Junior high school math finale, junior high school math finale

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There are no pictures, how to do it.

The intersection point of the primary function with the y-axis is a(-2,0),b(0,1). The answer is shown in the picture:

1.Because when x=0, y=6

When x=8, y=0

So we get the system of equations: b=6

8k+b=0

Solution, k = -3 4

b = 6 so y = -3 4x+6

2 Because the triangle APQ is similar to the triangle AOB.

So it should be discussed in two cases.

1) When the triangle APQ is similar to the triangle AOB.

Because the angle aob=90 degrees, there is the Pythagorean theorem, and we get ab=10, so ap ao=aq ab

So t 6 = 10 - 2 t 10

solution, t = 30 11

2) When the triangle AQP is similar to the triangle AOB.

So aq ao=ap ob

So 10-2t 6=t 8

The solution is t=40 11

In this question, we should pay attention to the problem of the correspondence of similar triangles, and consider two situations. Condition) 3QC vertical AO

Since the triangle ACQ = 90 degrees, ACQ is similar to AOB and cq = x

Then: aq qb=cq ob

So 10-2t 10=x 8

x=so the area of the triangle apq can be expressed as:

ap*qc=t*(

Because the area of the triangle APQ is twenty-four-fifths.

So t*(simplify, solution: t1 = 5 + root number 13 (rounded) t2 = 5 - root number 13 (the key to this problem is to use the algebraic formula containing t to express the high of apq) I did it in a bit of a hurry, maybe the result is not right, but the idea is definitely right, you will calculate it yourself.

Solution: (1) Let the analytic formula of the line ab be y=kx+b, and substitute the points a(0,6) and b(8,0) into $ left end}ight$, which is solved to get $ left } end}ight$, the analytic formula of the straight line ab is: y=-$ frac$x+6 (2) set the time of the points p and q to move in t seconds, oa=6, ob=8, the pythagorean theorem can be obtained, ab=10, ap=t, aq=10-2t

There are two cases, 1 when apq aob, frac= frac$$ frac= frac$$t= frac$, 2 when aqp aob $ frac= frac$$ frac= frac$, t=$ frac$, in summary, when t=$ frac$ or $t= frac$, triangle aob with vertices a, p, q is similar

The topic is not so.

As shown in Figure 3, the line y=a(a≠0) that is known to be parallel to the x-axis intersects with the function image of the function y=x, y=1 x, respectively. There is also p(2,0) (1) If two points are crossed and the vertex is on y=x, the line segment ab=8 3 is known, and on the left side of his axis of symmetry, y increases with the increase of x, try to find the parabolic analytic formula that satisfies the condition. (2) After knowing that the parabolic translation of three points obtains the image of y=(9 5)x, and finds the distance from the point p to the straight line ab.

Solution: a(a, a), b(1 a, a).

1 a-a) = 8 3, a = -1 3, a = 3 (rounded) because on the left side of his axis of symmetry, y increases with x, so a<0In the figure, if the symmetry axis of parabola A intersects y=a at the point d, we get ad=4 3, that is, d(5 3, 1 3), so x=-b 2a=5 3, b=5 2, 4ac-b squared 4a=5 3, and c=-145 48

The line y=a(a≠0) parallel to the x-axis and the function image of the function y=1 x intersect respectively, and these two lines will not have 2 intersection points! There is a problem with the topic!

Agree with the upstairs. Those two lines could not have intersected at two points. Is there a mistake? y=1 x 2 and y=a will intersect at two points. But I don't think I learned parabola in junior high school.

Khan: There is a problem with the topic, how can there be two intersections.

Triangles are similar, you can use equal angles, there is already a pair of vertex angles, so there is a pair of angles equal Here, for the sake of simplicity, choose equal right angles Do cm vertical ac cross parabola in m to get the straight line mc slope Substituting the straight line to get mc:y= synchronously find m (, or do mc vertical y-axis parabola in m, this is a good calculation m(0,6) So there is m (, or (0,6) in line with the topic The mobile phone party is not easy to score.

If APN is similar to CMP, cpm = APN, so PCN or CMP must have a right angle.

1 When PCN is at right angles, the addition point of the equation x-2y+14= with the parabola with the straight line cm is the point m (-1 4,55 8).

2 When CMP is a right angle, the angle between the straight line cm and the straight line AC is equal to cab, cm is parallel to the x-axis, and the point m is the symmetry point at point c on the parabola, that is, (0,6).

Glad to help you, I hope my answers are helpful to you.

(1) Solution: y=-1 4x2+6, y=-1 2x

The solution yields x1=6, y1=-3, x2=-4, y2=2

2) The perpendicular bisector of AB intersects the x-axis, the y-axis intersects the two points C and D, and the AB intersects M

1. It is known that the straight line y= - one-half x and the parabola y= - one-quarter +6-1 2x=-1 4x +6

x=6 x=-4

a(6 -3) b(-4 2)

Analysis: (1) The coordinates of point A are known to be (4,0) and the coordinates of point C are (1,2), and the analytical formula of the straight line ac can be obtained according to the "two-point method"; (2) After b is done as bh oa in h, the coordinates of point b can be obtained according to the properties of the isosceles trapezoid, the line segment pq can be represented by the analytical formula of the straight line ac, and the area of amq can be expressed by the known, and the maximum value is obtained according to the properties of the quadratic function; (3) When AMQ is an isosceles triangle with MQ as the waist, there are two cases: QM=QA, QM=MA, which can be solved according to the graph features and the Pythagorean theorem

Answer: Solution: Solution (1) Let the analytic formula of the straight line ac be:

y=kx+b, substituting the points a(4,0),c(1,2) gives 4k+b=0 k+b=2 k+b=2 solution k=-2 3 b=8 3 , y=-2 3 x+8 3 (2) through b as bh oa in h, c(1,2), by the property of the isosceles trapezoidal ah=1, then op=oa-ah-hp=4-1-bn=3-t point q is the point pq=-2 3 (3-t)+8 3 am=oa-om=4-2t s=1 2 am pq=1 2 （4-2t）（2 3 t+2 3 ）=-2 3 t +2 3 t+4 3 ； When t=1 2, s max=3 2

3) There are the following two cases: Qm=Qa, by the property of the three lines of the isosceles triangle, at this time MP=AP, that is, 3-3T=T+1, T=min) Qm=Ma, that is, Qm2=Ma2, from the Pythagorean theorem, MP2+PQ2=MA2, that is, (3-3T) +2 3 T+2 3 ) =(4-2T), T1=59 49 , T2=-1 (rounded) When T= or T1=59 49, AMQ is an isosceles triangle with Mq as the waist.

This question examines the method of finding the analytic formula of a straight line, the representation of the area of a triangle in a coordinate system, the problem of the maximum value of a quadratic function, and the conditions for finding an isosceles triangle).

To illustrate, ** is missing an answer, that is, the rounding of the penultimate line should be added, there are three t; t=4 3 is the answer!!

What area are you talking about?

<> In a planar Cartesian coordinate system, the coordinates of point A are (1, root number 3), and the area of the triangle AOB is root number 3

It's the above question.,This is the original question I found on the answer.,The red horizontal line is the answer**,Pro,Now what questions won't be answered.,We all use the answer.,There are a lot of original questions and similar questions above.,It's easy to search.,No need to log in and register.,Just search for the question OK.The address circled in red above, or directly enter the three words to search for the solution. Come on!

Find the coordinates of the point;

Find the analytic formula of the parabola of the point;

Is there a point on the axis of symmetry of the middle parabola that minimizes the perimeter? If so, find the coordinates of the point; If not, please explain the reason;

Whether there is a point on the parabola below the central axis, the point is the perpendicular line of the axis, the intersection line is at the point, and the line segment is divided into two triangles. Ratio the area of one of the triangles to the area of the quadrilateral? If so, find the coordinates of the point; If not, please explain the reason;

Found on the solution, give you a link to this question.

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