ОР-1
Вариант 1
Дано: A(2;3), B(5;0), C(–1;–6) и D(–4;–3); О – середина отрезка AD.
Определите вид ABC и четырехугольника ABCD.
Составьте уравнение окр.(O;AO).
Решение:
1) A(2;3)
B(5;0)
AB2=(2–5)2+(3–0)2=(–3)2+32=___+___=_____
AB=_____
2) B(5;0)
C(–1;–6)
BC2=(5–(–1))2+(0–(–6))2=(5+1)2+(0+6)2=62+62=___+___=_____
BC=_____
3) C(–1;–6)
D(–4;–3)
CD2=(–1–(–4))2+(–6–(–3))2=(–1+4)2+(–6+3)2=32+(–3)2=___+___=_____
CD=_____
4) A(2;3)
D(–4;–3)
AD2=(2–(–4))2+(3–(–3))2=(2+4)2+(3+3)2=62+62=___+___=_____
AD=_____
5) A(2;3)
C(–1;–6)
AC2=(2–(–1))2+(3–(–6))2=(2+1)2+(3+6)2=32+92=___+___=_____
AC=_____
6) B(5;0)
D(–4;–3)
BD2=(5–(–4))2+(0–(–3))2=(5+4)2+(0+3)2=92+32=___+___=_____
BD=_____
7) ABC – _______________________________________________
8) ABCD – _______________________________________________
9) A(2;3)
D(–4;–3)
O
=
=( ; ) – центр окружности
AO=AD:2=______
10) (х–____)2+(у–_____)2=(_____)2
_________________________–уравнение окр.(О;АО)
ОР-1
Вариант 2
Дано: A(–5;–1), B(3;3), C(9;1) и D(1;–3); ACBD=O.
Определите вид ABC и четырехугольника ABCD.
Составьте уравнение окр.(O;AO).
Решение:
1) A(–5;–1)
B(3;3)
AB2=(–5–3)2+(–1–3)2=(–8)2+(–4)2=___+___=_____
AB=_____
2) B(3;3)
C(9;1)
BC2=(3–9)2+(3–1)2=(–6)2+22=___+___=_____
BC=_____
3) C(9;1)
D(1;–3)
CD2=(9–1)2+(1–(–3))2=82+(1+3)2=___+___=_____
CD=_____
4) A(–5;–1)
D(1;–3)
AD2=(–5–1)2+(–1–(–3))2=(–6)2+(–1+3)2=(–6)2+22=___+___=_____
AD=_____
5) A(–5;–1)
C(9;1)
AC2=(–5–9)2+(–1–1)2=(–14)2+(–2)2=___+___=_____
AC=_____
6) B(3;3)
D(1;–3)
BD2=(3–1)2+(3–(–3))2=22+(3+3)2=22+62=___+___=_____
BD=_____
7) ABC – _______________________________________________
8) ABCD – _______________________________________________
9) A(–5;–1)
C(9;1)
O
=
=( ; ) – центр окружности
AO=AC:2=______
10) (х–____)2+(у–_____)2=(_____)2
_________________________–уравнение окр.(О;АО)
ОР-1
Вариант 3
Дано: A(–3;3), B(1;6), C(7;–2) и D(3;–5); О – середина отрезка AD.
Определите вид ABC и четырехугольника ABCD.
Составьте уравнение окр.(O;AO).
Решение:
1) A(–3;3)
B(1;6)
AB2=(–3–1)2+(3–6)2=(–4)2+(–3)2=___+___=_____
AB=_____
2) B(1;6)
C(7;–2)
BC2=(1–7)2+(6–(–2))2=(–6)2+(6+2)2=(–6)2+82=___+___=_____
BC=_____
3) C(7;–2)
D(3;–5)
CD2=(7–3)2+(–2–(–5))2=42+(–2+5)2=42+32=___+___=_____
CD=_____
4) A(–3;3)
D(3;–5)
AD2=(–3–3)2+(3–(–5))2=(–6)2+(3+5)2=(–6)2+82=___+___=_____
AD=_____
5) A(–3;3)
C(7;–2)
AC2=(–3–7)2+(3–(–2))2=(–10)2+(3+2)2=(–10)2+52=___+___=_____
AC=_____
6) B(1;6)
D(3;–5)
BD2=(1–3)2+(6–(–5))2=(–2)2+(6+5)2=(–2)2+112=___+___=_____
BD=_____
7) ABC – _______________________________________________
8) ABCD – _______________________________________________
9) A(–3;3)
D(3;–5)
O
=
=( ; ) – центр окружности
AO=AD:2=______
10) (х–____)2+(у–_____)2=(_____)2
_________________________–уравнение окр.(О;АО)
ОР-1
Вариант 4
Дано: A(–2;–4), B(–3;3), C(4;4) и D(5;–3); ACBD=O.
Определите вид ABC и четырехугольника ABCD.
Составьте уравнение окр.(O;AO).
Решение:
1) A(–2;–4)
B(–3;3)
AB2=(–2–(–3))2+(–4–3)2=(–2+3)2+(–7)2=12+(–7)2=___+___=_____
AB=_____
2) B(–3;3)
C(4;4)
BC2=(–3–4)2+(3–4)2=(–7)2+(–1)2=___+___=_____
BC=_____
3) C(4;4)
D(5;–3)
CD2=(4–5)2+(4–(–3))2=(–1)2+(4+3)2=(–1)2+72=___+___=_____
CD=_____
4) A(–2;–4)
D(5;–3)
AD2=(–2–5)2+(–4–(–3))2=(–7)2+(–4+3)2=(–7)2+(–1)2=___+___=_____
AD=_____
5) A(–2;–4)
C(4;4)
AC2=(–2–4)2+(–4–4)2=(–6)2+(–8)2=___+___=_____
AC=_____
6) B(–3;3)
D(5;–3)
BD2=(–3–5)2+(3–(–3))2=(–8)2+(3+3)2=(–8)2+62=___+___=_____
BD=_____
7) ABC – _______________________________________________
8) ABCD – _______________________________________________
9) A(–2;–4)
C(4;4)
O
=
=( ; ) – центр окружности
AO=AC:2=______
10) (х–____)2+(у–_____)2=(_____)2
_________________________–уравнение окр.(О;АО)
ПР-1
Вариант 1
Дано: A(–3;–4), B(–2;1), C(3;2) и D(2;–3); ACBD=O.
Определите вид ABC и четырехугольника ABCD.
Составьте уравнение окр.(O;AO).
Решение:
1) A(–3;–4)
B(–2;1)
AB2=(___–___)2+(___–___)2=___+___=_____
AB=_____
2) B(–2;1)
C(3;2)
BC2=(___–___)2+(___–___)2=___+___=_____
BC=_____
3) C(3;2)
D(2;–3)
CD2=(___–___)2+(___–___)2=___+___=_____
CD=_____
4) A(–3;–4)
D(2;–3)
AD2=(___–___)2+(___–___)2=___+___=_____
AD=_____
5) A(–3;–4)
C(3;2)
AC2=(___–___)2+(___–___)2=___+___=_____
AC=_____
6) B(–2;1)
D(2;–3)
BD2=(___–___)2+(___–___)2=___+___=_____
BD=_____
7) ABC – _______________________________________________
8) ABCD – _______________________________________________
9) A(–3;–4)
C(3;2)
O
= ( ; ) – центр окружности
AO=AC:2=______
10) (х–____)2+(у–_____)2=(_____)2
_________________________–уравнение окр.(О;АО)
ПР-1
Вариант 2
Дано: A(–3;–3), B(2;9), C(8;1) и D(3;–11); O – середина AD.
Определите вид ABC и четырехугольника ABCD.
Составьте уравнение окр.(O;AO).
Решение:
1) A(–3;–3)
B(2;9)
AB2=(___–___)2+(___–___)2=___+___=_____
AB=_____
2) B(2;9)
C(8;1)
BC2=(___–___)2+(___–___)2=___+___=_____
BC=_____
3) C(8;1)
D(3;–11)
CD2=(___–___)2+(___–___)2=___+___=_____
CD=_____
4) A(–3;–3)
D(3;–11)
AD2=(___–___)2+(___–___)2=___+___=_____
AD=_____
5) A(–3;–3)
C(8;1)
AC2=(___–___)2+(___–___)2=___+___=_____
AC=_____
6) B(2;9)
D(3;–11)
BD2=(___–___)2+(___–___)2=___+___=_____
BD=_____
7) ABC – _______________________________________________
8) ABCD – _______________________________________________
9) A(–3;–3)
D(3;–11)
O
= ( ; ) – центр окружности
AO=AD:2=______
10) (х–____)2+(у–_____)2=(_____)2
_________________________–уравнение окр.(О;АО)
ПР-1
Вариант 3
Дано: A(–3;–1), B(–5;7), C(3;5) и D(5;–3); ACBD=O.
Определите вид ABC и четырехугольника ABCD.
Составьте уравнение окр.(O;AO).
Решение:
1) A(–3;–1)
B(–5;7)
AB2=(___–___)2+(___–___)2=___+___=_____
AB=_____
2) B(–5;7)
C(3;5)
BC2=(___–___)2+(___–___)2=___+___=_____
BC=_____
3) C(3;5)
D(5;–3)
CD2=(___–___)2+(___–___)2=___+___=_____
CD=_____
4) A(–3;–1)
D(5;–3)
AD2=(___–___)2+(___–___)2=___+___=_____
AD=_____
5) A(–3;–1)
C(3;5)
AC2=(___–___)2+(___–___)2=___+___=_____
AC=_____
6) B(–5;7)
D(5;–3)
BD2=(___–___)2+(___–___)2=___+___=_____
BD=_____
7) ABC – _______________________________________________
8) ABCD – _______________________________________________
9) A(–3;–1)
C(3;5)
O
= ( ; ) – центр окружности
AO=AC:2=______
10) (х–____)2+(у–_____)2=(_____)2
_________________________–уравнение окр.(О;АО)
ПР-1
Вариант 4
Дано: A(–5;4), B(1;6), C(5;2) и D(3;–4); O – середина AD.
Определите вид ABC и четырехугольника ABCD.
Составьте уравнение окр.(O;AO).
Решение:
1) A(–5;4)
B(1;6)
AB2=(___–___)2+(___–___)2=___+___=_____
AB=_____
2) B(1;6)
C(5;2)
BC2=(___–___)2+(___–___)2=___+___=_____
BC=_____
3) C(5;2)
D(3;–4)
CD2=(___–___)2+(___–___)2=___+___=_____
CD=_____
4) A(–5;4)
D(3;–4)
AD2=(___–___)2+(___–___)2=___+___=_____
AD=_____
5) A(–5;4)
C(5;2)
AC2=(___–___)2+(___–___)2=___+___=_____
AC=_____
6) B(1;6)
D(3;–4)
BD2=(___–___)2+(___–___)2=___+___=_____
BD=_____
7) ABC – _______________________________________________
8) ABCD – _______________________________________________
9) A(–5;4)
D(3;–4)
O
= ( ; ) – центр окружности
AO=AD:2=______
10) (х–____)2+(у–_____)2=(_____)2
_________________________–уравнение окр.(О;АО)