he straight lines 2x + 3y = 6 and 3x – 2y = 12 are reflected in the lines y = x and y = –x respectively, to obtain lines L1 and L2 respectively. Find the acute angle (in degrees) between the lines L1 and L2.

The given lines are 2x + 3y = 6 and 3x - 2y = 12.

Step 1: Convert these lines into slope-intercept form (y = mx + c), where m is the slope of the line.

The first line becomes y = (-2/3)x + 2 and the second line becomes y = (3/2)x - 6.

Step 2: Find the slopes of the reflected lines.

The slope of a line reflected in the line y = x is -1/m (the negative reciprocal of the original slope).

So, the slope of the line reflected in y = x is -1/(-2/3) = 3/2.

The slope of a line reflected in the line y = -x is -m (the negative of the original slope).

So, the slope of the line reflected in y = -x is -(3/2) = -3/2.

Step 3: Find the angle between the reflected lines.

The formula to find the angle θ between two lines with slopes m1 and m2 is given by:

tan θ = |(m1 - m2) / (1 + m1*m2)|

Substitute m1 = 3/2 and m2 = -3/2 into the formula:

tan θ = |(3/2 - (-3/2)) / (1 + (3/2)*(-3/2))| = |3 / -1| = -3

So, θ = arctan(-3)

However, since we want the acute angle, we take the absolute value of θ, which is arctan(3).

Using a calculator, arctan(3) is approximately 71.57 degrees.

So, the acute angle between the lines L1 and L2 is approximately 71.57 degrees.