The coordinates of the midpoint M(X, y) of line segment A(X 1, Y 1)B(X2, Y2) satisfy the relation.
X=(X 1+X2)/2
Y=(Y 1+Y2)/2
That is, the abscissa of the midpoint is equal to half of the sum of the abscissas at both ends, and its ordinate is equal to half of the sum of the ordinate.
According to the requirements of friends, the process is simple, and the ideas quickly enter the third question.
(1) Let y=a(x-2)(x+4) and substitute it for B(0, -4).
Get a= 1/2, Y = 1/2x 2+x-4.
(2) The tangent point between a straight line parallel to AB and a parabola is the target.
(3) There are four situations, and the key is to take OB as the diagonal. The midpoint coordinate formula will be used.
PQBO is a parallelogram, so the diagonal is equally divided.
Let the intersection of diagonal OB and PQ be k, and the coordinate of k be (0, -2).
Let P(x, y), Q(x 1, y 1).
Then x+x 1=2*0 (1).
y+y 1 = 2 *(2)(2)
Because p is on a parabola and q is on a straight line,
So y = 1/2x 2+x-4 (3)
y 1=-x 1 (4)
From (1)(2)(3)(4)
x=-4，y = 0；
x 1=4，y 1 =-4；
That is, p (-4,0) coincides with point A.
Q(4，-4)
Perhaps the questioner wants to test the observation ability of the middle school students and find this problem directly through geometric understanding.
Because the triangle OAB is an isosceles right triangle and AB is parallel to the straight line y=-x, the intersection point B is the intersection point of the straight line parallel to OA and y=-x, which is also the Q point. But unfortunately, I thought of it later. I'll work it out first and then observe.
An important lesson: how important it is to observe geometric problems!
That's it. Please criticize and correct me.