The distance of a point (m, n) from the line ax + by + z =
0, is given by |am + bn + z|/ sqrt (m^2 + n^2).
The
distance of the point (x, y) from 3x + 4y + 8 = 0 is |3x + 4y + 8|/sqrt (3^2 + 4^2). And
the distance from 6x – 8y + 4 = 0 is |6x – 8y + 4|/sqrt (6^2 +
8^2).
From what we need to find: |3x + 4y + 8|/sqrt (3^2 +
4^2) = 2*|6x – 8y + 4|/sqrt (6^2 + 8^2)
=> |3x + 4y
+ 8|/sqrt (25) = 2*|6x – 8y + 4|/sqrt (100)
=> sqrt
100* |3x + 4y + 8| = 2* sqrt 25*|6x – 8y + 4|
=>
10|3x + 4y + 8| = 2* 5*|6x – 8y + 4|
=> |3x + 4y +
8| = |6x – 8y + 4|
Now |3x + 4y + 8|can be equal to 3x + 4y
+ 8 or – (3x + 4y + 8) based on the value of (3x + 4y + 8). And |6x –8 y + 4| can be
equal to 6x – 8y + 4 or - (6x – 8y + 4)
To include all
possibilities we have the equations:
(3x + 4y + 8) = (6x –
8y + 4)
=> 3x – 12y – 4 =
0
3x + 4y + 8 = - (6x – 8y +
4)
=> 3x + 4y + 8 = -6x + 8y –
4
=> 9x – 4y + 12 =
0
Therefore each point on the lines 3x – 12y
– 4 = 0 and 9x – 4y + 12 = 0 has a perpendicular distance from the line 3x + 4y + 8 = 0
that is twice the perpendicular distance from the line 6x – 8y + 4 =
0.
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