In a triangle PQR, the co-ordinates of the points P and Q are (-2, 4) and (4, -2) respectively. If the equation of the perpendicular bisector of PR is 2x - y + 2 = 0, then the centre of the circumcircle of the ΔPQR is :

Solution

Maths sir · 4 steps

The circumcentre sits on every perpendicular bisector, so build PQ's bisector and meet it with the given one.

1. Use what is given. The circumcentre is equidistant from all three vertices, so it lies on the perpendicular bisector of each side. We already have the one for PR, 2x - y + 2 = 0. R is unknown, so we build the bisector of PQ instead.
2. Bisector of PQ. The midpoint of PQ is ((-2+4)/2, (4-2)/2) = (1, 1). The slope of PQ is (-2-4)/(4-(-2)) = -1, so the perpendicular slope is 1. The bisector is y - 1 = x - 1, that is, y = x.
3. Meet the two lines. Put y = x into 2x - y + 2 = 0: 2x - x + 2 = 0, so x = -2 and y = -2. As a check, (-2, -2) is 6 units from P and 6 units from Q.

   Option	On 2x - y + 2 = 0?	On y = x?
   (a) (-1, 0)	yes	no
   (b) (1, 4)	yes	no
   (c) (0, 2)	yes	no
   (d) (-2, -2)	yes	yes

4. So. Every option lies on the given bisector, but only (-2, -2) also lies on the bisector of PQ. The circumcentre is (d) (-2, -2).

The idea to keep. Two perpendicular bisectors are enough to fix the circumcentre. Use the sides you know fully, and do not waste time hunting for a missing vertex.

Answer from the NTA final answer key. Solution by PrepWiser.