Answer:
Let P(h,k) be the point which is equidistant from the point (2,4) and the y-axis. The distance of point P(h,k) from the y-axis is h. ∴h=√(h−2)2+(k−4)2⇒h2−4h−4+k2−8k+16=h2⇒k2−4h−8k+20=0.
Hence,the locus of (h,k)is y2−4x−8y+20=0.
Step-by-step explanation:
