It is known that the primary function Y 2X 4 intersects with the X axis, and the Y axis intersects w

Solution: (1) Let x=0, then y=-2 0+4=4, let y=0, then x=1 2(4-0)=2

a（2，0），b（0，4）

2) P1(6,2), P2(4,6) can be found from the knowledge of congruent triangles, since P3 is the midpoint of bp1,ap2 p3(3,3)

s=sδoc2d2-sδoab=½×9×9/2-½×4×2=65/4=

or s= oc1d1- oab= 14 7- 4 2 = 45 <>

1) Let x=0, then y=-2 0+4=4, let y=0, then x=1 2(4-0)=2

2) The coordinates of point A are (2,0) and the coordinates of point B are (0,4), then the coordinates of point C in line segment AB are (1,2).The slope of the straight line PC is k=1 2 and the equation of the straight line PC is x-2y+3=0Let the coordinates of point p be (x,y), then [y-4] [x-0] [y-0] [x-2]=-1

From x=3,y=3(x=-1,y=1, rounded), the coordinates of p are (3,3).

3) The equation for the parallel line of the straight line ab through the point p is 2x+y-9=0, then c(0,9),d(,0)The area of the quadrilateral ABCD is s= ocd- oab=1 2

Summary. So 3-2x=0

x=3 2 points a(3

It is known that the primary function y=3-2x finds the coordinates of the intersection point a with the x-axis and the intersection point b with the y-axis.

Hello, the answer and process is as follows with the intersection of the x-axis, y=0

So 3-2x=0x=3 2 point a(3

Intersection with the y-axis, x=0 so y=3

Point b (I hope it will help you, if you don't know anything, you can consult the teacher, by the way, move Lao Min or move your little hand to give the teacher a thumbs up, thank you.)

For y=ax-2, let y=0, then ax-2=0, and the solution is to touch Changshen x=2a, so the coordinates of the intersection of the primary function y=ax-2 and the x-axis are (2a,0) Xiaojing, and for the xun, then y=bx-4, so that y=0, then bx-4=0, the solution is x=4b, so the coordinates of the intersection of the primary function y=bx-4 and the x-axis are (4b,0), and the primary function y=ax-2 and y=bx-4 and the x-axis.

a(2,0) b(0,4)

The second question should be asked: ABP as an equilateral triangle with AB as the side. If it is isosceles, there will be countless solutions.

It can be seen that the point p must be on the mid-vertical line of AB. The straight line perpendicular to ab can be set as y=1 2x+b because it is a perpendicular line, so the midpoint of ab is (1,2) point. Bring in the solution to get b = 3 2.

y=1/2x+。Let p(x,1 2x+ as pm x in m because it is equilateral, ab=bp=2 times the root number 5. In RT APM, the Pythagorean i.e. (x-2) + because in the first quadrant so x>0 is solved to get x=1+2 root number three, and then substituted to get p (1+2 times root number three, 2 + root number three).

With the foundation of the second question, the third question is easy to do, set the straight line cd: y=-2x+b and bring in p to get b=5, the root number three + 4

The coordinates of c and d are respectively (three + 2,0) and (0,5 three + 4).

Calculate s ocd = quarter (91 + 40 root number three) subtract s aob to get the quadrilateral area of 75 4 + 10 root number three.

As for the reason why the answer is so disgusting, here's why: it's possible that it's even wrong. It's possible that the question is wrong. It's possible that your teacher is wrong ...... head

Points are awarded for the perfect process!

It is better to teach a man to fish than to teach him to fish.

1. Because the function intersects with the x and y axes at two points of ab, i.e., a(2,0) b(0,4)2, let the p coordinate be (m,n) because ab is an edge, so there are the following situations: 1. |ab|=|ap|or |ab|=|bp|

Calculations are made according to coordinates (due to limited time and conditions, they will not be calculated one by one below.)

3. Calculate the coordinates of point P according to the formula of parallel lines in the coordinates, and then calculate the coordinates of points c and d according to point P, and the y of c in the coordinates of these two points is x, after finding the coordinates of these two points, you can find the area of the quadrilateral, and find the area of the quadrilateral, you can find the area of the three triangles respectively or find it directly, see which one is simple.

I know the answer but forget how to do it.

The second question should be an isosceles right triangle.

Very simple questions should be in high school.

y=1/2x^2+2

It is not difficult to conclude that a:(-2,0)b:(2,0)c:(0,2)ac is analytically y=x+2

The QC of PQC is the bottom, and the OP is high.

Let p:(x,0).

x-(-1)=t

x=t-1s=t*(t-1) 2,2 t 4 nexus ma,mb

oa=ob，∠aoc=∠boc=90°，oc=oc△aom≌△bom

aco=∠bco，ac=bc

cm=cm△acm≌△cbm

When ACM is an isosceles triangle, CBM is an isosceles triangle.

When ac=cm

ac=22, then cm=222

m1: (0,2+2 2) m2: (0,2-2 2) when ao=co

Then m: (0,0).

When am=cm

Let m: (0,m).

2^2+(m+2)^2=(m-2)^2

m=-1/2

Then m: (0,-1 2).

p:(t-1,0)q:(0,t+2)

The pq straight line is y=(2+t)x (1-t)+t+2 and y=x+2 and the intersection g is ( (t 2-1) (2t+1),(t 2+4t+1) (2t+1) )

e:(t-3)/2,(t+1)/2

eg=(5t+1)/(4t+1)√2

Change.

It is easy to find point A (-2,0), point B (2,0), point C (0,2) (1), so the AC line y=x

2.(2) When T is less than 2, the area of the triangle PCQ = triangle POQ - triangle POC = (2-t) T 2, when T is greater than 2 and less than or equal to 4, the area = , 2 + 2 root number 2) (0, 0) two points (4).

a(2,0)

b(0,4)

The second question should be: AB as the side to make an equilateral triangle to prepare the beard ABP. If it is isosceles, there will be countless solutions.

It can be seen that the point p must be on the mid-vertical line of AB. The straight line perpendicular to ab can be set as y=1 2x+b because it is a perpendicular line, so the midpoint of ab is (1,2) point. Bring in the solution to get b = 3 2.

y=1/2x+。Let p(x,1 2x+ be pm x in m

Because it is equilateral, ab=bp=2 times the root number 5. Pythagorean in RT APM.

i.e. (x-2)+

Because in the first quadrant.

Therefore, x>0 solves x=1+2 root number three, and then substitutes it to understand p(1+2 times root number three, 2 + root number three).

With the foundation of the second question, the third question is easy to do, set the straight line cd: y=-2x+b and bring in p to get b=5, the root number three + 4

Solve the coordinates of c and d respectively (three times the root number + 2,0) and (0,5 times the root number three + 4), and calculate s ocd = four vertical rolls (91 + 40 root number three) minus s aob

The quadrilateral area is 75 4+10 root number three.

As for the reason why the answer is so disgusting, here's why: it's possible that it's even wrong.

It's possible that the question is wrong.

It's possible that you, Teacher Yu Rolling, have made a mistake in your head......

Points are awarded for the perfect process!

It is better to teach a man to fish than to teach him to fish.

1. Because the function intersects with the x and y axes at two points ab, i.e., a(2,0)b(0,4).

2. Let the p coordinate be (m,n) because ab is the edge, so there is the following situation: 1. |ab|=|ap|or |ab|=|bp|

In the calculation according to the coordinates (due to the limited time and conditions, in the next will not be calculated one by one.

3. Calculate the coordinates of point P according to the formula of the missing line of the parallel stool in the jujube smash standard, and then calculate the coordinates of the two points of c and d according to the point of p, and the y of c is 0 in the coordinates of these two points, and after finding the coordinates of these two points, you can find the area of the quadrilateral, and find the area of the quadrilateral, you can find the area of the three triangles respectively or find it directly, see which one is simple.

1) Let y=0, get x=2, let x=0, get y=4, so the coordinates of a and b are a(2,0) and b(0,4) respectively, the slope of the straight line ab k=4 (-2)=-2, and the straight line ab is y=-2 (x-2), that is, 2x+y-4=0

o The distance to 2x+y-4=0 is:

2) Let the parabola be y=a(x-2) (vertex y=a(x-h) +k, where (h,k) is the vertex of the parabola).

Substituting b(0,4) into the parabolic equation gives 4=a(0-2) and a=1, so the analytical formula of Wu Xiao for finding the parabola is y=(x-2).

1) When x=0, y=2 0+4=4 ∴b（0,4）

When y=0, 0= 2x+4, the solution is x=2 a(2,0).

2) p(6,2)【bp=ba】regret p(4,6)【ap=ab】 p(3,3)【bp=ap】

3) When the point p coordinate is (3,3), the analytic formula of the straight line cd is y= 2x+b.(Because CD and AB are Pingbi Shen Hall, K is the same, B is different).

Substitute coordinates (3,3) into the analytic formula.

3 = 2 3 + b, the solution is b = 9 ∴y=－2x+9.

When y=0, 0= 2x+9, the solution is x=, that is, oc=

When x=0, y=2 0+9=9, i.e., od=9

s quadrilateral abcd=s doc s aob=9

Let the analytic formula of the straight line cd be y= 2x+b. when the point p coordinate is (6,2) or (4,6).

Substitute (6,2) into the analytic formula.

2 = 2 6 + b, the solution is b = 14 ∴y=－2x+14.

When y=0, 0= 2x+14, the solution is x=7, i.e., oc=7

When x=0, y=2 0+14=14, i.e., od=14

S quadrilateralabcd=s doc s aob=14 7 2 4 2 2=45

1)y=-2x+4,y=0,x=2;x-=0,y=4, a(2,0),b(04);

2),ab=ap,bp x-axis,p(4,4),3)k-with -2,p(4,4),yp=-2x+12,c(6,0),d(012).

s quadrilateral abcd=s ocd-s ab=1 2(12*6-2*4)=32

1）a(2，0);b(0,4) (2) p(6,2) (3) passes the lead starvation p point and is parallel to the straight line ab, so the slope of the straight line is the same as y=-2x+4, that is, the straight line y=-2x+b, bring the p point in, and the straight line is y=-2x+14, so the point c(7,0) and the point d(0,14).