It is known that the vertices of the quadrilateral ABCD are A 2,m ,B 2,2 ,C 0,n ,D 4, 2, and find th

In fact, the words of this drawing are very clear, make a coordinate system, point c is a point on the y axis, and a, which is a point on the line of x=2, is .|bcad) There are about four kinds of such states, rectangles, first of all, the length is equal, the parallel is pit at that time (you can find the axisymmetric one), take one, a 2 (a squared) b 2 c 2 d 2, all like this!

Point D can be seen as translating the point C by 1 unit length upwards and 2 units to the left, so as long as the point B is also translated accordingly, the coordinates of point A can be obtained: A(1,-1).Thus, m=1, n=-1

Two possibilities.

1) Take point B as the vertical foot and make a straight line L1 vertical BC

From b(6,1)c(3,3), the analytic formula of the straight line bc is: y=-2x 3+5, that is, the slope of the straight line bc is, k=-2 3

Because the straight line BC is perpendicular to the straight line L1.

So the slope of l1 is k=3 2

Because of point B.

So the analytic formula for the straight line l1 is: y=3x 2-8

Make a straight line L2 and a vertical L1 over point D

Because l2 passes the point d, the slope is, k=-2 3

So the analytic formula of the straight line l2 is: y=-2x 3+19 3 solution system of equations l1, l2

The solution yields x=86 13, y=25 13

That is, the intersection point of L1 and L2, that is, point A, A(86 13, 25 13) (2) passes through point D to make a straight line L1 perpendicular DC

Because c(3,3)d(2,5).

So the analytic formula for the straight line dc is: y=-2x+9

That is, the slope of the straight line cd is: k=-2

Because the straight line DC is perpendicular to the straight line L1.

So the slope of l1 is: k=1 2

Since L1 crosses the point D, the analytic formula for the straight line L1 is: y=x 2+4

Make a straight line L2 and a vertical L1 through point B

So the slope of the straight line l2 is: k=-2

Because the straight line L2 crosses point B, the analytic formula of the straight line L2 is: rate y=-2x+13 to solve the equation line L1, L2

The solution is x=18 5, y=29 5

That is, the intersection of L1 and L2, that is, point A, A(18 5, 29 5).

The simplest solution to this type of problem is to draw a diagram, you draw the coordinate axis to determine the three points b(6,1),c(3,3),d(2,5), and then connect the coordinates of point a. If it is a fill-in-the-blank multiple-choice question, you can make the picture more accurate, and you can see the answer on the way, which greatly improves your answering time. If it is a question and answer question, you need to solve the problem step as follows:

As can be seen from the figure, to make the quadrilateral ABCD a right-angled trapezoid, then the line segment AB and BC are perpendicular to B(6,1)C(3,3), and the analytical formula of the straight line BC is: y=-2x 3+5, that is, the slope of the straight line BC is, k=-2 3

Because of the point b, the analytic formula of the straight line AB is: y=3x 2-8 and because AD is perpendicular to AB, the slope of AD is k=-2 3,d(2,5), so the analytical formula of the straight line AD is: y=-2x 3+19 3 and AB and AD intersect with point A, then.

3x/2-8=-2x/3+19/3

The solution yields x=86 13, y=25 13

i.e. intersection point a (86 13, 25 13).

2、s=s△ade+s△cde+s△cbe=4×5/2+5×3/2+4×3/2

3. The abscissa minus 2 is equivalent to moving all points to the left by 2 units of ordinate plus 3, which is equivalent to moving all points up by 3 units.

Therefore, the problem is to move the quadrilateral 2 units to the left and 3 units to the top, so the area remains the same.

(1) If you have a title, you can draw the following picture.

2) There is a diagram that divides the diagram into two right triangles and a trapezoid, and the area that can be found is .

3) Same as (2) method, the area is 23

Let d(a,k a),c(b,k b).

From a(-4,0),b(0,-2) gives the straight line ab:y=-1 2x-2

The line cd is obtained from d(a,k a),c(b,k (-4,0),b(0,-2): y-k b=((k a-k b) (a-b))x-b

From ab cd, (k a-k b) (a-b) = -1 2 solution k=ab 2

From a(-4,0),d(a,k a),b(0,-2),c(b,k b), ad:y=k (a(a+4))*x+4k(a(a+4)),bc:y=(k+2b) b 2*x-2

From ad bc, k (a(a(a+4))=k+2b) b 2, substitute k = ab 2, get b = a+4

Straight line AD:y=2x+2,BC:y=2x-2 e(0,2) The distance between two parallel lines AD and BC d= 5*(8 5).

ABE area s1 = 1 2 * oa * be = 8

The area of the quadrilateral BCDE S2 is 5 times that of S1 The area of the quadrilateral ABCD is S=S1+S2=6S1=48

s=ad*d∴ad=6√5

From a(-4,0),d(a,k a), ad 2=(a+4) 2+(k a) 2=5 4*b 2=(6 5) 2=180

The solution is b=12, a=8, k=48

kab=(2-0)(4-0)=1/2

kad=(2-0)(-1-0)=-2

kbc=(4-2)/(3-4)=-2

kcd=(2-4) (-1-3)=1 2kab=kcd

ab||cd

kad=kbc

ad||bc

The quadrilateral ABCD is a parallelogram.

kab*kad=1/2*(-2)=-1

ab ad parallelogram abcd is rectangular.

ab= coincidence[(4-0) 2+(2-0) 2]=2 5ad= [2-0) 2+(-1-0) 2]= 5ab is equal to ad

Excludes the possibility that ABCD is square).

The parallelogram ABCD is rectangular.

This is a small topic.

Just use the idea of panning and you can do it.

B is to translate A 2 units to the right.

Then pan upwards by 4 units to grind the field shot.

Then d is to translate c to the right by 2 ridge lead units.

Then move 4 units upwards.

becomes (0,5).

Solution: Two possibilities.

1) Take point B as the vertical foot and make a straight line L1 vertical BC

From b(6,1)c(3,3), the analytic formula of the straight line bc is: y=-2x 3+5, that is, the slope of the straight line bc is, k=-2 3

Because the straight line BC is perpendicular to the straight line L1.

So the slope of l1 is k=3 2

Because of point B.

So the analytic formula for the straight line l1 is: y=3x 2-8

Make a straight line L2 and a vertical L1 over point D

Because l2 passes the point d, the slope is, k=-2 3

So the analytic formula of the straight line l2 is: y=-2x 3+19 3 solution system of equations l1, l2

The solution yields x=86 13, y=25 13

That is, the intersection point of L1 and L2, that is, point A, A(86 13, 25 13) (2) passes through point D to make a straight line L1 perpendicular DC

Because c(3,3)d(2,5).

So the analytic formula for the straight line dc is: y=-2x+9

That is, the slope of the straight line cd is: k=-2

Because the straight line DC is perpendicular to the straight line L1.

So the slope of l1 is: k=1 2

Since L1 crosses the point D, the analytic formula for the straight line L1 is: y=x 2+4

Make a straight line L2 and a vertical L1 through point B

So the slope of the straight line l2 is: k=-2

Because the straight line L2 crosses point B, the analytic formula of the straight line L2 is: rate y=-2x+13 to solve the equation line L1, L2

The solution is x=18 5, y=29 5

That is, the intersection of L1 and L2, that is, point A, A(18 5, 29 5).