CO= 14

2

T=7

The first two questions are very simple, and the key is the third question.

three

When P is on AB, D (0 0,8), let P (x x,8), and get Q (x/2,4) or (x/2, 12).

Line OA: y = 4x/3

Then 4=4*X/2/3, X=6, and then p (12,8).

Or 12=4*X/2/3, X= 18 (irrelevant, don't).

When P is on the side of BC, the vertical line of Y axis passes through P, the vertical line of X axis passes through Q, and the two lines intersect at point M by constructing congruence.

QM goes through AB to N.

Note: this method still adopts the coordinate idea of "horizontal and vertical"

D (0 0,8), let P( 14, y), Q(X, 4X/3).

The right triangle DNQ is equal to QMP.

X=4X/3-Y

3X/3-8=Y-X

solve an equation

X=24/5，Y=8/5

At this time, P( 14, 8/5)

Note: The process and calculation data are not necessarily accurate and are for reference only. The key is to understand the method.