As shown in the figure, in the plane rectangular coordinate system, point O is the coordinate origin, a (- 4,0), B (0,2) C (6,0). The abscissa and ordinate of line AB and CD are the same as that of point D 1 find D coordinate Starting from O, point P moves at a speed of one unit per second along the positive half axis of x-axis at a constant speed. The vertical line passing through point P and the straight line AB CD intersect at two points E and f respectively. Let P move for T seconds and EF length be y (Y > 0). Find the functional relationship between Y and T and write out the value range of T 3 under the condition of 2, is there a point Q on the straight line CD such that △ bpq is an isosceles right triangle with point P as the right vertex? If so, ask for the coordinates of point Q The graph is the intersection of the first-order function ad through the 123 quadrant and the first-order function DC through the 124 quadrant at the first quadrant (D, a on the negative half axis of X, C on the positive half axis of X, B on the positive half axis of Y)

As shown in the figure, in the plane rectangular coordinate system, point O is the coordinate origin, a (- 4,0), B (0,2) C (6,0). The abscissa and ordinate of line AB and CD are the same as that of point D 1 find D coordinate Starting from O, point P moves at a speed of one unit per second along the positive half axis of x-axis at a constant speed. The vertical line passing through point P and the straight line AB CD intersect at two points E and f respectively. Let P move for T seconds and EF length be y (Y > 0). Find the functional relationship between Y and T and write out the value range of T 3 under the condition of 2, is there a point Q on the straight line CD such that △ bpq is an isosceles right triangle with point P as the right vertex? If so, ask for the coordinates of point Q The graph is the intersection of the first-order function ad through the 123 quadrant and the first-order function DC through the 124 quadrant at the first quadrant (D, a on the negative half axis of X, C on the positive half axis of X, B on the positive half axis of Y)

1. The analytic formula of line AB y = KX + B takes a (- 4,0), B (0,2) into 0 = - 4K + B, and K = 1 / 2 of 2 = B has y = x / 2 + 2
The abscissa and ordinate of point D are the same, so y = x has x = x / 2 + 2 and x = 4,
The coordinates of point d (4,4)
2. A. when point e coincides with point D, the coordinates are (4,4)
b. When point F coincides with point C, the coordinate is (6,0)
The x-coordinate of E is equal to that of C, which is 6
Take x = 6 into y = x / 2 + 2 to get y = 5, i.e. coordinates of point E (6,5)
So the minimum length of EF is 0, the maximum length is 5, the time starts at 0 and ends at 6-4 = 2
There are start coordinates (0,0), end coordinates (2,6)
Let the first-order function be y = KX + B and bring in the coordinates to get b = 0 and K = 3
So find the functional relationship between Y and T, y = 3T, the value range of T is 0

Place the two sides OA and OC of the rectangular ABCO in the rectangular coordinate system, OA = 4, OC = four times the root sign, and fold the angle B with the point D on the diagonal AC, the crease
For CE, find the analytic expression of inverse scale function of point D

In fact, this problem is not difficult, very simple, can mainly find out the coordinates of point D, you can solve it. Find the coordinates of point D, the most important sentence is that the crease is CE, which means that the crease must pass through point C, I drew a picture for you. You can fold any rectangular paper by yourself, if you want to fold point B to AC, and the crease passes through

On the distance between two points in rectangular coordinate system
Who knows the formula~
(x^1-x^2)^2
This formula is about
A(x^1,y^1),B(x^2,y^2)
The method of finding the length between ab~

AB = radical [(x1-x2) square + (y1-y2) square]

As shown in the figure, in △ ABC, ab = AC, CD is the height on the edge of AB, and the proof is: ∠ BCD = 12 ∠ a

It is proved that a is AE ⊥ BC in E, Cd in F, and ∫ BAE + ⊥ B = 90 ° and ab = AC, ∫ BAE = 12 ⊥ BAC. And ∫ CD ⊥ AB, ∫ BCD + ⊥ B = 90 ° and ∫ BAE = ∫ BCD. ∫ BCD = 12 ∫ a