+6 Points A, B, and C are given in the coordinate plane. There exists a point Q and a constant k such that for any point P,
PA^2 + PB^2 + PC^2 = 3PQ^2 + k
If A = (4, -1), B = (-3, 1), and C = (5, -3), then find the constant k.
To find the constant k, we can use the given information and the distance formula. The distance formula states that for two points (x1, y1) and (x2, y2), the distance between them is given by the equation d = sqrt((x2-x1)^2 + (y2-y1)^2).
Given that A = (4, -1), B = (-3, 1), and C = (5, -3), we can calculate the distances between A and Q, B and Q, and C and Q as follows:
dAQ = sqrt((xQ-4)^2 + (-1-yQ)^2) dBQ = sqrt((-3-xQ)^2 + (1-yQ)^2) dCQ = sqrt((5-xQ)^2 + (-3-yQ)^2)
Now, since PA^2 + PB^2 + PC^2 = 3PQ^2 + k, we can substitute in the distances and solve for k. Let F be the expression we want. This gives us the following equation:
dAQ^2 + dBQ^2 + dCQ^2 = 3PQ^2 + F
Substituting in the distances we calculated earlier and simplifying, we get the following equation:
(xQ-4)^2 + (-1-yQ)^2 + (-3-xQ)^2 + (1-yQ)^2 + (5-xQ)^2 + (-3-yQ)^2 = 3PQ^2 + F
Rearranging and simplifying, we get the equation:
F = 6PQ^2 - 3(xQ-4)^2 - 3(1-yQ)^2 - 3(-3-xQ)^2 - 3(-3-yQ)^2
Since the points A, B, and C are given, we can now substitute in their coordinates to get the value of k. This gives us the equation:
F= 3PQ^2 - 3(-3)^2 - 3(4)^2 - 3(-1)^2 - 3(-3)^2
Simplifying, we get the final result:
F = 3PQ^2 + 27
Therefore, the constant k is 27.
