Let A equiv (0, 1),B equiv (2, 0),and point P | vedclass

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(D) For any point $P$ on the line,the value of $|PA - PB|$ is at most $AB$,which occurs when $P, A, B$ are collinear.To maximize $|PA - PB|$,$P$ must

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$\left( -\frac{24}{5}, \frac{17}{5} \right)$

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(D) For any point $P$ on the line,the value of $|PA - PB|$ is at most $AB$,which occurs when $P, A, B$ are collinear.To maximize $|PA - PB|$,$P$ must be the intersection of the line $AB$ and the given line $4x + 3y + 9 = 0$.The equation of line $AB$ passing through $(0, 1)$ and $(2, 0)$ is $\frac{x}{2} + \frac{y}{1} = 1$,which simplifies to $x + 2y = 2$.We solve the system of equations:$4x + 3y = -9$ $(1)$$x + 2y = 2$ $(2)$From $(2)$,$x = 2 - 2y$. Substituting into $(1)$:$4(2 - 2y) + 3y = -9$$8 - 8y + 3y = -9$$-5y = -17 \implies y = \frac{17}{5}$.Substituting $y = \frac{17}{5}$ into $x = 2 - 2y$:$x = 2 - 2\left(\frac{17}{5}\right) = 2 - \frac{34}{5} = -\frac{24}{5}$.Thus,the point $P$ is $\left( -\frac{24}{5}, \frac{17}{5} \right)$.

Let $A \equiv (0, 1)$,$B \equiv (2, 0)$,and point $P$ be a point on the line $4x + 3y + 9 = 0$. Find the coordinates of point $P$ such that $|PA - PB|$ is maximized.

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