Question

determine o valor de a para que os pontos A(a,1), B(3,a) e C(5,4), pertençam a mesma reta:​

Answer 1

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$\Large\boxed{\begin{array}{l}\sf Sejam~A(x_A,y_A),B(x_B,y_B)~e~C(x_C,y_C)\\\sf pontos~quasiquer~do~plano~cartesiano.\\\sf Dizemos~que~estes~pontos~est\tilde ao~alinhados~quando\\\begin{vmatrix}\sf x_A&\sf y_A&\sf1\\\sf x_B&\sf y_B&\sf1\\\sf x_C&\sf y_C&\sf1\end{vmatrix}=0\end{array}}$

$\boxed{\begin{array}{l}\begin{vmatrix}\sf a&\sf1&\sf1\\\sf 3&\sf a&\sf1\\\sf 5&\sf4&\sf1\end{vmatrix}=0\\\sf a\cdot(a-4)-1\cdot(3-5)+1\cdot(12-5a)=0\\\sf a^2-4a+2+12-5a=0\\\sf a^2-9a+14=0\\\sf \Delta=b^2-4ac\\\sf\Delta=(-9)^2-4\cdot1\cdot14\\\sf\Delta=81-56\\\sf\Delta=25\\\sf a=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf a=\dfrac{-(-9)\pm\sqrt{25}}{2\cdot1}\\\sf a=\dfrac{9\pm5}{2}\begin{cases} \sf a_1=\dfrac{9+5}{2}=\dfrac{14}{2}=7\\\sf a_2=\dfrac{9-5}{2}=\dfrac{4}{2}=2\end{cases}\\\sf a=7~~ou~~a=2\end{array}}$