If A equiv (2i + 3j), B equiv (pi + 9j), and | vedclass

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(C) Given points are $A(2, 3)$, $B(p, 9)$, and $C(1, -1)$.For points $A, B$, and $C$ to be collinear, the vector $\vec{AB}$ must be parallel to the ve

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$7$

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(C) Given points are $A(2, 3)$, $B(p, 9)$, and $C(1, -1)$.For points $A, B$, and $C$ to be collinear, the vector $\vec{AB}$ must be parallel to the vector $\vec{AC}$.$\vec{AB} = (p - 2)i + (9 - 3)j = (p - 2)i + 6j$.$\vec{AC} = (1 - 2)i + (-1 - 3)j = -i - 4j$.Since $\vec{AB} \parallel \vec{AC}$, the ratio of their components must be equal:$\frac{p - 2}{-1} = \frac{6}{-4}$.$\frac{p - 2}{-1} = -\frac{3}{2}$.$p - 2 = \frac{3}{2}$.$p = 2 + \frac{3}{2} = \frac{4 + 3}{2} = \frac{7}{2}$.

If $A \equiv (2i + 3j)$, $B \equiv (\pi + 9j)$, and $C \equiv (i - j)$ are collinear, then the value of $p$ is: (in $/2$)

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