# A moving point remains equidistant from the point (a,0)and the line y=x. Find the equation of its locus.

justaguide | Certified Educator

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We need to determine the locus of a point that is equidistant from the line y = x and the point (a , 0)

Now the relation for determining the distance d of a point (x1, y1) from the line ax+by +c = 0, is:

d = |ax1+by1+c|/ sqrt (a^2+b^2)

and the relation for the distance between two points (x1, y1) and (x2, y2) is sqrt [ (x2 - x1)^2 + (y2 - y1)^2]

Let the required point be (X, Y). Substituting the values we have:

| X - Y | / sqrt ( 1^2 + 1^2) = sqrt [( X - a)^2 + Y^2]

=> |X - Y| / sqrt 2 = sqrt [( X - a)^2 + Y^2]

=> |X - Y| = sqrt [ 2* (X - a)^2 + 2*Y^2]

Therefore the required locus is:

x - y = sqrt [2(x - a)^2 + 2y^2]

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## Related Questions

neela | Student

Let (x1,y1) be the moving point which is equidistant from (a,0) and  y =  x.

Then distance d of the point (x1,y1) from, (a,0) is  d^2 = (x1-a)^2+(y1-0)^2...(1)

Therefore  distance d of the point  (h,k) from a line ax+by+c = 0 is given by:

d = (ah+by+c)^2|/(a^2+b^2)

The distance d of the point from y = x , or x-y+0 = 0 from a point (x1,y1) is given by : d^2 = (x1-y1)^2/(1^2+1^2)^(1/2)}....(2).

Therefore from (1) and (2)  the locus of the point (x1,y1) is given by:

(x1-a)^2 +y1^2 = (x1-y1)^2/2

2(x1^2-2ax1+a^2 + y1^2 = x1^2 -2x1y1+y1^2.

x1^2-4ax1+2a^2 +2x1y1 = 0

x1^2 +2(y1-2a)x1+ 2a^2 = 0

So by dropping the suffixes, we get the equation of the locus.

x^2+2(y-2a)+2a^2 = 0.

Therefore the equation of the locus is  x^2+2(y-2a)+2a^2 = 0.

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