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MPM 2D1 Assignment Analytic Geometry: Line Segments and Circles 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.

Ask by Kelley Deleon. in Canada
Mar 15,2025

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Answer

The shortest distance from The Big Pumpkin to the highway is approximately 3.1 kilometers.

Solution

1. Determine the slope of the highway line passing through the points \( (-4,-2) \) and \( (8,4) \). The slope \( m \) is given by \[ m = \frac{4 - (-2)}{8 - (-4)} = \frac{6}{12} = \frac{1}{2}. \] 2. Write the equation of the line in slope-intercept form. Starting with the point-slope form for the point \( (-4,-2) \): \[ y - (-2) = \frac{1}{2}(x - (-4)), \] which simplifies to \[ y + 2 = \frac{1}{2}(x + 4). \] Expanding the right-hand side: \[ y + 2 = \frac{1}{2}x + 2. \] Subtract \(2\) from both sides: \[ y = \frac{1}{2}x. \] 3. Convert the line equation to standard form \( Ax + By + C = 0 \). The equation \( y = \frac{1}{2}x \) can be rewritten as \[ \frac{1}{2}x - y = 0. \] Multiplying through by 2 to eliminate the fraction, we have \[ x - 2y = 0. \] 4. Compute the perpendicular distance from The Big Pumpkin at \( (-1,3) \) to the highway. The distance \( d \) from a point \( (x_0,y_0) \) to the line \( Ax + By + C = 0 \) is given by \[ d = \frac{\left| Ax_0 + By_0 + C \right|}{\sqrt{A^2+B^2}}. \] Here, \( A = 1 \), \( B = -2 \), \( C = 0 \), and \( (x_0,y_0) = (-1,3) \). Substitute these values: \[ d = \frac{\left| 1(-1) + (-2)(3) + 0 \right|}{\sqrt{1^2 + (-2)^2}} = \frac{\left| -1 - 6 \right|}{\sqrt{1 + 4}} = \frac{7}{\sqrt{5}}. \] 5. Calculate the numerical value and round to the nearest tenth. \[ d \approx \frac{7}{2.2361} \approx 3.1305 \text{ km}. \] Rounding to the nearest tenth gives \[ d \approx 3.1 \text{ km}. \] The shortest distance from The Big Pumpkin to the highway is approximately \( 3.1 \) km.

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To find the shortest distance from The Big Pumpkin at \((-1,3)\) to the highway defined by the points \((-4,-2)\) and \((8,4)\), we first derive the equation of the line that represents the highway. 1. **Finding the Slope of the Line:** The slope \(m\) between the points \((-4, -2)\) and \((8, 4)\) is given by: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-2)}{8 - (-4)} = \frac{6}{12} = \frac{1}{2} \] 2. **Equation of the Line:** Using point-slope form, we can find the equation of the line: \[ y - y_1 = m(x - x_1) \] Using point \((-4,-2)\): \[ y + 2 = \frac{1}{2}(x + 4) \\ y = \frac{1}{2}x - 2 + 2 \\ y = \frac{1}{2}x - 2 \] Thus, the equation of the highway is \(y = \frac{1}{2}x - 2\). 3. **Finding the Perpendicular Line:** The slope of the perpendicular line to the highway is the negative reciprocal of \(\frac{1}{2}\), which is \(-2\). Using the point \((-1, 3)\), the equation for the perpendicular line is: \[ y - 3 = -2(x + 1) \\ y - 3 = -2x - 2 \\ y = -2x + 1 \] 4. **Finding the Intersection Point:** Set the equations equal to find the intersection point: \[ \frac{1}{2}x - 2 = -2x + 1 \\ \frac{1}{2}x + 2x = 1 + 2 \\ \frac{5}{2}x = 3 \\ x = \frac{6}{5} \approx 1.2 \] Substitute \(x\) back into either equation to find \(y\): \[ y = \frac{1}{2}\left(\frac{6}{5}\right) - 2 = \frac{3}{5} - 2 = \frac{3}{5} - \frac{10}{5} = -\frac{7}{5} \approx -1.4 \] Thus, the intersection point is \((\frac{6}{5}, -\frac{7}{5})\). 5. **Calculating Distance:** Now to find the distance between The Big Pumpkin at \((-1, 3)\) and the intersection point: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{\left(\frac{6}{5} - (-1)\right)^2 + \left(-\frac{7}{5} - 3\right)^2} \] \[ = \sqrt{\left(\frac{6}{5} + \frac{5}{5}\right)^2 + \left(-\frac{7}{5} - \frac{15}{5}\right)^2} = \sqrt{\left(\frac{11}{5}\right)^2 + \left(-\frac{22}{5}\right)^2} \] \[ = \sqrt{\frac{121}{25} + \frac{484}{25}} = \sqrt{\frac{605}{25}} = \sqrt{24.2} \approx 4.1 \text{ km}. \] Thus, the shortest distance from The Big Pumpkin to the highway is approximately **4.1 km**.

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