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.

1. **Find the equation of the highway.** The highway passes through \( (-4,-2) \) and \( (8,4) \). First, compute the slope: \[ m=\frac{4-(-2)}{8-(-4)}=\frac{6}{12}=0.5. \] Using the point-slope form with point \( (-4,-2) \): \[ y-(-2)=0.5(x-(-4)) \quad \Rightarrow \quad y+2=0.5(x+4). \] Simplify the equation: \[ y+2=0.5x+2 \quad \Rightarrow \quad y=0.5x. \] 2. **Write the distance formula from a point to a line.** For a point \( (x_0, y_0) \) and the line given by \( ax+by+c=0 \), the distance \( d \) is: \[ d=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}. \] 3. **Express the highway in standard form.** The equation \( y=0.5x \) can be rewritten in standard form: \[ 0.5x-y=0. \] Here, \( a=0.5 \), \( b=-1 \), and \( c=0 \). 4. **Insert the coordinates of The Big Pumpkin.** The Big Pumpkin is located at \( (-1,3) \). Substitute \( x_0=-1 \) and \( y_0=3 \) into the formula: \[ d=\frac{\left|0.5(-1)-1(3)+0\right|}{\sqrt{(0.5)^2+(-1)^2}}. \] 5. **Compute the numerator and the denominator.** Compute the numerator: \[ 0.5(-1)-3 = -0.5-3 = -3.5. \] The absolute value is: \[ | -3.5 | = 3.5. \] Compute the denominator: \[ \sqrt{(0.5)^2+1}=\sqrt{0.25+1}=\sqrt{1.25}. \] 6. **Simplify the distance expression.** The distance becomes: \[ d=\frac{3.5}{\sqrt{1.25}}. \] Recognize that: \[ \sqrt{1.25}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}. \] Thus: \[ d=\frac{3.5}{\frac{\sqrt{5}}{2}}=\frac{3.5 \times 2}{\sqrt{5}}=\frac{7}{\sqrt{5}}. \] 7. **Evaluate the numerical value and round to the nearest tenth.** Compute: \[ d=\frac{7}{\sqrt{5}}\approx \frac{7}{2.236}\approx 3.130. \] Rounding to the nearest tenth of a kilometre, \( d\approx 3.1 \) km.