Is there a simple way to explain how to find the distance between the parallel lines of y=2x + 4 and y=2x + 1?
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ANSWERS: 2
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2x+1 has a Y-intercept of 1 while the 2x+4 has a Y-intercept of 4. Therefore the 2x+1 has a point at (0,1) and 2x+4 has a point at (0,4). To find the vertical distance, subtract Y-values (one from each equation) that have the same X-value. In this problem 4-1=3
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I assume you mean the distance between the parallel lines measured along a perpendicular. For a line with a slope of m, the perpendicular has a slope of -1/m. So in this case a perpendicular line will have the form y = (-1/2)x + b. It would be nice to have a perpendicular going through the point (0,4), because that would form a right triangle with a hypotenuse of 3. (Draw a diagram). So you get 4 = (-1/2)*0 + b So b=4 so the desired perpendicular is given by y=-1/2x + 4. Where does this intersect the other line? y = 2x+1 and y = -1/2x + 4 So 2x+1 = -1/2 x + 4 (5/2)x = 3 x = 6/5 y = 2x+1 = 12/5 + 1 = 17/5. Intersection point is (6/5, 17/5) So now there's a right triangle with vertices: (0,1); (0,4); and right angle at (6/5,17/5) All that remains is to find the distance D between points (0,4) and (6/5,17/5) D = √[(∆x)²+(∆y)²] D = √[(6/5)²+(3/5)²] D = √(45/25) = (3√5)/5 = 1.34... Sorry I don't know a simpler way to do it!
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