Skip to main contentSkip to solution

Find a relation between x and y such that the points (x, y) is equidistant from points (7, 3) and (5, 2).

Solution

✅ Correct Option: 3

Equate the squared distances: (x−7)2+(y−3)2=(x−5)2+(y−2)2(x-7)^2 + (y-3)^2 = (x-5)^2 + (y-2)^2.

Expanding, −14x+49−6y+9=−10x+25−4y+4-14x + 49 - 6y + 9 = -10x + 25 - 4y + 4, so −4x−2y+29=0-4x - 2y + 29 = 0.

Hence 4x+2y=294x + 2y = 29.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question