Odpowiedź :
P =(1;4), Q=(2;7), R =(-3;5) środki boków AB,BC,CA
trójkąta ABC.Znajdź A,B,C
A =(x1;y1) , B =(x2;y2), C =(x3;y3)
[x1 +x2]/2 = 1 oraz [y1 + y2}/2 = 4 czyli
x1 +x2 =2 oraz y1 +y2 =8
[x2 +x3[/2 =2 oraz [y2 + y3]/2 =7
czyli x2 +x3 =4 oraz y2 + y3 = 14
[x1+x3]/2 = -3 oraz [y1 + y3]/2 =5
czyli x1 + x3 =-6 oraz y1 + y3 = 10
Mamy układ 6 równań z 6 niewiadomymi :x1,y1,x2,y2,x3,y3
x1 +x2 =2 oraz y1 +y2 =8
x2 +x3 =4 oraz y2 + y3 = 14
x1 +x3 = -6 oraz y1 + y3 = 10
----------------------------------------
x2 -x3 = 8
x2 +x3 = 4
-------------
2x2 =12
x2 = 6
x1 =2 -x2 = 2 -6 =-4
x3 =4 -x2 =4-6 =-2
---------------------
x1 =-4;x2 =6;x3 = -2
----------------------------
y2 - y3 =8-10 = -2
y2 + y3 =14
---------------------
2y2 = 14 -2 = 12
y2 = 6
y1 = 8 -y2 = 8 - 6 = 2
y3 = 14 - y2 = 14 - 6 = 8
-------------------------------
y1 =2; y2 = 6; y3 =8
Odp.
A =( -4;2)
B =( 6;6)
C =(-2;8)
trójkąta ABC.Znajdź A,B,C
A =(x1;y1) , B =(x2;y2), C =(x3;y3)
[x1 +x2]/2 = 1 oraz [y1 + y2}/2 = 4 czyli
x1 +x2 =2 oraz y1 +y2 =8
[x2 +x3[/2 =2 oraz [y2 + y3]/2 =7
czyli x2 +x3 =4 oraz y2 + y3 = 14
[x1+x3]/2 = -3 oraz [y1 + y3]/2 =5
czyli x1 + x3 =-6 oraz y1 + y3 = 10
Mamy układ 6 równań z 6 niewiadomymi :x1,y1,x2,y2,x3,y3
x1 +x2 =2 oraz y1 +y2 =8
x2 +x3 =4 oraz y2 + y3 = 14
x1 +x3 = -6 oraz y1 + y3 = 10
----------------------------------------
x2 -x3 = 8
x2 +x3 = 4
-------------
2x2 =12
x2 = 6
x1 =2 -x2 = 2 -6 =-4
x3 =4 -x2 =4-6 =-2
---------------------
x1 =-4;x2 =6;x3 = -2
----------------------------
y2 - y3 =8-10 = -2
y2 + y3 =14
---------------------
2y2 = 14 -2 = 12
y2 = 6
y1 = 8 -y2 = 8 - 6 = 2
y3 = 14 - y2 = 14 - 6 = 8
-------------------------------
y1 =2; y2 = 6; y3 =8
Odp.
A =( -4;2)
B =( 6;6)
C =(-2;8)