Question

Napisz równanie prostych zawierających boki trójkąta o wierzchołkach A, B, C.
A = (-3, -3) B= (3,-1) C= (1.5).
Jeśli moge to poprosiłabym także rysunek. Z góry dziękuje :)

Answer 1

A =(-3;-3)
B = (3; -1)
C =(1; 5)
pr AB
y = ax + b
-3 = -3a +b
-1 =3a + b
--------------- dodaję stronami
-4 = 2b --> b = -4 : 2 = -2
-1 = 3a -2
3a = 2-1 = 1 ----> a = 1 : 3 = 1/3
pr AB: y = (1/3) x - 2
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pr BC
y = ax +b
-1 = 3a +b
5 = a + b ----> a = 5 -b
-1 = 3(5-b) +b = 15 - 3b +b = 15 -2b
2b = 15 +1 = 16
b = 16: 2 = 8
a = 5 - b = 5 -8 = -3
pr BC : y = -3x + 8
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pr AC
y = ax +b
-3 = -3a + b
5 = a + b ---> a =5 - b
-3 = -3(5 -b) + b = -15 +3b +b = -15 + 4b
4b -15 = -3
4b = -3 + 15 = 12
b = 12 : 4 = 3
a = 5 - b = 5 - 3 = 2
pr AC : y = 2x + 3
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