It is known that in the parallelogram ABCD, AC is diagonal and a (0,7), B (1,5), C (- 2,3), then the coordinates of point D are

It is known that in the parallelogram ABCD, AC is diagonal and a (0,7), B (1,5), C (- 2,3), then the coordinates of point D are

The midpoint of AC and BD are the same
Let D (x, y)
((0-2)/2,(7+3)/2)=（(1+x)/2,(5+y)/2）
1+x=-2 x=-3
5+y=10 y=5
D（-3,5）

A. The positions of B and C are 3, 7, 7, 7, 8 and 5 respectively. Please add a point D and link ABCD in turn to form a parallelogram
A. The positions of B and C are 3, 7, 7, 7, 8 and 5 respectively. Please add a point D and link ABCD in turn to form a parallelogram. Point D has three different positions. Where are they?

Let D point coordinate: (x, y) four points be divided into two groups, each group of two points as the vertex of the diagonal can form a parallelogram. Because the coordinate of the intersection of the diagonal is equal to half of the sum of the corresponding coordinates of the vertex, so the sum of the corresponding coordinates of the two groups must be equal, so that: 3 + 8 = 7 + x 7 + 5 = 7 + y solution: x = 4, y = 5 or: 3 + 7 = 8 + X

It is known that ABCD is a parallelogram, and a (4,1,3), B (2, - 5,1), C (3,7, - 5)
It is known that ABCD is a parallelogram, and a (4,1,3), B (2, - 5,1), C (3,7, - 5), then the vertex D is sitting
Mark as. The most stupid way to use it, I just learned here, only some fur

Let the coordinates of d be (x, y, z)
The midpoint of AC coincides with the midpoint of BD,
So:
x+2=4+3
y-5=1+7
z+1=3-5
So:
x=5, y=13, z=-3
The coordinates of D are (5,13, - 3)

ABCD × 9 = DCBA, then a =? B =? C =? D =?

a=1 b=0 c=8 d=9
4 digits multiplied by 9 = 4 digits, so a must be 1
From a = 1, d = 9
Because a * 9 = 1 * 9 = 9, B must be less than or equal to 1, otherwise 5-digit will appear after the hundreds are progressed to thousands
And because a = 1, B can only be 0
So 10c9 * 9 = 9c01
Because d * 9 = 9 * 9 = 81, that is to say, if you go to ten, you will go to eight
We also know that B = 0, so the number of C * 9 must be 2, so C = 8
Now the answer is 1089 * 9 = 9801
I don't know if you can understand it. In fact, it's not difficult. It's simple multiplication. You can only make a four letter equation call the master. If you analyze it carefully, there are many hidden conditions. Just like what I analyzed above

ABCD * 9 = DCBA, what are a, B, C and D respectively?
It's hard,

A = 1, B = 0, C = 8, d = 9, 4 digits multiplied by 9 = 4 digits, so a must be = 1, and d = 9 can be obtained from a = 1, because a * 9 = 1 * 9 = 9, so B must be less than or equal to 1, otherwise 5-digit will appear after the hundreds are progressed to thousands. And because a = 1, B can only be = 0, so there is 10c9 * 9 = 9c01, because d * 9 = 9 * 9 = 81

ABCD multiplied by 9 = DCBA, how many are a, B, C and D?
Eight hundred miles urgent

If B = 1, then C = 0 (because this four digit number is less than 1112), then ABCD = 1109, which is not suitable. If B = 0, from 9 * 9 = 81, then c * 9 =? 2, then C = 8

ABCD multiplied by 9 is equal to DCBA, how much is ABCD

Four digit * 9 is still four digit, so a can only be 11bcd * 9 = DCB1, so d = 91bc9 * 9 = 9cb1, because 9b can't carry, so B can only be 1 or 0. Suppose B is 1,9c + 8, the number of digits is 1,9 * C is 3, so c is 7, but c * 9 can't carry, so B can only be 010c9 * 9 = 9c01,9c * 9 + 8

ABCD × 9 = DCBA to find the value of ABCD

First of all, a must be 1
Because DCBA is still four digit
So a must be 1,
Otherwise ABCD × 9 will not be a four digit number
Because the number of D × 9 is 1,
So d must be 9
Write the formula as
1 BC9
× 9
9CB 1
Because B × 9 has no carry (otherwise a × 9 + carry product is not four digits)
So B must be 0 (because a = 1)
Because the number of C × 9 + 8 is 0,
So C must be 8
That is, ABCD = 1089
Checking calculation: 1089 × 9 = 9801

ABCD × 9 = DCBA find ABCD?

1089×9=9801
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(2004, Beibei District) as shown in the figure, there is a plastic rectangular template ABCD, which is 10cm long and 4cm wide. Place the right angle vertex P of PHF on the side of AD (not coincident with a and D), and move the triangle vertex P properly on AD
(1) Can you make the two right angle sides of your triangle pass through point B and point C respectively? If you can, please find out the length of AP at this time; if not, please explain the reason;
(2) Move the position of the triangle board again, so that the vertex P of the triangle board moves on ad, the right angle side pH always passes through point B, and the right angle side PF and DC extension line intersect at point Q, and BC intersect at point E. can CE be 2 cm? If yes, please find out the length of AP at this time; if not, please explain the reason
Can the second question be used as Er ⊥ AD and R to prove △ ABR ∽ pre again

As shown in the figure, there is a plastic rectangular template ABCD, the length is 10cm, the width is 4cm, your hand is big enough right angle triangle PHF rectangular vertex P waterfall ad edge (does not coincide with a, d), appropriate movement ad triangle vertex P: (1) can make your triangle two right angle sides through B point to C point, if you get ap