1. The problem involves finding the intersection points of two pairs of lines given by points on their graphs. 2. For each pair, we first find the equations of the lines using the slope formula $m=\frac{y_2-y_1}{x_2-x_1}$ and point-slope form $y-y_1=m(x-x_1)$. 3. For lines E and F: - Line E passes through $(0,10)$ and $(10,0)$. Calculate slope: $$m_E=\frac{0-10}{10-0} = \frac{-10}{10} = -1$$ Equation: $$y-10 = -1(x-0) \Rightarrow y = -x + 10$$ - Line F passes through $(0,0)$ and approximately $(15,10)$. Calculate slope: $$m_F=\frac{10-0}{15-0} = \frac{10}{15} = \frac{2}{3}$$ Equation: $$y-0 = \frac{2}{3}(x-0) \Rightarrow y = \frac{2}{3}x$$ 4. Find intersection of E and F by solving system: $$\begin{cases} y = -x + 10 \\ y = \frac{2}{3}x \end{cases}$$ Set equal: $$-x + 10 = \frac{2}{3}x$$ Add $x$ to both sides: $$10 = \frac{2}{3}x + x = \frac{2}{3}x + \frac{3}{3}x = \frac{5}{3}x$$ Multiply both sides by $\cancel{\frac{3}{5}}$: $$10 \times \cancel{\frac{3}{5}} = \cancel{\frac{5}{3}}x \times \cancel{\frac{3}{5}} \Rightarrow 10 \times \frac{3}{5} = x$$ Simplify: $$x = 6$$ Substitute back to find $y$: $$y = \frac{2}{3} \times 6 = 4$$ Intersection point: $(6,4)$ 5. For lines G and H: - Line G passes through $(0,5)$ and approximately $(15,-5)$. Calculate slope: $$m_G=\frac{-5-5}{15-0} = \frac{-10}{15} = -\frac{2}{3}$$ Equation: $$y-5 = -\frac{2}{3}(x-0) \Rightarrow y = -\frac{2}{3}x + 5$$ - Line H passes through $(0,-3)$ and approximately $(20,2)$. Calculate slope: $$m_H=\frac{2-(-3)}{20-0} = \frac{5}{20} = \frac{1}{4}$$ Equation: $$y+3 = \frac{1}{4}(x-0) \Rightarrow y = \frac{1}{4}x - 3$$ 6. Find intersection of G and H by solving system: $$\begin{cases} y = -\frac{2}{3}x + 5 \\ y = \frac{1}{4}x - 3 \end{cases}$$ Set equal: $$-\frac{2}{3}x + 5 = \frac{1}{4}x - 3$$ Add $\frac{2}{3}x$ to both sides: $$5 = \frac{1}{4}x + \frac{2}{3}x - 3$$ Add 3 to both sides: $$8 = \frac{1}{4}x + \frac{2}{3}x$$ Find common denominator 12: $$8 = \frac{3}{12}x + \frac{8}{12}x = \frac{11}{12}x$$ Multiply both sides by $\cancel{\frac{12}{11}}$: $$8 \times \cancel{\frac{12}{11}} = \cancel{\frac{11}{12}}x \times \cancel{\frac{12}{11}} \Rightarrow 8 \times \frac{12}{11} = x$$ Simplify: $$x = \frac{96}{11} \approx 8.727$$ Substitute back to find $y$: $$y = \frac{1}{4} \times \frac{96}{11} - 3 = \frac{96}{44} - 3 = \frac{24}{11} - 3 = \frac{24}{11} - \frac{33}{11} = -\frac{9}{11} \approx -0.818$$ Intersection point: $\left(\frac{96}{11}, -\frac{9}{11}\right)$ or approximately $(8.73, -0.82)$ Final answers: - Intersection of E and F: $(6,4)$ - Intersection of G and H: $\left(\frac{96}{11}, -\frac{9}{11}\right)$