Because S△ABC=S△ABP, the distance between P and AB is equal to the length of AC, which is 2.
The formula for the distance from a point to a straight line is | a+√ 3 *1/2-√ 3 |/√ [1? +(√3)? ]=2, the solution is: a = √ 3/2 4.
Because P is in the second quadrant, which is A.
5.( 1) Because the area of the rectangle is to be equally divided, the straight line L must pass through the symmetrical center (3,2) of the rectangle.
Then, the linear equation passing through (3,2) and (0,6) is (y-6)/(x-0)=(2-6)/(3-0), that is, 4x+3y- 18=0.
(2) Obviously, the straight line will not be x=0, so let the straight line equation be y=kx+6.
After the split, it will be discussed in two situations:
I) If left: right = 1:2, it is obviously two trapezoids, and the ratio of the sum of the upper and lower bottoms of the two trapezoids is 1:2.
When y=0, x =-6/k; When y=4, x =-2/k.
Therefore, the equation is: [(-2/k)+(-6/k)]/[12-(-2/k)] =1/2, and the solution is: k=-2.
So the equation of the straight line L at this time is 2x+y-6=0.
Ii) If left: right =2: 1, first determine whether the straight line L intersects OA or AB.
If a straight line passes through point A (6,0), the equation of the straight line is x+y-6=0, and its intersection with y=4 is (2,4).
The sum of the upper and lower bottoms of the left trapezoid is 2+6=8, and the bottom of the right triangle is 6-2=4, just the ratio of the bottoms of the left and right graphs is 2: 1, and the height is 4, which meets the conditions! ! X+y-6=0 is exactly what you want.
To sum up, the equation of the straight line L is 2x+y-6=0 or x+y-6=0.