( 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