Find an equation of the line that bisects the acute angle formed by two lines...?

Question

Find the equation of the line that bisects the acute angle formed by the lines y = (sq. root 3)x and y = 2.

Answer 1

if you have y = mx + b

then tanA = m

where A is the angle formed by the horizontal.

y = 2 is a horizontal line

y = √3 x .. . . is a line where tanA = √3

thus A = 60°

the line bisecting it will have an angle of 30° with the horizontal.

thus its corresponding slope will be tan(30°) = (1/√3)

to get the point of intesection

y = 2

2 = √3 x

x = 2 / √3

thus

y = (1/√3)(x - 2 / √3) + 2

y = (1/√3)x - 2/3 + 2

y = (1/√3)x + 4/3

Answer 2

This bisecting line will go through the intersection point of the two lines

Intersection point: 2 = SQRT(3)x

x = 2SQRT(3)/3 and y = 2

There are different ways to do this but let us find the angle between the lines and take half of it. Use this to determine another point on the bisector line.

tan(Angle) = SQRT(3) .. tangent of the angle is just the slope of the line

Angle = 60 degrees

So we want 30 degrees which bisects the angle

So slope of bisector = tan(30) = SQRT(3)/3 = m

The line is:

y = mx + b

y = [SQRT(3)/3]x + b

2 = [SQRT(3)/3][2SQRT(3)/3] + b

2 = 2/3 + b

b = 4/3

The line that bisects the angle is:

y = [SQRT(3)/3]x + 4/3

Answer 3

y = √3.x

y = 2

Point of intersection: (2/√3, 2)

Slope of y=2 is 0

This line makes an angle of 0° with the x-axis

Slope of y=√3.x is √3

This line makes an angle of tanֿ¹(√3) = 60° with the x-axis

The acute angle between these lines is 60°

The bisector will make an angle of 30° with either of these lines (and also with the x-axis)

The slope of the bisector will be tan(30°) = 1/√3

y = x/√3 + c

It passes through (2/√3, 2)

2 = 2/3 + c

c = 4/3

y = x/√3 + 4/3

3y = √3x + 4

√3x - 3y + 4 = 0

The requried equation is

√3x - 3y + 4 = 0