7)x + 45/77

y = (-9/7)x + 115/77
or
9x + 7y = 115/11

The second bisector is at right angles to this line.

m' = -1/m = -1/(-9/7) = 7/9

The point slope equation of the second bisector is:

y - 10/11 = (7/9)(x - 5/11) = (7/9)x - 35/99

y = (7/9)x + 5/9
or
7x - 9y = -5

The equations of the two lines that bisect the two angles formed by the two given lines are:

9x + 7y = 115/11
7x - 9y = -5