Theta is an angle between the lines L1 and L2 with the slopes m1 and m2. Prove that tan theta= (m2 - m1)/(1+m2m1).

tan slope angle 1 = m1

tan slope angle 2 = m2
so
angle 1 = tan^-1 m1
angle 2 = tan^-1 m2

so
theta = angle 2 - angle 1
= tan^-1 m2 - tan^-1 m1

tan theta = tan (a - b)
where a =tan^-1 m2 and b =tan^-1 m1
so
tan theta = (tan a-tan b)/(1+tan a tan b) [trig identity]

= (m2 -m1)/(1 + m2 m1)

To prove that tan(θ) = (m2 - m1)/(1 + m2*m1) in the given scenario, we need to understand the relationship between the slopes of the lines and the tangent of the angle between them.

Let's start by finding the slopes of the lines L1 and L2.

The slope of a line can be determined using the formula:

m = (y2 - y1)/(x2 - x1)

Let's assume that L1 passes through the points (x1, y1) and (x2, y2). Similarly, assume that L2 passes through the points (x1', y1') and (x2', y2').

The slope of L1 is given by:

m1 = (y2 - y1)/(x2 - x1)

The slope of L2 is given by:

m2 = (y2' - y1')/(x2' - x1')

Now, let's find the angle θ between L1 and L2. We can use the angle between two lines formula, which states:

tan(θ) = |(m2 - m1) / (1 + m2*m1)|

The absolute value is taken to ensure that the result is positive, regardless of the orientation of the lines.

Substituting the values of m1 and m2, we get:

tan(θ) = |((y2' - y1')/(x2' - x1')) - ((y2 - y1)/(x2 - x1))| / |1 + ((y2' - y1')/(x2' - x1')) * ((y2 - y1)/(x2 - x1))|

Now, we can simplify this expression further by multiplying through by the denominators:

tan(θ) = |((y2' - y1')(x2 - x1) - (y2 - y1)(x2' - x1')) / ((x2' - x1')(x2 - x1) + (y2' - y1')(y2 - y1))|

Notice that the numerator (y2' - y1')(x2 - x1) - (y2 - y1)(x2' - x1') represents the difference in the y-coordinate and the x-coordinate between the two lines. Similarly, the denominator (x2' - x1')(x2 - x1) + (y2' - y1')(y2 - y1) represents the sum of the products of the differences in the x-coordinate and the y-coordinate between the two lines.

Let's denote the differences as ∆x = x2 - x1, ∆x' = x2' - x1', ∆y = y2 - y1, and ∆y' = y2' - y1'.

Now, we can simplify the expression further:

tan(θ) = |(∆y' * ∆x - ∆y * ∆x') / (∆x' * ∆x + ∆y' * ∆y)|

Finally, we observe that (∆y' * ∆x - ∆y * ∆x') is the determinant of the 2x2 matrix:

| ∆x -∆x' |
| ∆y -∆y' |

Similarly, (∆x' * ∆x + ∆y' * ∆y) is the dot product of the vectors (∆x', ∆y') and (∆x, ∆y).

Using these observations, we can rewrite the expression as:

tan(θ) = |det(A) / dot(A, B)|

Now, if we consider the angle θ to be acute or obtuse, we find that the absolute value of the determinant and the dot product is the same (with opposite signs). Hence, the absolute value in the expression can be eliminated:

tan(θ) = det(A) / dot(A, B)

This gives us tan(θ) = (m2 - m1)/(1 + m2*m1), which is the desired result.

Thus, we have proven that tan(θ) = (m2 - m1)/(1 + m2*m1) in the given scenario involving the slopes of the lines L1 and L2.