Let A(-1,1) and B(2,3) be two points and P(x, | vedclass

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(A) The area of $	riangle PAB$ with vertices $P(x,y)$,$A(-1,1)$,and $B(2,3)$ is given by the determinant formula:$\frac{1}{2} |x(1-3) + (-1)(3-y) + 2

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$-\frac{12}{5}$

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(A) The area of $	riangle PAB$ with vertices $P(x,y)$,$A(-1,1)$,and $B(2,3)$ is given by the determinant formula:$\frac{1}{2} |x(1-3) + (-1)(3-y) + 2(y-1)| = 10$$\frac{1}{2} |-2x - 3 + y + 2y - 2| = 10$$|-2x + 3y - 5| = 20$Since $P$ is above the line $AB$,we consider the case $-2x + 3y - 5 = 20$,which gives $-2x + 3y = 25$.We need the locus in the form $ax + by = 15$. Dividing the equation $-2x + 3y = 25$ by $\frac{25}{15} = \frac{5}{3}$:$-\frac{2x}{5/3} + \frac{3y}{5/3} = 15$$-\frac{6}{5}x + \frac{9}{5}y = 15$Comparing this with $ax + by = 15$,we get $a = -\frac{6}{5}$ and $b = \frac{9}{5}$.Now,calculate $5a + 2b = 5(-\frac{6}{5}) + 2(\frac{9}{5}) = -6 + \frac{18}{5} = \frac{-30+18}{5} = -\frac{12}{5}$.

Let $A(-1,1)$ and $B(2,3)$ be two points and $P(x,y)$ be a variable point above the line $AB$ such that the area of $\triangle PAB$ is $10$. If the locus of $P$ is $ax+by=15$,then $5a+2b$ is:

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