2. Find the analytical formula of linear function:
(1) Let a point on a straight line be (x, y) and use the two-point formula:
[y-( 1/3)]/[( 1/3)-3]=(x- 1)/(-3- 1)。
Simplify, and get the analytical formula of 2x-3y- 1 = 0- straight line;
(2) y=kx+b, and substitute the coordinates of (8,0) and (0,4) into the equation to obtain:
8k+b=0，
4=0*k+b，b=4，
8k+4=0，k=- 1/2。
∴ y =-( 1/2) x+4,-is the analytical formula for finding a straight line;
Turn it into a general formula: x+2y-8=0.
(3) Given the coordinates of two points on a straight line (3, 1), (-3, -3), use the two-point formula:
(y- 1)/(-3- 1)=(x-3)/(-3-3)。
y- 1=(-4)(x-3)(-6)。
(-3)(y- 1)=2(x-3)。
2x-3y-3 = 0-that is, the required analytical formula.
Because k = 2/3 >: 0, so the straight line AB passes through the first, fourth and third quadrants, which meets the requirements of the question.
3. Let the analytical formula of the straight line be: y=kx+b,
∵ The found straight line is parallel to the known straight line y=-x+ 1, and the slopes of the two straight lines are equal, that is, k=- 1.
Then substitute the coordinates (2,5) of each point on the straight line into y=kx+b, and get:
(- 1)*2+b=5，b=7，
∴y =-x+7- Analytical formula of straight line.
4.y=(m-3)x+n+4
(1) When m-3
(2) When -n+4
(3) When m-3 = 1m = 4, -n+4 = 0 and n = 4, the straight line passes through the origin, and the analytical formula of the straight line at this time is: y = x.
Of course, as long as "-n+4=0", the straight line also passes through the origin. The analytical formula of the straight line at this time is: y = (m-3) x 。
5. Point P(x, y) is in the first quadrant, x+y=8, and the areas of point A (6 6,0) and △ OPA are s:
(1) Find the analytical expressions of s and x;
The problem is that the area of △OPA is s =( 1/2)| OA | * py(py- the ordinate of point py).
S=( 1/2)*6*y
=3y
=3(8-x) [x+y=8，y=8-x]
∴S=-3x+24 x∈(0，6)。
(2) When x=5, the area of the triangle is -3 * 5+24 = 9 (area unit).
(3) The area of triangle OPA cannot be greater than 24,
Because the abscissa of a (6 6,0). It's 6