Question: Draw the line of reflection that reflects $\triangle ABC$ onto $\triangle A'B'C'$
1. **Problem Statement:** We need to find the line of reflection that maps $\triangle ABC$ onto $\triangle A'B'C'$. This line is the axis of symmetry between the two triangles.
2. **Key Concept:** The line of reflection is the perpendicular bisector of the segment joining each pair of corresponding points, e.g., $AA'$, $BB'$, and $CC'$. All these segments are reflected across the same line.
3. **Given Information:** The reflection line passes approximately through points $(2,6)$ and $(5,-8)$.
4. **Find the equation of the line passing through $(2,6)$ and $(5,-8)$:**
Calculate the slope $m$:
$$m = \frac{-8 - 6}{5 - 2} = \frac{-14}{3} = -\frac{14}{3}$$
5. **Use point-slope form to find the line equation:**
$$y - 6 = -\frac{14}{3}(x - 2)$$
6. **Simplify the equation:**
$$y - 6 = -\frac{14}{3}x + \frac{28}{3}$$
$$y = -\frac{14}{3}x + \frac{28}{3} + 6$$
$$y = -\frac{14}{3}x + \frac{28}{3} + \frac{18}{3}$$
$$y = -\frac{14}{3}x + \frac{46}{3}$$
7. **Final equation of the line of reflection:**
$$\boxed{y = -\frac{14}{3}x + \frac{46}{3}}$$
This line is the axis of reflection that maps $\triangle ABC$ onto $\triangle A'B'C'$.