# Find the point (x, y) such that the perpendicular distance of the point from 3x + 4y + 8 = 0 is twice its distance from 6x –8 y + 4 = 0?

justaguide | College Teacher | (Level 2) Distinguished Educator

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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.

neela | High School Teacher | (Level 3) Valedictorian

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The perpendicular distance d of the point (x1,y1) from the line ax+by +c = 0 is given by:

d = |ax1+by+c|/(a^2+b^2)^(1/2).

Therefore the perpendicular distance d of the point (x,y) from the lines: 3x + 4y + 8 = 0  and 6x –8 y + 4 = 0 is given by:

d1 = |3x + 4y + 8|/((3^2+4^2)^(1/2)

d2 = |6x –8 y + 4|/{(6^2+(-8)^2}^(1/2).

It is given that d1 = 2*d2,

|3x+4y+8|/5 = 2|6x-8y+4|/10

3x+4y+8 = (6x-8y+4) , Or  3x+4y+8 = -(6x-8y+4)

=> 3x-12y -4= o, or 9x-4y+12 = 0.

=> Therefore  any point (x,y) on the lines 2x-12y-4 = 0, or 9x+4y+12= 0 satisfy the given condition.