Current location - Training Enrollment Network - Mathematics courses - The math problem of analytic function in the second day of junior high school is solved quickly. thank you
The math problem of analytic function in the second day of junior high school is solved quickly. thank you
Solution:

( 1)

Because a two-point formula of (0,4) c (4,0) can establish a linear equation.

Therefore, the analytical formula of the straight line where AC is located is Y=-X+4.

(2)

Extend BA to point e so that AE=AB. At this time, it is easy to know E (4,8) and PB=PE to prove the congruence of triangles.

So PB+PO=PE+PO

From the fact that the sum of any two sides of a triangle is greater than the third side, it can be seen that when P is the intersection of EO and AC, PB+PO is the smallest.

Because e (4,8) o (0,0), we know that the analytical formula of EO is Y=2X.

At the same time, EO and AC can get X=4/3.

That is, when PB+PO is minimum, the coordinates of point P are (4/3, 8/3).

(3)

PB=PC+CQ

Prove:

ABC is an isosceles right triangle.

PB & gtAB = AC & gtPC, so you must find a point D on PB, so that PC=PD.

Because PQ is the bisector of the angle BPC, all triangles PCQ are equal to triangles PDQ(SAS is an edge axiom).

So CQ=QD, angle PDQ= angle PCQ=45 degrees.

So BDQ angle = 135 degrees.

Because PB is the bisector of the angle ABC, the angle DBQ=22.5 degrees.

Because the sum of the internal angles of the triangle is 180 degrees, the DQB angle =22.5 degrees.

So DBQ is an isosceles triangle, that is, DB=DQ.

So PB=PD+BD=PD+DQ=PD+CQ=PC+CQ.

Certificate of completion

thank you