Discussion
The distance between the parallel lines \(3x + 4y = 9\) and \(6x + 8y = 15\) is:
Answer: A
To find the distance, the coefficients of x and y in both equations must be the same. Let's rewrite the second equation by dividing by 2: \(3x + 4y = 7.5\).
Now we have two lines in the form Ax + By + C₁ = 0 and Ax + By + C₂ = 0.
Line 1: 3x + 4y - 9 = 0
Line 2: 3x + 4y - 7.5 = 0
The distance is given by \(\frac{|C_1 - C_2|}{\sqrt{A^2+B^2}}\).
Distance = \(\frac{|-9 - (-7.5)|}{\sqrt{3^2+4^2}} = \frac{|-1.5|}{\sqrt{25}} = \frac{1.5}{5} = \frac{3/2}{5} = \frac{3}{10}\).
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