What is the angle between the lines 2x + 3y - 6 = 0 and 4x - y + 1 = 0?

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Q: What is the angle between the lines 2x + 3y - 6 = 0 and 4x - y + 1 = 0?

Solution: The slopes of the lines are -2/3 and 4. The angle θ can be found using tan(θ) = |(m1 - m2) / (1 + m1*m2)|.

Steps: 9

Step 1: Write down the equations of the lines: Line 1 is 2x + 3y - 6 = 0 and Line 2 is 4x - y + 1 = 0.
Step 2: Rearrange Line 1 into slope-intercept form (y = mx + b). Start with 3y = -2x + 6, then divide by 3 to get y = -2/3 x + 2. The slope (m1) of Line 1 is -2/3.
Step 3: Rearrange Line 2 into slope-intercept form. Start with -y = -4x - 1, then multiply by -1 to get y = 4x + 1. The slope (m2) of Line 2 is 4.
Step 4: Use the formula for the angle θ between two lines: tan(θ) = |(m1 - m2) / (1 + m1*m2)|.
Step 5: Substitute m1 and m2 into the formula: tan(θ) = |(-2/3 - 4) / (1 + (-2/3)*4)|.
Step 6: Calculate the numerator: -2/3 - 4 = -2/3 - 12/3 = -14/3. So, the numerator is |-14/3| = 14/3.
Step 7: Calculate the denominator: 1 + (-2/3)*4 = 1 - 8/3 = 3/3 - 8/3 = -5/3. So, the denominator is -5/3.
Step 8: Now, tan(θ) = (14/3) / (-5/3) = -14/5. Since we want the absolute value, tan(θ) = 14/5.
Step 9: To find θ, use the arctan function: θ = arctan(14/5).