We know that, every point on the X-axis in the form (x,0).
Let the point on the X-axis be P(x,0) .
Given: P(x,0) is at a distance of 2√5 units from the point Q(7,-4).
By given condition,PQ=2√5
[∵Distance formula√(x2−x1)2+(y2−y1)2]
⇒(PQ)2=4×5⇒(x−7)2+(0+4)2=20⇒x2+49−14x+16=20⇒x2−14x+65−20=0⇒x2−14x+45=0⇒x2−9x−5x+45=0[by factorisation method]⇒x(x−9)−5(x−9)=0⇒(x−9)(x−5)=0∴x=5 and 9
Hence, there are two such points i.e., point P can be (5,0) or (9,0).
[∵ Both the points are at a distance of 2√5 from (7, -4) ]