# Find the locus of a moving point equidistant from the line 2x+y=10 and 3x+4y=6.

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A point that is equidistant from 2 lines lies on the angle bisector of the angle formed by the lines. (A point equidistant from the sides of an angle lies on the angle bisector.)

Thus the locus of points will be an "X" formed by 2 intersecting lines.

Let line 1 be 2x+y=10. The angle that line 1 forms with the x-axis is `tan^(-1)(-2)=116.565^@` .

Let line 2 be 3x+4y=6. The angle that line 2 forms with the x-axis will be `tan^(-1)(-3/4)=143.130^@`

The angle that the angle bisector of the angle formed by the two lines will be the arithmetic mean of the two angles at the x-axis. So the angle bisector forms an angle of `129.848^@` with the x-axis.

`tan(129.848^@)~~-1.198` . This is the slope of the angle bisector line.

The angle bisector goes through the point of intersection of line 1 and line 2. Solving for the intersection we get (6.8,-3.6).

The equation of the line is y+3.6=-1.198(x-6.8) or y=-1.198x+4.548

The other angle bisector is perpendicular to this line. Its slope will be `m=-1/(-1.198)=.835` ; it also goes through the point of intersection (6.8,-3.4) so the equation is y+3.6=.835(x-6.8) or y=.835x-9.278

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The locus of points equidistant from the given lines in a plane are the lines y=-1.198x+4.548 and y=.835x-9.278

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The graphs of the given lines (in black) and the angle bisectors (in red):

Let P(a,b) be point equidistance from from lines 2x+y=10 and 3x+4y=6 .

Thus distance of the point P(a,b) from line 2x+y-10=0 be

`p=+-(2a+b-10)/sqrt(2^2+1^2)`

`p=+-(2a+b-10)/sqrt(5)`

Also the distance of the point P(a,b) from line 3x+4y-6=0 be

`p=+-(3a+4b-6)/sqrt(3^2+4^2)`

`=+-(3a+4b-6)/5`

Thus ,we have

`+-(2a+b-10)/sqrt(5)=+-(3a+4b-6)/5`

`So`

`` either

`(2a+b-10)/sqrt(5)=(3a+4b-6)/5`

`2sqrt(5)a+sqrt(5)b-10=3a+4b-6`

`(2sqrt(5)-3)a+(sqrt(5)-4b)-4=0`

`or`

`2sqrt(5)a+sqrt(5)b-10=-3a-4b+6`

`(2sqrt(5)+3)a+(sqrt(5)+4)b-16=0`

`Thus `

**Locus of point P(a,b) be**

`(2sqrt(5)-3)x+(sqrt(5)-4)y-4=0`

`or`

`(2sqrt(5)+3)x+(sqrt(5)+4)y-16=0`