Find the distance between the parallel lines whose equations are below. y = 1/3x + 1 y = 1/3x − 2 step 1: find the equation of the line that is perpendicular to both parallel lines. use the opposite reciprocal slope and one of the y-intercepts to write your line. step 2: now, use the equation from step 1 and the parallel line whose y-intercept you did not use, to set up a system of equations and determine the other endpoint. write your answer as an ordered pair. enter your answers as decimals, rounded to the nearest tenth. step 3: now, find the distance between the two endpoints (the y-intercept you used in step 1, and the point you found in step 2). use the distance formula and enter your answer as a decimal, rounded to the nearest hundredth.

1. The slope of the given lines is the x-coefficient, 1/3. The opposite reciprocal of that is -1/(1/3) = -3. We simply need to replace the slope in one of the slope-intercept equations. Using the first one, we find the equation of the perpendicular line to be ...

... y = -3x +1

2. The above equation together with the second given equation form the system of interest:

Subtracting the first equation from the second, we get ...

... (1/3x -2) -(-3x +1) = 0 = (3 1/3)x -3

... 3 = (10/3)x . . . . . . . . add 3

... 9/10 = x . . . . . . . . . . multiply by 10/3

We can substitute this value into either above equation to find y. Using the first, we have ...

... y = -3(9/10) +1 = -2.7 +1 = -1.7

The solution to the system is (x, y) = (0.9, -1.7).

3. The distance formula tells us the distance can be found from the sum of squares of the differences in coordinates.

The points on either line where the perpendicular intersects are ...

... (0, 1) and (0.9, -1.7)

The differences between these points are (∆x, ∆y) = (0.9, -2.7), so the distance between the points is ...

... d = √(0.9² +(-2.7)²) = √8.1 = 0.9√10

In decimal form, d ≈ 2.8460499

... d ≈ 2.85

Comment on distance between parallel lines

The distance from a point to a line can be found using a formula based on the general-form equation of a line. For line ...

... ax +by +c = 0

The distance from point (x, y) to the line is given by

... d = |ax +by +c|/√(a²+b²)

The two lines we have can be rearranged to general form as ...

We know that any point on the first line will satisfy x -3y = -3. Substituting this value into the distance equation for the distance to the second line, we get ...

... d = |-3 -6|/√(1²+(-3)²) = 9/√10 = 0.9√10

In my opinion, this is a much easier way to find the distance between parallel lines.

a ruler? idk

step-by-step explanation: