What is the distance between the lines: x/2 + y + 2 = 0 and 4x + 8y = -13

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justaguide | College Teacher | (Level 2) Distinguished Educator

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It is possible to find the distance between two lines only if they are parallel to each other.

The lines we have are: x/2 + y + 2 = 0 and 4x + 8y = -13.

Convert them to the slope-intercept form y = mx + c

x/2 + y + 2 = 0

=> y = -x/2 - 2

4x + 8y = -13

=> 8y = -4x - 13

=> y = -x/2 - 13/8

The slope of both the lines is -1/2, they are parallel.

The product of the slope of perpendicular lines is equal to -1. Take a line perpendicular to the two given lines like y = 2x

Find the point of intersection of y = 2x with y = -x/2 - 2

2x = -x/2 - 2

=> 5x/2 = -2

=> x = -4/5

y = -8/5

The point of intersection is (-4/5, -8/5)

Find the point of intersection of y=2x and y = -x/2 - 13/8

2x = -x/2 - 13/8

=> 5x/2 = -13/8

=> x = -13/20

y = -13/10

TheĀ  point of intersection is (-13/20, -13/10)

The distance between (-4/5, -8/5) and (-13/20, -13/10) is

sqrt[(-13/20 + 4/5)^2 + (-13/10 + 8/5)^2]

=> 0.3354

The distance between the two lines is 0.3354

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