The distance between the parallel lines y = x + 3 and y = x + 1, rounded to the nearest hundredth, is 1.41 units.
To find the correct distance, we need to consider the perpendicular lines that intersect the parallel lines.
Identify a point on either line: Let's choose the point (0, 3) on the line y = x + 3.
Find the equation of the perpendicular line: The slope of the parallel lines is 1, and its negative reciprocal is -1. So, the perpendicular line passing through (0, 3) will have a slope of -1 and an equation of the form:
y = -x + b
Substitute the point (0, 3) to find b:
3 = -0 + b
=> b = 3
Therefore, the equation of the perpendicular line is y = -x + 3.
Find the intersection point with the other parallel line: Solve the system of equations:
y = x + 1
y = -x + 3
Adding the equations, we get:
2y = 2
=> y = 1
Substituting y back into one of the equations, we find x = 0.
Therefore, the intersection point is (0, 1).
Calculate the distance: Now, we have the point (0, 3) on one line and the intersection point (0, 1) on the other line. The distance between these two points is simply the absolute difference of their y-coordinates:
|3 - 1| = 2
However, this is not the final answer. We need to consider that this line segment is not perpendicular to the parallel lines, and its length is a diagonal across the rectangle formed by the parallel lines and the perpendicular line.
Divide by the square root of 2 to find the perpendicular distance:
The diagonal of a square is longer than any side by a factor of the square root of 2. Therefore, to find the actual perpendicular distance between the parallel lines, we need to divide the length of the diagonal by the square root of 2:
2 / √2 ≈ 1.41 (rounded to the nearest hundredth)
Therefore, the distance between the parallel lines y = x + 3 and y = x + 1, rounded to the nearest hundredth, is 1.41 units.
Question
What is the distance between parallel lines y = x+3 and y=x+1, rounded to the nearest hundredth?