Figure 1:

1, let a straight line y = kx+b.

Substituting b and c, we get k= 1/2, and b=-2.

The linear equation is y= 1/2x-2.

For p,-1= 1/2x-2, x=2.

PQ//x axis, ordinate, etc. , q ordinate-1,

Substitute Y=3/X and the abscissa is -3.

P(2，- 1)

Q(-3，- 1)

2. The hyperbola Y=5/X is symmetrical about the origin.

Suppose P(x, y)Q(-x, -y)

According to symmetry, there are:

s△PQB = 4×I( 1/2)xyI = 5/2 x4 = 10

Figure 2:

1, ① Find the value of k.

K= 1*3=3

② Since the abscissa of point E is m, substituting y=k/x to get the ordinate of point E is k/m..

So E(m, k/m)

E is also the midpoint of BD, so the ordinate of A is twice that of E, which is 2k/m, and thus A(m/2, 2k/m) is obtained.

It is also known that half of the sum of the abscissa of C and A is equal to M (because E is between B and C, and the abscissa of B is the abscissa of A).

So the abscissa of c is 3m/2.

③ When the angle ABD=45 degrees, find the value of m..

That is, ABCD is a square.

Then AB = BC = 2OB.

6/M=(M/2)*2=M

M squared =6

The square root of M = 6 or the square root of M =-6.

2.( 1) Substitutes X=-2 to get Y=4 ∴A(-2,4).

Substitute Y=-2 to get X=4 ∴B(4,-2).

Substituting a and b into Y=KX+B gives k=- 1 b=2.

∴ y=-x+2

(2) let Y=0, then X=2 ∴y=-x+2 and the intersection of x axis c (2,0).

S△AOB = S△AOC+S△BOC = 2 * 4 * 1/2+2 * 2 * 1/2 = 6

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