1. **Stating the problem:** We have a geometric figure with line segments forming kite shapes and a central square. We will find the equations of 5 selected line segments in slope-intercept form $y=mx+b$. 2. **Selecting 5 line segments and finding their equations:** - Line segment AB (top kite left side): Points A $(0,1)$ and B $(0,13)$. Slope $m=\frac{13-1}{0-0}$ is undefined (vertical line), so equation is $x=0$. - Line segment BC (top kite top side): Points B $(0,13)$ and C $(5,13)$. Slope $m=\frac{13-13}{5-0}=0$. Equation: $y=13$. - Line segment CD (top kite right side): Points C $(5,13)$ and D $(10,1)$. Slope $m=\frac{1-13}{10-5}=\frac{-12}{5}=-2.4$. Using point-slope form with point C: $$y-13=-2.4(x-5)$$ $$y=-2.4x+12+13=-2.4x+25$$ - Line segment DE (top kite bottom right): Points D $(10,1)$ and E $(0,1)$. Slope $m=\frac{1-1}{0-10}=0$. Equation: $y=1$. - Line segment EF (right kite horizontal): Points E $(0,1)$ and F $(14,0)$ (approximate right kite end). Slope $m=\frac{0-1}{14-0}=-\frac{1}{14}$. Equation using point E: $$y-1=-\frac{1}{14}(x-0)$$ $$y=-\frac{1}{14}x+1$$ 3. **Selecting 3 points of intersection for linear systems:** - Intersection 1: Lines AB ($x=0$) and BC ($y=13$) intersect at $(0,13)$. - Intersection 2: Lines CD ($y=-2.4x+25$) and DE ($y=1$) intersect. - Intersection 3: Lines EF ($y=-\frac{1}{14}x+1$) and DE ($y=1$) intersect. 4. **Solving linear systems:** **System 1: AB and BC** - Equations: $$x=0$$ $$y=13$$ - Solution is $(0,13)$ directly. **System 2: CD and DE** - Equations: $$y=-2.4x+25$$ $$y=1$$ - Substitution method: Substitute $y=1$ into first: $$1=-2.4x+25$$ $$-2.4x=1-25=-24$$ $$x=\frac{-24}{-2.4}=10$$ Solution: $(10,1)$. - Elimination method: Write system: $$y+2.4x=25$$ $$y=1$$ Subtract second from first: $$(y+2.4x)-y=25-1$$ $$2.4x=24$$ $$x=10$$ Substitute back $y=1$. **System 3: EF and DE** - Equations: $$y=-\frac{1}{14}x+1$$ $$y=1$$ - Substitution method: Substitute $y=1$: $$1=-\frac{1}{14}x+1$$ $$-\frac{1}{14}x=0$$ $$x=0$$ Solution: $(0,1)$. - Elimination method: Write system: $$y+\frac{1}{14}x=1$$ $$y=1$$ Subtract second from first: $$(y+\frac{1}{14}x)-y=1-1$$ $$\frac{1}{14}x=0$$ $$x=0$$ 5. **Final answers:** - Line equations: 1. $x=0$ 2. $y=13$ 3. $y=-2.4x+25$ 4. $y=1$ 5. $y=-\frac{1}{14}x+1$ - Intersection points: 1. $(0,13)$ 2. $(10,1)$ 3. $(0,1)$