wo lines L1 and L2 has the equations 6y + 3x = 1 and 3y − 2x = 4 respectively. Calculate(a) the angle between L1 and L2(b) the distance between the point (2, 3) from the line L2(c) the distance between the point (1,1) from the line L1

To calculate the angle between two lines, L1 and L2, we can use the formula:

θ = arctan |(m1 - m2) / (1 + m1 * m2)|

where m1 and m2 are the slopes of L1 and L2, respectively.

(a) First, let's find the slopes of L1 and L2. The equation of L1 is 6y + 3x = 1. We can rewrite it in slope-intercept form (y = mx + b) by solving for y:

6y = -3x + 1 y = (-1/2)x + 1/6

So, the slope of L1 is -1/2.

Similarly, for L2, the equation is 3y - 2x = 4. Solving for y:

3y = 2x + 4 y = (2/3)x + 4/3

The slope of L2 is 2/3.

Now, we can substitute the slopes into the formula to find the angle:

θ = arctan |((-1/2) - (2/3)) / (1 + (-1/2) * (2/3))| = arctan |(-3/6) / (1 - 1/3)| = arctan |-1/2|

Using a calculator, we find that the angle between L1 and L2 is approximately 26.57 degrees.

(b) To find the distance between the point (2, 3) and the line L2, we can use the formula for the distance between a point (x1, y1) and a line Ax + By + C = 0:

d = |Ax1 + By1 + C| / sqrt(A^2 + B^2)

For L2, the equation is 3y - 2x = 4. Plugging in the values (x1, y1) = (2, 3):

d = |3(3) - 2(2) - 4| / sqrt(3^2 + (-2)^2) = |9 - 4 - 4| / sqrt(9 + 4) = |1| / sqrt(13) = 1 / sqrt(13)

So, the distance between the point (2, 3) and the line L2 is approximately 0.27 units.

(c) Similarly, to find the distance between the point (1, 1) and the line L1, we can use the same formula. For L1, the equation is 6y + 3x = 1. Plugging in the values (x1, y1) = (1, 1):

d = |6(1) + 3(1) - 1| / sqrt(6^2 + 3^2) = |6 + 3 - 1| / sqrt(36 + 9) = |8| / sqrt(45) = 8 / sqrt(45)

So, the distance between the point (1, 1) and the line L1 is approximately 1.06 units.