Question

The perpendicular distance between the lines: 9x+40y-20=0 and 9x+40y+21=0 is

Answer 1

GIVEN:

9x + 40y - 20 = 0

9x + 40y + 21 = 0

TO FIND:

The distance between the lines 9x + 40y - 20 = 0 and

        9x + 40y + 21 = 0

SOLUTION:

Distance between two parallel lines y = mx + c1 and y = mx + c2 is given by

D = |C1–C2|/√A^2 + B^2

c1 = -20 & c2 = 21

A^2 = 81 & B^2 = 1600

Applying values in equation,

D = I -20 - 21 I / √81 + 1600

D = I -41 I / √1681    [ Modulus of any number is positive ( I -41 I = 41 )]

D = 41 / 41  [ Root of 1681 is 41 ]

D = 1 unit

Therefore, The distance between the lines 9x + 40y - 20 = 0 and 9x + 40y + 21 = 0 is 1 unit.