Δ ABC is a regular triangle a (1,2) B (3, - 4) to find the coordinate of point C

Δ ABC is a regular triangle a (1,2) B (3, - 4) to find the coordinate of point C

The midpoint of AB is d (2, - 1), the slope of AB is k = (- 4 - 2) / (3 - 1) = - 3AB, the slope of the middle vertical line is 1 / 3, the equation of the middle vertical line is y + 1 = (x - 2) / 3, x = 3Y + 5, let C (3b + 5, b) Ca & # 178; = AB & # 178; (3b + 5-1) &# 178; + (b-2) &# 178; = (3 - 1) &# 178; + (- 4-2) &# 178; B &

Urgent! A high school mathematics problem, to process
It is known that f is the focus of the parabola C: Y2 (square) = 4x, P.Q is two points on the parabola C, and the midpoint of the line PQ is a (2,2). The area of the triangle PQF is obtained
Write it all intact.

A: 2
Let P (x1, Y1), q (X2, Y2), then
4x1=y12; 4x2=y22; x1+x2=4; y1+y2=4;
① If Y 1 (y 1-4) = 0, then Y 1 = 0 or 4, i.e. P (0,0), q (4,4). F is (1,0) ‖ s = 1 / 2 x 1 x 4 = 2

Let y = f (x) be a decreasing function defined on R + and satisfy f (XY) = f (x) + F (y), f (1 / 3) = 1,
(1) : find the value of F (1)
(2) If f (x) + F (2-x) < 2, find the value range of X

1. F (1) = f (1) + F (1) = 2F (1), so f (1) = 02. F (1 / 9) = f (1 / 3) + F (1 / 3) = 2, so the original formula = f (x) + F (2-x) < f (1 / 9) = f (x multiplied by 2-x) < f (1 / 9) 0 to positive infinity? If so, first we need to take the intersection of 2-x > 0 and x > 0. Finally, because it is a subtractive function, so x multiplied by 2-x ＞