Answer:
Slopes: ²/₃ and -³/₂
Quadrilateral: Rectangle
Step-by-step explanation:
Given equations:
[tex]\begin{cases}3(y+1)=2x\\2y-8=-3x\\2x+3=3y\\-2(y-12)=3x\end{case}[/tex]
Slope-intercept form of a linear equation
[tex]\boxed{y=mx+b}[/tex]
Where:
To determine the slopes of the given equations, rearrange them to isolate y.
[tex]\begin{aligned}\textsf{Equation 1}: \quad 3(y+1)&=2x\\\dfrac{3(y+1)}{3}&=\dfrac{2x}{3}\\y+1&=\dfrac{2}{3}x\\y+1-1&=\dfrac{2}{3}x-1\\y&=\dfrac{2}{3}x-1\\\end{aligned}[/tex]
[tex]\begin{aligned}\textsf{Equation 2}: \quad \quad2y-8&=-3x\\2y-8+8&=-3x+8\\2y&=-3x+8\\\dfrac{2y}{2}&=\dfrac{-3x}{2}+\dfrac{8}{2}\\y&=-\dfrac{3}{2}x+4\\\end{aligned}[/tex]
[tex]\begin{aligned}\textsf{Equation 3}: \quad 2x+3&=3y\\3y&=2x+3\\\dfrac{3y}{3}&=\dfrac{2x}{3}+\dfrac{3}{3}\\y&=\dfrac{2}{3}x+1\end{aligned}[/tex]
[tex]\begin{aligned}\textsf{Equation 4}: \quad -2(y-12)&=3x\\\dfrac{-2(y-12)}{-2}&=\dfrac{3x}{-2}\\y-12&=-\dfrac{3}{2}x\\y-12+12&=-\dfrac{3}{2}x+12\\y&=-\dfrac{3}{2}x+12\end{aligned}[/tex]
The slopes of Equation 1 and Equation 3 are the same: ²/₃
The slopes of Equation 2 and Equation 4 are the same: -³/₂
If two lines are parallel, the slopes are the same.
Therefore, the opposite sides of the quadrilateral bounded by the lines are parallel.
-³/₂ is the negative reciprocal of ²/₃.
If two lines are perpendicular (intersect at a right angle), the slopes are negative reciprocals (multiply to -1).
Therefore, all the interior angles of the quadrilateral bounded by the lines are 90°.
As the opposite sides of the quadrilateral are parallel and its interior angles are 90°, the quadrilateral bounded by the sidewalks is a rectangle.
