Oblicz obwód trójkąta, którego boki zawierają się w prostych o podanych równaniach.
1) 7x + y + 16 = 0
3x + 4y - 36 = 0
4x - 3y +2 = 0

2) x + 2y - 6 = 0
x - y - 3 = 0
x + 4 = 0

3) 4x + 3y - 2 = 0
3x + 4y +2y = 0
x - y + 3 = 0

4) 3x - 4y - 15 = 0
2x - 5y - 10 = 0
4x - 3y - 6 = 0

+ rozwiązania, błagam chociaż ze 2 przykłady :(

wiem, że w przykładach trzeba obliczyć układy równań na zasadzie, że w 1 przykładzie a z b i b z c.

1

Odpowiedzi

Najlepsza Odpowiedź!
2010-03-29T07:41:00+02:00
2)
x+2y - 6 = 0
x - y - 3 = 0
x +4 = 0
--------------------
x+2y = 6
x-y = 3 --->y = x - 3
--------------------
x + 2*(x -3) = 6
x +2x - 6 = 6
3x = 12
x = 4
y = 4- 3 = 1
===============
x+2y = 6
x + 4 = 0 ---> x = -4
----------
-4 +2y = 6
2y = 6+4 = 10
y = 10:2 = 5
x = - 4 oraz y = 5
=======================
x-y = 3
x + 4 = 0 ---> x = -4
-----------------
-4 - y = 3
y = -4-3 = -7
x = -4 oraz y = -7
Mamy punkty:
A = (- 4; -7), B = (4; 1), C = (-4; 5)
I AB I² = (4 -(-4))² + (1 -(-7))² = 8² +8² = 2*8²
I AB I =√(8²*2) = 8√2
I AC I² = (-4 -(-4))² + ( 5 -(-7))² = 0 + 12² = 144
I AC I = √144 = 12
I BC I² = (-4 -4)² + (5 -1)² = (-8)² + 4² =64 +16 = 80 = 16*5
I BC I = √16 * √5 = 4√5
L - obwód Δ ABC
L = 8√2 + 12 + 4√5 = 4*(2√2 +3 + √5)
================================================
4)
3x -4y -15 = 0
2x - 5y - 10 = 0
4x - 3y - 6 = 0
==============
3x - 4y = 15,mnożymy przez 2
2x - 5y = 10 , mnożymy przez (-3)
----------------------
6x -8y = 30
-6x +15 y = -30, dodajemy stronami
----------------
7y = 0 ---> y = 0
3x = 15 +4*0 = 15 ---> x = 5
Mamy x = 5 oraz y = 0
-----------------------------------

3x - 4y = 15 / * 4
4x - 3y = 6 / *(-3)
------------------------
12 x -16y = 60
-12x + 9y - -18, dodajemy stronami
-----------------------
-7y = 42 --> y = -6
3x = 15 +4*(-6) = 15 - 24 = -9 --> x = -3
Mamy x = -3 oraz y = -6
===================================
2x - 5y = 10 / * (-2)
4x - 3y = 6
-----------------------------
-4x + 10y = -20
4x - 3y = 6 , dodajemy stronami
---------------------
7y = -14 ---> y = -2
2x = 10 +5*(-2) = 10 - 10 = 0 --> x = 0
Mamy x = 0 oraz y = -2
=============================
Mamy punkty : A =(-3; -6), B = (5; 0), C = ( 0; -2)
I AB I² = (5-(-3))² + (0 -(-6))² = 8² + 6² = 64 + 36 = 100
I AB I = √100 = 10
I BC I² = (0-5)² + (-2 -0 )² = (-5)² + (-2)² = 25 + 4 = 29
I BC I = √29
I AC I² = (0-(-3))² + (-2 -(-6))² = 3² + 4² = 9 +16 = 25
I AC I = √25 = 5
L = 10 + √29 + 5 = 15 + √29
==========================

1)
7x +y + 16 = 0
3x +4y - 36 = 0
4x - 3y + 2 = 0
=================
7x + y = -16 / * (-4)
3x + 4y = 36
-----------------------
-28x -4y = 64
3x + 4y = 36, dodajemy stronami
-25x = 100 ---> x = -4
y = -7x - 16 = -7*(-4) -16 = 28 - 16 = 12
Mamy x = -4 oraz y = 12
---------------------------------
7x + y = -16 / *3
4x - 3y = -2
----------------------
21x +3y = -48
4x - 3y = -2 , dodajemy stronami
--------------------
25x = -50 ---> x = -2
y = -16 -7x= -16 -7*(-2) = -16 +14 = -2
Mamy x = -2 oraz y = -2
-------------------------------
3x +4y = 36 / *3
4x - 3y = -2 / *4
------------------------
9x +12y = 108
16x -12y = -8 , dodajemy stronami
-----------------
25x = 100 --> x = 4
4y = 36 - 3x = 36 -3*4 = 36 - 12 = 24 ---> y = 6
Mamy x = 4 oraz y = 6
-------------------------------
Mamy A = (-2; -2), B = (4; 6), C = (-4; 12)
I AB I² = (4-(-2))² + (6 -(-2))² = 6² + 8² = 36 + 64 = 100
I AB I = √100 = 10
I AC I² = (-4 -(-2))² + (12-(-2))² = (-2)² + 14² = 4 + 196 = 200
I AC I = √100 *√2 = 10√2
I BC I² = (-4-4)² + (12 -6)² = (-8)² + 6² = 64 +36 = 100
I BC I = √100 = 10

L = 10 + 10√2 + 10 = 20 +10√2
===============================
============================
4 3 4