Find the distance between lines y plus mx plus c and y equals mx plus b?

Answer 1

y = mx + c

y = mx + b

We see that the lines have the same slope m.

If the lines have slopes with the same sign, then the lines are parallel.

Soppose that the slope is positive, m > 0.

Case 1: If c > 0, b> 0, and c > b, or c < 0, b < 0, and c < b, then we have to work in the same way to find the distance between lines, which is also the same distance.

Let work for the first possibility where c > 0, b > 0, and c > b.

If we draw the lines, we see that from the interception of the lines with x and y-coordinate axis, are formed two similar right triangles on the second quadrant.

The bigger triangle vertices are (0, c), (0, 0), and (-c, 0)

The midpoint of the hypotenuse is (-c/2, c/2).

The distance between the origin and the midpoint is (c√2)/2.

√[(-c/2)2 + (c/2)2] = √(c2/4 + c2/4) = √ (2c2/4) = (c√2)/2

The smaller triangle vertices are (0, b), (0, 0), and (-b, 0)

The midpoint of the hypotenuse is (-b/2, b/2).

The distance between the origin and the midpoint is (b√2)/2.

√[(-b/2)2 + (b/2)2] = √(b2/4 + b2/4) = √ (2b2/4) = (b√2)/2

The distance between the lines is (√2)/2)(c - b).

(c√2)/2 - (b√2)/2 = (√2)/2)(c - b)

Case 2: If c > 0, b< 0 or c < 0, b > 0.

If we draw the lines, we see that from the interception of the lines with x and y-coordinate axis are formed two similar right triangles on the second and fourth quadrant. In this case the distance between lines is (√2)/2)(c + b).

If the slope is negative for both lines, then we find the same result as above, but the formed right triangles are in the first and third quadrant.

If the lines have slopes with different sign, then the lines intersect.