# geometry

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## Perpendicular distance between two parallel lines

(1) The distance between two parallel lines is the distance between the points of intersection of a third line that is perpendicular to both these lines.

(2) The slopes of two perpendicular lines are negative reciprocals of one another.

The two blueish triangles are copies of each other, one of which has been rotated 90 degrees about point A (the origin). This means that the line segment c1 is perpendicular to line segment c2.

Since line segment c2 is parallel to line segment e, the perpendicular distance between both line segments is the distance between the two points where the line segment c1 intersects c2 and e.

For non-visual proofs of 1 and 2, see below.

The givens:

Line segment c1 is perpendicular to line segment c2.

Line segment c2 is parallel to line segment e.

Coordinates of point A = (0, 0).

Length of line segment b = b.

Length of line segment a = a.

Length of line segment c = c1 = c = sqrt(b^2 + a^2) (by the Pythagorean theorem).

Length of line segment e = c.

The slope of line c1 = m1 = a/b.

The slope of line c2 = m2 = -b/a.

Coordinates of point B1 = (b, a) = (b, m1b).

Coordinates of point B2 = (-a, b) = (-a, m2*-a).

Length of line segment d = sqrt(c1^2 + c2^2) = sqrt(2c^2) = sqrt(sqrt(2)*sqrt(2)*sqrt(c^2)*sqrt(c^2)) = sqrt(sqrt(2)*sqrt(c^2)*sqrt(2)*sqrt(c^2)) = sqrt(2)sqrt(c^2) = sqrt(2)*c = sqrt(2)*sqrt(b^2 + a^2) (by the Pythagorean theorem). Respectively, d = sqrt((b-(-a))^2 +  (a-b)^2) = sqrt((b – (-a))^2 + (m1b – m2*-a)^2).

Proof that the slopes of two perpendicular lines are negative reciprocals of one another:

d^2 = (sqrt(2)*sqrt(b^2 + a^2))^2 = sqrt(2)*sqrt(2)*sqrt(b^2 + a^2)*sqrt(b^2 + a^2) = 2(b^2+a^2) = (b^2 + (m1b)^2)+(-a^2 + (m2*-a)^2) = (b^2 + (m1b)^2) + (a^2 + (m2a)^2)

d^2 = (b – (-a))^2 + (m1b – m2*-a)^2

(b + a)^2 + (m1b – m2*-a)^2 = b^2 + (m1b)^2 + a^2 + (m2a)^2

b^2 + 2ab + a^2 + (m1b)^2 + 2m1m2ab + (m2a)^2 = b^2 + (m1b)^2 + a^2 + (m2a)^2

2ab + 2m1m2ab = 0

2m1m2ab = -2ab

m1m2 = -1

m1m2 = (a/b)(-b/a) = -1

Proof that the distance between two parallel lines is the distance between the points of intersection of a third line that is perpendicular to both these lines:

Consider any two parallel lines,

f(x) = y1 = mx + b1

g(x) = y2 = mx + b2,

and a third line that is perpendicular to both these lines,

h(x) = y3 = (-1/m)x = -x/m.

Then the intersection point of y1 and y3 is the solution to the system of linear equations,

f(x) = h(x)

mx + b= -x/m

(m^2)x + mb1 = -x

(m^2)x + x = -mb1

((m^2) + 1)x = -mb1

x = -mb1 / ((m^2) + 1)

h(-mb1 / ((m^2) + 1)) =  y1 = y3 = (-1/m)(-mb1 / ((m^2) + 1)) = b1 / ((m^2) + 1)

Intersection point 1: (x1, y1) = (-mb1 / ((m^2) + 1), b1 / ((m^2) + 1))

By the same logic, the intersection point of y2 and y3 is,

g(x) = h(x)

Intersection point 2: (x2, y2) = (-mb2 / ((m^2) + 1), b2 / ((m^2) + 1))

By the Pythagorean theorem, the distance between intersection point 1 and 2 is,

d = sqrt((x– x2)^2 + (y1 – y2)^2) = sqrt(((-mb1 – (-mb2)) / ((m^2) + 1))^2 + ((b1 – b2) / ((m^2) + 1))^2) = sqrt((-mb1 + mb2)^2 / (m^2 + 1)^2 + (b1 – b2)^2 / (m^2 + 1)^2)  = sqrt((m^2)(-b1 + b2)^2 / (m^2 + 1)^2 + (b1 – b2)^2 / (m^2 + 1)^2) = sqrt(((m^2)(-b1 + b2)^2 + (b1 – b2)^2) / (m^2 + 1)^2) = sqrt(((m^2)(b1^2 – 2b1b+ b2^2) + (b1^2 – 2b1b+ b2^2)) / (m^2 + 1)^2) = sqrt((m^2 + 1)(b1^2 – 2b1b+ b2^2) / (m^2 + 1)^2) = sqrt((b1^2 – 2b1b+ b2^2) / (m^2 + 1)) = sqrt((b1 – b2)^2 / (m^2 + 1)) = sqrt((b1 – b2)^2) / sqrt(m^2 + 1) = |(b1 – b2)| / sqrt(m^2 + 1)

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