Stepwise
Start practisingPractise

JAMB 2012 · UTME · Q31

The locus of a point equidistant from the intersection of the lines 3x−7y+7=03x - 7y + 7 = 0 and 4x−6y+1=04x - 6y + 1 = 0 is a

Worked solution (try it first)
  1. Two lines that are not parallel meet at one point.
  2. Solving 3x−7y=−73x - 7y = -7 and 4x−6y=−14x - 6y = -1 gives (3.5,2.5)(3.5, 2.5).
  3. Points all at the same distance from one fixed point form a circle centred on that point.
  4. So the locus is a circle, option B.

Report a problem with this question