Question

已知点A（8，0）及在第一象限的动点P（X，Y），且X+Y=10，△OPA的面积为S。求；

Answer 1

由图知：S=底*高/2=8*Y/2=4Y
因为X+Y=10，所以S=4（10-X）
因为点P在第1象限，所以0〈X〈10
当S=12时。由1问知，X=7，Y=3
∴P(7,3)

Answer 2

1.
OA
=
8,
OA上高
=
P纵坐标
=Y
=
10
-
X
S
=
(1/2)*8*Y
=
4Y
=
4(10
-
X)
2.
S
=
3(10
-
X)
=
12,
X
=
6,
P(6,
4)
3.
(a)
OP
=
OA
=
8
OP²
=
X²
+
(10
-
X)²
=
64
X
=
5
±
√7
P(5
+
√7,
5
-
√7)
P(5
-
√7,
5
+√7)
(b)
OA
=
AP
=
8
AP²
=
(X
-
8)²
+
(10
-
X)²
=
64
X
=
9
±
√31
P(9
+
√31,
1
-
√31)
(第象限舍去)
P(9
-
√31,
1
+
√31)
(c)
OP
=
PA
OP²
=
PA²
X²
+
(10
-
X)²
=
(X
-
8)²
+
(10
-
X)²
X
=
32/9
P(32/9,
58/9)
共有4