28) \( \mathrm{S}(0,-6), \mathrm{R}(-10,-6), \mathrm{Q}(1,0) \)

**Step 1. Plot the Points and Identify the Triangle** We are given three points: \( \mathrm{S}(0,-6) \) \( \mathrm{R}(-10,-6) \) \( \mathrm{Q}(1,0) \) Plotting these points shows that points \( \mathrm{S} \) and \( \mathrm{R} \) lie on the same horizontal line (both have \( y = -6 \)). Thus, the segment \( \overline{SR} \) is horizontal and can be used as the base of the triangle. --- **Step 2. Calculate the Length of the Base \( \overline{SR} \)** Since \( \mathrm{S}(0,-6) \) and \( \mathrm{R}(-10,-6) \) have the same \( y \)-coordinate, the length of \( \overline{SR} \) is simply the horizontal distance between them: \[ \text{Length of } \overline{SR} = |0 - (-10)| = 10 \] --- **Step 3. Find the Height of the Triangle** The height is the perpendicular distance from the third vertex \( \mathrm{Q}(1,0) \) to the line containing \( \overline{SR} \) (which is the line \( y = -6 \)). The vertical distance from \( \mathrm{Q} \) (with \( y = 0 \)) to the line \( y = -6 \) is: \[ \text{Height} = |0 - (-6)| = 6 \] --- **Step 4. Calculate the Area of the Triangle** The area \( A \) of a triangle with base \( b \) and height \( h \) is given by: \[ A = \frac{1}{2} b h \] Substitute the values obtained: \[ A = \frac{1}{2} \times 10 \times 6 = 30 \] --- **Final Answer** The area of the triangle with vertices \( \mathrm{S}(0,-6) \), \( \mathrm{R}(-10,-6) \), and \( \mathrm{Q}(1,0) \) is \( 30 \) square units.