The linear analytical formula y=ax+b B, and the c coordinate brings into the solvable a=- 1 b=- 1. The linear analytical formula y=-x- 1.
(2) From the title, we can see that
OB=OA=OC= 1
So BC=CA=√2 AB=2.
So △ABC is an isosceles right triangle.
∠ ABC=∠ BAC=45
∫BC∨AP
∴∠PAB=45
So ∠ cap = 90.
CA⊥AP
∴AP's linear analytical formula is y=-x- 1+2=-x+ 1.
∵ Parabolic formula y=x? - 1
The coordinates of the two intersections of AP- parabola are A( 1.0)P(-2.3) respectively.
S△ABP=3*2*( 1/2)=3
s△ABC = 1/2 * 1 * 2 = 1
The area of quadrilateral ACBP =S△ABP+S△ABC=4.
(3) If it exists, the m coordinate is (a.a? -1)(M coordinate satisfies parabolic formula)
According to the two questions, P(-2.3) is AP=3√ 2 (let p be PK⊥X axis, then △PKA is an isosceles right triangle with angle pab = 45).
In △APC, ∠ PAC = 90, so PC=2√5.
∫PCA∽△AMG
∴MG:AP=AG:AC
That is (a? - 1):3√ 2=(a- 1):√ 2
Get a= 1 or a=2.
When a= 1, the coincidence of point m and point a does not hold.
So a=2
The coordinate of point M is (2.3)