The area of the quadrilateral bounded by the $Y$-axis, the line $x=5$, and the lines $|x-y|-|x-5|=2$, is

We need to find the area of a quadrilateral with these boundaries:

Left side: Y-axis ($x = 0$)

Right side: Line $x = 5$

Top and bottom: The equation $|x-y| - |x-5| = 2$ $\newline\newline$ The key insight is that since our region is between $x = 0$ and $x = 5$, we know $x \leq 5$ always. This means $|x-5| = 5-x$. $\newline\newline$

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$\newline\newline$ Simplifying the equation:

$|x-y| - |x-5| = 2$

$|x-y| - (5-x) = 2$

$|x-y| = 2 + 5 - x = 7 - x$ $\newline\newline$

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$\newline\newline$ Removing the remaining absolute value gives us two cases: $\newline\newline$ Case A: When $x-y \geq 0$ (meaning $x \geq y$):

$x - y = 7 - x$

$2x - y = 7$

$y = 2x - 7$ $\newline\newline$ Case B: When $x-y < 0$ (meaning $x < y$):

$-(x-y) = 7 - x$

$-x + y = 7 - x$

$y = 7$ $\newline\newline$

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$\newline\newline$ Finding the vertices where these lines meet $x = 0$ and $x = 5$: $\newline\newline$ At $x = 0$:

Line $y = 2x - 7$: $y = 2(0) - 7 = -7$ → Point $(0, -7)$

Line $y = 7$: $y = 7$ → Point $(0, 7)$ $\newline\newline$ At $x = 5$:

Line $y = 2x - 7$: $y = 2(5) - 7 = 3$ → Point $(5, 3)$

Line $y = 7$: $y = 7$ → Point $(5, 7)$ $\newline\newline$

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$\newline\newline$ Our quadrilateral has vertices: $(0, -7)$, $(0, 7)$, $(5, 7)$, $(5, 3)$ $\newline\newline$ This forms a trapezoid with:

Left side height: from $y = -7$ to $y = 7$, so height $= 14$

Right side height: from $y = 3$ to $y = 7$, so height $= 4$

Width: $5 - 0 = 5$ $\newline\newline$ Using trapezoid formula: $\newline\newline$ $\text{Area} = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{width}$ $\newline\newline$ $\text{Area} = \dfrac{1}{2} \times (14 + 4) \times 5 = \dfrac{1}{2} \times 18 \times 5 = 45$