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