Two vertices of a triangle ABC are A(3, -1) and B(-2, 3), and its orthocentre is P(1, 1). If the coordinates of the point C are ( , ) and the centre of the circle circumscribing the triangle PAB is (h, k), then the value of ( + ) + 2(h + k) equals
- 81
- 15
- 51
- 5
Answer
(4) 5
Solution
[figure]
P is the orthocentre, so each altitude passes through P.
Slope of AP = 1 + 11 - 3 = -1; since BC ⟂ AP, BC: y - 3 = 1(x + 2), i.e. y = x + 5.
Slope of BP = 1 - 31 + 2 = - 23; since AC ⟂ BP, AC: y + 1 = 32(x - 3), i.e. 2y = 3x - 11.
Solving, C = (21, 26), so + = 47.
Centre (h, k) of circle PAB is equidistant from P, A, B:
PA: (h - 1)2 + (k - 1)2 = (h - 3)2 + (k + 1)2 h - k = 2
PB: (h - 1)2 + (k - 1)2 = (h + 2)2 + (k - 3)2 6h - 4k = -11
h = - 192, k = - 232, so 2(h + k) = -42
( + ) + 2(h + k) = 47 - 42 = 5
Answer: (4)