Current location - Training Enrollment Network - Mathematics courses - The detailed explanation of junior high school math problems (if there is a simple one, it should be simple) is a bit difficult for me to make do with. Thank you.
The detailed explanation of junior high school math problems (if there is a simple one, it should be simple) is a bit difficult for me to make do with. Thank you.
Solution: (1)① Let the analytical formula of straight line AB be y=kx+3,

Substituting x=-4 and y=0 gives: -4k+3=0,

∴k=,

The analytical formula of the straight line is y=x+3,

② It is known that the coordinate of point P is (1, m),

∴m=× 1+3=;

(2)∫PP′∨AC,

△PP′D∽△ACD,

= =, that is =,

∴a=;

(3) When the point p is in the first quadrant,

1) If ∠ AP ′ c = 90, P ′ a = P ′ c (as shown in figure 1).

Passing point p' is the axis of P'H⊥x at point H.

∴pp′=ch=ah=p′h=ac.

∴2a=(a+4),

∴a=,

2) If ∠ p ′ ac = 90 and p ′ a = c,

Then PP'= AC,

∴2a=a+4,

∴a=4,

3) If ∠ p ′ ca = 90,

Then points p' and p are in the first quadrant, which contradicts the condition.

∴△ p ′ ca cannot be an isosceles right triangle with C as the right vertex.

∴a All values that meet the conditions are a=4 or.