What is the distance from the point (1, 2) to the line 3x + 4y - 12 = 0?

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Question: What is the distance from the point (1, 2) to the line 3x + 4y - 12 = 0?

Options:

Correct Answer: 3

Solution:

Distance = |Ax1 + By1 + C| / √(A² + B²) = |3(1) + 4(2) - 12| / √(3² + 4²) = |3 + 8 - 12| / 5 = | -1 | / 5 = 1/5.

What is the distance from the point (1, 2) to the line 3x + 4y - 12 = 0?

What is the distance from the point (1, 2) to the line 3x + 4y - 12 = 0?

Step 1: Identify the point and the line. The point is (1, 2) and the line is given by the equation 3x + 4y - 12 = 0.
Step 2: Rewrite the line equation in the form Ax + By + C = 0. Here, A = 3, B = 4, and C = -12.
Step 3: Use the distance formula from a point (x1, y1) to a line Ax + By + C = 0, which is Distance = |Ax1 + By1 + C| / √(A² + B²).
Step 4: Substitute the values into the formula. Here, x1 = 1, y1 = 2, A = 3, B = 4, and C = -12.
Step 5: Calculate the numerator: |3(1) + 4(2) - 12| = |3 + 8 - 12| = |-1| = 1.
Step 6: Calculate the denominator: √(A² + B²) = √(3² + 4²) = √(9 + 16) = √25 = 5.
Step 7: Divide the numerator by the denominator: Distance = 1 / 5.

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