Provide a complete solution to achieve full marks. Problem: The Big Pumpkin, a restaurant and pie shop, is a short distance from a major highway. The highway passes, in a straight line, through \( (-4,-2) \) and \( (8,4) \) on a map. The Big Pumpkin is located at \( (-1,3) \). What is the shortest distance from The Big Pumpkin to the highway, to the nearest tenth of a kilometre, if 1 unit on the map represents 1 km ? Consider a sketch.

To find the shortest distance from The Big Pumpkin at point \((-1, 3)\) to the highway represented by the line passing through the points \((-4, -2)\) and \((8, 4)\), we first need to determine the equation of the line. 1. **Find the slope of the line:** \[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-2)}{8 - (-4)} = \frac{6}{12} = \frac{1}{2} \] 2. **Using point-slope form, find the line equation:** Using point \((-4, -2)\): \[ y - (-2) = \frac{1}{2}(x - (-4)) \\ y + 2 = \frac{1}{2}(x + 4) \] \[ y + 2 = \frac{1}{2}x + 2 \implies y = \frac{1}{2}x + 0 \] The line equation is \(y = \frac{1}{2}x\). 3. **Now rearranging this into standard form:** \[ -\frac{1}{2}x + y = 0 \implies x - 2y = 0 \] 4. **Apply the point-to-line distance formula:** The formula to find the distance \(D\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is: \[ D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \] Here, \(A = 1\), \(B = -2\), \(C = 0\) and the point is \((-1, 3)\): \[ D = \frac{|1(-1) + (-2)(3) + 0|}{\sqrt{1^2 + (-2)^2}} \\ = \frac{|-1 - 6|}{\sqrt{1 + 4}} = \frac{|-7|}{\sqrt{5}} = \frac{7}{\sqrt{5}} \approx 3.14 \text{ km} \] 5. **Round to the nearest tenth:** Therefore, the shortest distance from The Big Pumpkin to the highway is approximately **3.1 km**. Remember to visualize and sketch this layout on a graph to see how the point, line, and distance interact! This can be helpful for verifying your solution!