find the equations of the lines (in 2-space) bisecting the angles formed between the lines x+3y=10 & 3x+y=14?

Answer 1

First find the point of intersection (which both lines you want to find will pass through) by making the equations simultaneous and solving.
Usually finding the equation of angle bisectors is very difficult but in this case it is particularly easy as you can see from a diagram that the gradients will be 1 and -1.
Try to finish it from there.

Answer 2

Find the equations of the lines (in 2-space) bisecting the angles formed between the lines:

x + 3y = 10
3x + y = 14

The first thing we want to do is find the point of intersection of the given lines. Subtract three times the second equation from the first.

-8x = -32
x = 4

Plug the value for x back into the first equation.

x + 3y = 10
4 + 3y = 10
3y = 6
y = 2

The point of intersection is P(4, 2).

Now we want to find the slope of the lines that bisect the angles formed by the given lines. Remember the slope of a line is the tangent of the angle the line makes with the positive x-axis.

Let
tanα = slope of first line
tanβ = slope of the second line

tanα = -1/3
tanβ = -3/1 = -3

The slope of one of the bisecting lines will be the average of these. First find tan(α + β). Use the formula for angle addition.

tan(α + β) = [tanα + tanβ] / [1 - (tanα)(tanβ)]
tan(α + β) = [-1/3 - 3] / [1 - (-1/3)(-3)] = (-10/3)/(1 - 1)
tan(α + β) = (-10/3) / 0 is undefined

α + β = π/2
(α + β)/2 = π/4

tan[(α + β)/2] = tan(π/4) = 1

With the slope and a point on the line P(4, 2) we can write the equation of one of the bisecting lines.

y - 2 = 1(x - 4) = x - 4
y = x - 2

The second bisecting line will be perpendicular to the first so its slope m, is the negative reciprocal.

m = -1/1 = -1

With the slope and a point on the line P(4, 2) we can write the equation of the other bisecting line.

y - 2 = -1(x - 4) = -x + 4
y = -x + 6

Mathsmanretired is right about the slopes of the two bisecting lines in this particular case. The methods outlined above will always work. And I don't consider them to be "very difficult".

Answer 3

This bisecting line will bypass in the process the intersection factor of the two lines Intersection factor: 2 = SQRT(3)x x = 2SQRT(3)/3 and y = 2 There are diverse common approaches to do this yet enable us to locate the attitude between the lines and take a million/2 of it. Use this to verify yet another factor on the bisector line. tan(attitude) = SQRT(3) .. tangent of the attitude is in basic terms the slope of the line attitude = 60 levels So we want 30 levels which bisects the attitude 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 attitude is: y = [SQRT(3)/3]x + 4/3