1. We are asked to find the equation of the red straight line passing approximately through points (-3,-3) and (3,5). 2. The formula for the equation of a line given two points $(x_1,y_1)$ and $(x_2,y_2)$ is: $$y - y_1 = m(x - x_1)$$ where the slope $m$ is calculated as: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ 3. Calculate the slope $m$: $$m = \frac{5 - (-3)}{3 - (-3)} = \frac{5 + 3}{3 + 3} = \frac{8}{6} = \frac{4}{3}$$ 4. Use point-slope form with point (-3,-3): $$y - (-3) = \frac{4}{3}(x - (-3))$$ which simplifies to: $$y + 3 = \frac{4}{3}(x + 3)$$ 5. Distribute the slope: $$y + 3 = \frac{4}{3}x + 4$$ 6. Subtract 3 from both sides to isolate $y$: $$y = \frac{4}{3}x + 4 - 3$$ $$y = \frac{4}{3}x + 1$$ 7. Therefore, the equation of the line is: $$y = \frac{4}{3}x + 1$$ This means the line has a slope of $\frac{4}{3}$ and crosses the y-axis at 1. This completes the solution for the first problem (c).