A (1, 2) and B (2, 3) are the given points. F | vedclass

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(D) Given that $C - A - B$ and $2 AC = 3 AB$.This implies that $A$ divides the segment $\overline{CB}$ externally or $A$ lies on the line segment $\ov

Vedclass · Accepted Answer

$(-0.5, 0.5)$

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(D) Given that $C - A - B$ and $2 AC = 3 AB$.This implies that $A$ divides the segment $\overline{CB}$ externally or $A$ lies on the line segment $\overline{CB}$ such that $C$ is outside. Specifically,$A$ is between $C$ and $B$ is not possible here as $C-A-B$ means $A$ is between $C$ and $B$. Wait,the condition $C-A-B$ implies $A$ is between $C$ and $B$. Let's re-evaluate: $CA + AB = CB$. Given $2 AC = 3 AB$,so $AC = 1.5 AB$.Since $C-A-B$,the distance $CB = CA + AB = 1.5 AB + AB = 2.5 AB$.Using the section formula where $A(1, 2)$ divides $\overline{CB}$ in the ratio $CA:AB = 3:2$.Let $C = (x_1, y_1)$ and $B = (2, 3)$. $A(1, 2)$ divides $CB$ in ratio $m:n = 3:2$.$1 = \frac{3(2) + 2(x_1)}{3+2} \implies 5 = 6 + 2x_1 \implies 2x_1 = -1 \implies x_1 = -0.5$.$2 = \frac{3(3) + 2(y_1)}{3+2} \implies 10 = 9 + 2y_1 \implies 2y_1 = 1 \implies y_1 = 0.5$.Thus,$C = (-0.5, 0.5)$.

$A (1, 2)$ and $B (2, 3)$ are the given points. Find the point $C \in \overleftrightarrow{AB}$ such that $C - A - B$ and $2 AC = 3 AB$.

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