1. **Stating the problem:** We have a quadrilateral with vertices A(-8,-3), B(13,-3), C(8,9), D(-3,9). We need to represent it on a Cartesian plane, identify the polygon, calculate area and perimeter with unit $u=1cm$, analyze the line $y=\frac{3}{4}x+3$ relative to the polygon, find equations of lines parallel and perpendicular to it through (0,2), find the support line of side AB, and solve two given equations verifying solutions equal the $x$-coordinate of point C. 2. **Plotting the quadrilateral:** Plot points A(-8,-3), B(13,-3), C(8,9), D(-3,9) and connect in order A-B-C-D. 3. **Polygon type:** Connecting points in order forms a quadrilateral. Since AB and DC are horizontal lines (both $y=-3$ and $y=9$ respectively), and AD and BC are vertical segments (x=-8 to -3 and x=13 to 8), the polygon is a trapezoid or more specifically a trapezium. 4. **Calculating area and perimeter:** - Length AB: distance between A(-8,-3) and B(13,-3) is $|13 - (-8)| = 21$ cm. - Length BC: distance between B(13,-3) and C(8,9) is $\sqrt{(8-13)^2 + (9+3)^2} = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ cm. - Length CD: distance between C(8,9) and D(-3,9) is $|8 - (-3)| = 11$ cm. - Length DA: distance between D(-3,9) and A(-8,-3) is $\sqrt{(-8+3)^2 + (-3-9)^2} = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = 13$ cm. Perimeter $P = 21 + 13 + 11 + 13 = 58$ cm. Area can be calculated using the shoelace formula: $$\text{Area} = \frac{1}{2} |x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$$ Substituting: $$= \frac{1}{2} |(-8)(-3) + 13 \cdot 9 + 8 \cdot 9 + (-3)(-3) - (-3 \cdot 13 + (-3) \cdot 8 + 9 \cdot (-3) + 9 \cdot (-8))|$$ $$= \frac{1}{2} |24 + 117 + 72 + 9 - (-39 -24 -27 -72)|$$ $$= \frac{1}{2} |222 - (-162)| = \frac{1}{2} (222 + 162) = \frac{384}{2} = 192 \text{ cm}^2$$ 5. **Position of line $y=\frac{3}{4}x+3$ relative to polygon:** Check if line intersects polygon by testing vertices: - At A(-8,-3), line value: $y=\frac{3}{4}(-8)+3 = -6+3 = -3$ matches A's y, so A lies on the line. - At B(13,-3), line value: $y=\frac{3}{4}(13)+3=9.75+3=12.75$ which is above B's y (-3), so B is below the line. - At C(8,9), line value: $y=\frac{3}{4}(8)+3=6+3=9$ matches C's y, so C lies on the line. - At D(-3,9), line value: $y=\frac{3}{4}(-3)+3=-2.25+3=0.75$ which is below D's y (9), so D is above the line. The line passes through points A and C, so it crosses the polygon diagonally. 6. **Equation of line parallel to $y=\frac{3}{4}x+3$ through (0,2):** Parallel lines have same slope $m=\frac{3}{4}$. Equation: $$y = \frac{3}{4}x + b$$ Passing through (0,2): $$2 = \frac{3}{4} \cdot 0 + b \Rightarrow b=2$$ So, $$y = \frac{3}{4}x + 2$$ 7. **Equation of line perpendicular to $y=\frac{3}{4}x+3$ through (0,2):** Slope perpendicular is negative reciprocal: $$m_{\perp} = -\frac{4}{3}$$ Equation: $$y = -\frac{4}{3}x + b$$ Passing through (0,2): $$2 = -\frac{4}{3} \cdot 0 + b \Rightarrow b=2$$ So, $$y = -\frac{4}{3}x + 2$$ 8. **Equation of support line of side AB:** Points A(-8,-3) and B(13,-3) have same y-coordinate, so line is horizontal: $$y = -3$$ 9. **Solving equations and verifying solutions equal $x$-coordinate of C (which is 8):** Equation 1: $$5x + 3(12 - x) = 9x - 28 + x$$ Expand: $$5x + 36 - 3x = 10x - 28$$ Simplify left: $$2x + 36 = 10x - 28$$ Bring terms together: $$36 + 28 = 10x - 2x$$ $$64 = 8x$$ Divide both sides by 8: $$\cancel{8}x = \frac{64}{\cancel{8}}$$ $$x = 8$$ Equation 2: $$\frac{x + 4}{6} - \frac{4 + x}{3} = 2 - \frac{x}{2}$$ Find common denominators and simplify: $$\frac{x+4}{6} - \frac{2(4+x)}{6} = 2 - \frac{x}{2}$$ $$\frac{x+4 - 2(4+x)}{6} = 2 - \frac{x}{2}$$ $$\frac{x + 4 - 8 - 2x}{6} = 2 - \frac{x}{2}$$ $$\frac{-x - 4}{6} = 2 - \frac{x}{2}$$ Multiply both sides by 6: $$-x - 4 = 12 - 3x$$ Bring terms together: $$-x + 3x = 12 + 4$$ $$2x = 16$$ Divide both sides by 2: $$\cancel{2}x = \frac{16}{\cancel{2}}$$ $$x = 8$$ Both solutions equal 8, which matches the $x$-coordinate of point C. **Final answers:** - Polygon is a quadrilateral with vertices A-B-C-D. - Perimeter = 58 cm. - Area = 192 cm$^2$. - Line $y=\frac{3}{4}x+3$ passes through points A and C. - Parallel line through (0,2): $y=\frac{3}{4}x+2$. - Perpendicular line through (0,2): $y=-\frac{4}{3}x+2$. - Support line of AB: $y=-3$. - Solutions to equations: $x=8$ matching $x$ of point C.