A (1, -2) and B (-7, 1) are the given points. | vedclass

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(A) Given points are $A(1, -2)$ and $B(-7, 1)$.Let the coordinates of $P$ be $(x, y)$.The condition $3AP = 5AB$ implies $AP = \frac{5}{3} AB$.Since $P

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$(-12.33, 3)$

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(A) Given points are $A(1, -2)$ and $B(-7, 1)$.Let the coordinates of $P$ be $(x, y)$.The condition $3AP = 5AB$ implies $AP = \frac{5}{3} AB$.Since $P, A, B$ are collinear and $P-A-B$ implies $A$ is between $P$ and $B$,the distance $PB = PA + AB = \frac{5}{3} AB + AB = \frac{8}{3} AB$.Using the section formula for external division or vector approach:Vector $\vec{P} = \vec{A} + \frac{5}{3}(\vec{A} - \vec{B})$.$x = 1 + \frac{5}{3}(1 - (-7)) = 1 + \frac{5}{3}(8) = 1 + \frac{40}{3} = \frac{43}{3}$.$y = -2 + \frac{5}{3}(-2 - 1) = -2 + \frac{5}{3}(-3) = -2 - 5 = -7$.Thus,$P = (\frac{43}{3}, -7)$.

$A (1, -2)$ and $B (-7, 1)$ are the given points. Find $P$ on the line segment $AB$ such that $P$ lies on the extension of $AB$ (as indicated by $P-A-B$ order,though the condition $3AP = 5AB$ implies $P$ is outside the segment $AB$) and $3AP = 5AB$.

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