Calcular o valor de ''p'' para que os pontos A (3, 5),B (1, 0) e C (2, p) estejam alinhados
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Se os pontos estão alinhados, entende-se que:
![\large\boxed{ \left[\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right] =0} \large\boxed{ \left[\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right] =0}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%26amp%3By_1%26amp%3B1%5C%5Cx_2%26amp%3By_2%26amp%3B1%5C%5Cx_3%26amp%3By_3%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%3D0%7D)
x1 = 3 ; y1 = 5 ; x2 = 1 ; y2 = 0 ; x3 = 2 ; y3 = p
Substituindo:
![\left[\begin{array}{ccc}3&1&1\\1&0&1\\2&p&1\end{array}\right] =0\\ \\ (3*0*1) + (1*1*2)+(1*1*p)\\ -(2*0*1)-(1*1*1)-(3*p*1)=0\\ \\ 0+2+p - 0-1-3p=0\\ \\ -2p+1=0\\ \\ 2p=1\\ \\ \boxed{p=\frac{1}{2}} \left[\begin{array}{ccc}3&1&1\\1&0&1\\2&p&1\end{array}\right] =0\\ \\ (3*0*1) + (1*1*2)+(1*1*p)\\ -(2*0*1)-(1*1*1)-(3*p*1)=0\\ \\ 0+2+p - 0-1-3p=0\\ \\ -2p+1=0\\ \\ 2p=1\\ \\ \boxed{p=\frac{1}{2}}](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26amp%3B1%26amp%3B1%5C%5C1%26amp%3B0%26amp%3B1%5C%5C2%26amp%3Bp%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%3D0%5C%5C+%5C%5C+%283%2A0%2A1%29+%2B+%281%2A1%2A2%29%2B%281%2A1%2Ap%29%5C%5C+-%282%2A0%2A1%29-%281%2A1%2A1%29-%283%2Ap%2A1%29%3D0%5C%5C+%5C%5C+0%2B2%2Bp+-+0-1-3p%3D0%5C%5C+%5C%5C+-2p%2B1%3D0%5C%5C+%5C%5C+2p%3D1%5C%5C+%5C%5C+%5Cboxed%7Bp%3D%5Cfrac%7B1%7D%7B2%7D%7D)
x1 = 3 ; y1 = 5 ; x2 = 1 ; y2 = 0 ; x3 = 2 ; y3 = p
Substituindo:
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