Question

Alguém me ajuda por favor

Answer 1

1) Verifique se os pontos são colineares

1-b) D(1, -3), E(1, 3), F(6,1) => Não colineares

1-c) G(-4, 5), H(-1, 2), I(2,0) => Não colineares

1-d) J(-7, 1), K(5, -5), L(3,-4) => Colineares

2) Dados os pontos colineares A(-6,p), B(2,-5)e C(4,-7), determine o valor de p

p = 3

1-b)

$\begin{array}{l} \large \text{ \textsf{\textbf{D(1, -3), E(1, 3), F(6,1)}} }\normalsize\\\\ \qquad \quad \qquad \qquad \red{\tt 18}\quad \red{\tt \!1}\quad \red{\tt \!\!-3}\\ \bf D = \left|\begin{array}{ccc|cc} \bf 1&\bf \!-\!3&\bf 1&\bf 1&\bf \!-\!3\\ \bf 1&\bf 3&\bf 1&\bf 1&\bf 3\\ \bf 6&\bf 1&\bf 1&\bf 6&\bf 1\\ \end{array}\right.\\ \qquad \qquad \qquad \quad \orange{\tt 3}\quad \orange{\tt \!\!\!-\!18}\quad \orange{\tt \!1}\\ \bf D = \orange{(\bf 3-18+1)} - \red{(18+1-3)}\\ \bf D = \orange{\bf -14} - \red{\bf 16}\\ \bf D = -30\\ \boxed{\large \bf N\tilde{a}o~colinear} \end{array}$

1-c)

$\begin{array}{l} \large \text{ \textsf{\textbf{G(-4, 5), H(-1, 2), I(2,0)}} }\normalsize\\\\ \qquad \quad \qquad \qquad \red{\tt 4}\quad~ \red{\tt \!-\!5}\quad \red{\tt \!\!-3}\\ \bf D = \left|\begin{array}{ccc|cc} \bf \!-\!4&\bf 5&\bf 1&\bf \!-\!4&\bf 5\\ \bf \!-\!1&\bf 2&\bf 1&\bf \!-\!1&\bf 2\\ \bf 2&\bf 0&\bf 1&\bf 2&\bf 0\\ \end{array}\right.\\ \qquad \qquad \qquad \quad \orange{\tt \!-\!8}\quad \orange{\tt 10}\quad \orange{\tt \!-\!1}\\ \bf D = \orange{(\bf 8-10+1)} - \red{(4-5-3)}\\ \bf D = \orange{\bf -1} + \red{\bf 4}\\ \bf D = 3\\ \boxed{\large \bf N\tilde{a}o~colinear}\end{array}$

1-d)

$\begin{array}{l} \large \text{ \textsf{\textbf{J(-7, 1), K(5, -5), L(3,-4)}} }\normalsize\\\\ \qquad \quad \qquad \qquad ~~ \red{\tt \!-\!\!15}~~~ \red{\tt 28}~~~ \red{\tt 5}\\ \bf D = \left|\begin{array}{ccc|cc}\bf \!-\!7&\bf 1&\bf 1&\bf \!-\!7&\bf 1\\ \bf 5&\bf \!-\!5&\bf 1&\bf 5&\bf \!-\!5\\ \bf 3&\bf \!-\!4&\bf 1&\bf 3&\bf \!-\!4\\ \end{array}\right.\\ \qquad \qquad \qquad \quad~~~ \orange{\tt 35}\quad \orange{\tt 3}\quad \orange{\tt -\!20}\\ \bf D = \orange{(\bf 35+3-20)} - \red{(-15+28+5)}\\ \bf D = \orange{\bf 18} + \red{\bf 18}\\ \bf D = 0\\ \boxed{\large \bf Colinear} \end{array}$

2)

$\begin{array}{l} \large \text{ \textsf{\textbf{A(-6, p), B(2, -5), C(4,-7)}} }\normalsize\\\\ \qquad \quad \qquad~~ \red{\tt \!-\!\!20}~~~ \red{\tt 42}~~~ \red{\tt 2p}\\ \left|\begin{array}{ccc|cc} \bf \!-\!6&\bf p&\bf 1&\bf \!-\!6&\bf p\\ \bf 2&\bf \!-\!5&\bf 1&\bf 2&\bf \!-\!5\\ \bf 4&\bf \!-\!7&\bf 1&\bf 4&\bf \!-\!7\\ \end{array}\right.\\ \qquad \qquad \quad~~ \orange{\tt 30}\quad \orange{\tt \!\!4p}\quad \!\orange{\tt -\!\!14}\\ \bf \orange{(\bf 30+4p-14)} - \red{(-20+42+2p)}\\ \bf \orange{\bf 16+4p} - \red{\bf 22 -2p}\\ \bf 16-22+2p\\ \bf -6+2p\\\bf 2p=6\\\bf p=\dfrac{6}{2}\\ \boxed{\large \bf p=3}\end{array}$

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