The questions of the second year of junior high school should be written in the book of the second y

This needs to be drawn, so I can't draw it. Tell you how to do it: take x = any real number substitution, of course, pull, it is definitely more convenient to choose 0 and 1 substitution.

If you substitute 0 into x, then y = 5, and if you substitute 1 into x, then y = 3, and then draw it in Cartesian coordinates. I thought: you should know the Cartesian coordinate system.

It is a line segment that is perpendicular to the intersection of the abscissa x-axis and the ordinate y-axis. On the x-axis, there is a y-axis separated, so the x-axis can be divided into 2 parts, and the left one is a negative number; The one on the right is a positive number. You write from right to left on the left:

1,-2,-3,-4,-5.On the right, write from left to right: 1, 2, 3, 4, 5

Then on the y-axis, there is an x-axis to divide the y-axis into 2 parts, the top is a positive number, and the bottom one is a negative number. Write the above ones from bottom to top: 1, 2, 3, 4, 5

The following should be written from top to bottom: -1, -2, -3, -4, -5.

Okay, the Cartesian coordinate system is written. Then put x=0, y=5;x=1 and y=3 are written into the Cartesian coordinate system (the intersection of the x-axis and y-axis is the origin). x=0, y=5, write the coordinates (0,5).

Then, x=1, y=3, write the coordinates (1,3). In this way, 2 points appear on the y-axis and in the first quadrant.

Finally, connect these 2 points. (The first quadrant is the place in the upper right corner, the second quadrant is the place in the upper left corner, the third quadrant is the place in the lower left corner, and the fourth quadrant is the place in the lower right corner.) ）

Well, that's the answer.

The inner angles of the parallelogram are 360 degrees, so the angle bae + angle adc = 180 degrees, the angle bisector, so the angle fad + angle adf = 90 degrees, so the angle afd = 90 degrees, the same way proves that the remaining three angles are all 90 degrees, so they are rectangular.

The second question is whether there are few conditions?

Angle abc + angle bcd = 180, so angle bec = 90Angular ebc = angular edf, angular ebc + angular deb = 180 = angular edf + angular deb, so be parallel df, so rectangular.

Don't look anywhere else, just look at the picture I made with a red pen.

The requirement of the question is actually to go from point B to a point of AC to let the back go to point E, is it a little familiar, the teacher should have talked about this is the problem of the general drinking horses.

This diagram can be made from this.

The symmetry of the rhombus shows that the symmetry point of point D and point B about ac is returned, and the symmetry point of point b is named b'

Now the shortest distance of EP + PB is equal to DE, but in this problem E is a moving point, then the minimum value of DE is also required, the essence of DE is the distance from point D to AB, the distance from point D to a straight line should be learned in the first year of junior high school, and the perpendicular segment is the shortest, so when DB is perpendicular to AB. de the shortest.

So now you just need to find the length of de, and set the length of de to x. Now if we look at the triangle adb alone, we can find s= db* using the area formula of the triangle. However, a triangle has more than one side, and now s can be found using the same method as the ab side and the high on this edge in the same way, since s are equal, then the equation can be listed.

Finally, solve.

Establish a Cartesian coordinate system. Use equations to find the minimum solution.

The symmetry point of b is d, and the minimum value can be obtained by connecting ed.

a√(1-b²)=1-b√(1-a²)

Both sides are squared at the same time.

a(1-b)=1+b(1-a)-2b(1-a).

b²+1-a²=2b√(1-a²)

Again squared gets.

b²+1-a²)²=4b²(1-a²)

b²)²2b²(1-a²)+1-a²)²=4b²(1-a²)∴b²)²2b²(1-a²)+1-a²)²=0∴(b²+a²-1)²=0

b²+a²-1=0

b²+a²=1

Solution: (1): abc and def congruence.

ABC= def BC=EFBC is parallel to EF (the inner wrong angles are equal, the two lines are parallel) Quadrilateral BCEF is a parallelogram (the opposite side is a parallelogram) (2): when the quadrilateral BCEF is a diamond, BCE=CE BCE is an isosceles triangle.

and abc=60 degrees so bce is an equilateral triangle be=bc bc=ab*sin30=2t=[bc-(ab-ae)] 1=1(s)(3): probably.

When EF and CE are perpendicular to the quadrilateral BCEF is rectangular, CEB+ BEF = 90 degrees.

and bef=60 degrees so ceb=30 degrees, i.e., the point e coincides with the point a.

t=(ab-ae) 1=3 (seconds).

1；According to the Pythagorean theorem, the triangle ABC and the triangle EFD are two identical triangles, then the angle CB=angle BEF, so CB is parallel to EF, the inner wrong angle is equal, and the two days are parallel to the straight line, BC=EF, so the quadrilateral, CBFE is a parallelogram, and the opposite side of the parallel and equal quadrilateral is a parallelogram

2；Connect FE to intersect the remainder

Suppose the quadrilateral cefb is a diamond, then cf is perpendicular to be cos60=bc:ab out of bc the length of cos60=bo:bc out of bo t=2*bo "Do the math yourself, I haven't played this thing for a long time".

3: CEFB is a rectangle, when CE and AC overlap, it is a rectangle, t=3, it is a rectangle, and the area is very simple.

= =1 cba= fed=60° cb=ef cefb is a parallelogram.

When 2t=1, you think that ceb= cbe= bef= bfe=60° is a diamond, ceb and ebf are regular triangles, and all three sides are 2

3. Isn't E a rectangle when it coincides with A? Area = 2 2 times the root number 3 = 4 times the root number 3