Determine a equação reduzida da reta que passa pelos pontos A e B.
a) A (1,5) e B (3, 9)
b) A (-1,2 ) e B (0,-2)
c) A(8,-1 ) e B (24,5)
Soluções para a tarefa
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Para isso vamos utilizar determinantes:
![\left[\begin{array}{ccc}X&Y&1\\1&5&1\\3&9&1\end{array}\right] \left[\begin{array}{ccc}X&Y&1\\1&5&1\\3&9&1\end{array}\right]](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DX%26amp%3BY%26amp%3B1%5C%5C1%26amp%3B5%26amp%3B1%5C%5C3%26amp%3B9%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+)
Utilizando a regra de Sarrus:
![\left[\begin{array}{ccccc}X&Y&1&X&Y\\1&5&1&1&5\\3&9&1&3&9\end{array}\right] \left[\begin{array}{ccccc}X&Y&1&X&Y\\1&5&1&1&5\\3&9&1&3&9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7DX%26amp%3BY%26amp%3B1%26amp%3BX%26amp%3BY%5C%5C1%26amp%3B5%26amp%3B1%26amp%3B1%26amp%3B5%5C%5C3%26amp%3B9%26amp%3B1%26amp%3B3%26amp%3B9%5Cend%7Barray%7D%5Cright%5D)
det = (5x +3y +9) - (y +9x +15)
-4x +2y - 6 = 0
Isolando o Y:
2y = 6 + 4x
y = 2x + 3
________________________________________________________
As seguintes são o mesmo procedimento:
![\left[\begin{array}{ccccc}X&Y&1&X&Y\\-1&2&1&-1&2\\0&-2&1&0&-2\end{array}\right] \left[\begin{array}{ccccc}X&Y&1&X&Y\\-1&2&1&-1&2\\0&-2&1&0&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7DX%26amp%3BY%26amp%3B1%26amp%3BX%26amp%3BY%5C%5C-1%26amp%3B2%26amp%3B1%26amp%3B-1%26amp%3B2%5C%5C0%26amp%3B-2%26amp%3B1%26amp%3B0%26amp%3B-2%5Cend%7Barray%7D%5Cright%5D)
det = (2x + 2) - (y -2x)
4x + y + 2 = 0
y = -4x-2
__________________________________________
![\left[\begin{array}{ccccc}X&Y&1&X&Y\\8&-1&1&8&-1\\24&5&1&24&5\end{array}\right] \left[\begin{array}{ccccc}X&Y&1&X&Y\\8&-1&1&8&-1\\24&5&1&24&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7DX%26amp%3BY%26amp%3B1%26amp%3BX%26amp%3BY%5C%5C8%26amp%3B-1%26amp%3B1%26amp%3B8%26amp%3B-1%5C%5C24%26amp%3B5%26amp%3B1%26amp%3B24%26amp%3B5%5Cend%7Barray%7D%5Cright%5D)
det = (-x+24y+40)-(8y+5x-24)
-6x +16y+64 =0
16y=-64+6x
y=-4+(3/8)x
Utilizando a regra de Sarrus:
det = (5x +3y +9) - (y +9x +15)
-4x +2y - 6 = 0
Isolando o Y:
2y = 6 + 4x
y = 2x + 3
________________________________________________________
As seguintes são o mesmo procedimento:
det = (2x + 2) - (y -2x)
4x + y + 2 = 0
y = -4x-2
__________________________________________
det = (-x+24y+40)-(8y+5x-24)
-6x +16y+64 =0
16y=-64+6x
y=-4+(3/8)x
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