If the distance between circumcentre and orthocentre of triangle whose vertices are (1,5) (2,3) (4,4) is k/√10 then sumof abscissa of points which are k distance away from the point (3,2) on the line y=x is

A (1,5)
B (2,3)
C (4,4)
AB= √5 units
BC=√5 units
AC= √10 units
Hence given triangle is right angled triangle ( AB² + BC² = AC²) at B
Orthocenter is B only
CIrcumcenter for rt triangle is mid point of hypotenuse i.e. ( (1+4)/2 , (5+4)/2 )
= ( 5/2, 9/2)

Distance between orthocenter and circumcenter is √10/2 which is k/√10
k= 100/2= 50

Any point on line y=x will be of form P (a,a)
Q(3,2)
PQ=k=50
Solve to get a=37.85 and -32.85
Sum of abscissa =5