- Find the equation for each line using the slope formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\) and the point-slope form \(y - y_1 = m(x - x_1)\).
- Set the two line equations equal to each other to solve for the \(x\)-coordinate of the intersection.
- Substitute the \(x\)-coordinate into either line equation to determine the corresponding \(y\)-coordinate.
- The intersection points are \((-4, -7)\), \((4, 7)\), \((-4, -5)\), and \((1, -1)\) for the rows 1 to 4 respectively. \(\boxed{(-4, -7), (4, 7), (-4, -5), (1, -1)}\).
Explanation
- Analysis and Methodology
To find the intersection point of two lines given by sets of points, we first determine the equation of each line in the slope-intercept form \(y = mx + b\). The slope \(m\) is given by \(\frac{y_2 - y_1}{x_2 - x_1}\), and the \(y\)-intercept \(b\) is found by substituting one point into the equation \(y = mx + b\). After finding the equations for both lines, we set them equal to each other to solve for the \(x\)-coordinate of the intersection, and subsequently find the \(y\)-coordinate by substituting \(x\) back into one of the equations.
- Calculating Intersection Points (Rows 1 & 2)
Row 1: Line 1 passes through \((2, 5)\) and \((-3, -5)\). Slope \(m_1 = \frac{-5 - 5}{-3 - 2} = \frac{-10}{-5} = 2\). The equation is \(y - 5 = 2(x - 2) \Rightarrow y = 2x + 1\). Line 2 passes through \((3, 0)\) and \((0, -3)\). Slope \(m_2 = \frac{-3 - 0}{0 - 3} = 1\). The equation is \(y = x - 3\). Setting \(2x + 1 = x - 3\), we get \(x = -4\). Substituting \(x = -4\) into \(y = x - 3\), we get \(y = -7\). The intersection is \((-4, -7)\).
Row 2: Line 1 passes through \((1, 1)\) and \((2, 3)\). Slope \(m_1 = \frac{3 - 1}{2 - 1} = 2\). The equation is \(y - 1 = 2(x - 1) \Rightarrow y = 2x - 1\). Line 2 passes through \((0, 3)\) and \((2, 5)\). Slope \(m_2 = \frac{5 - 3}{2 - 0} = 1\). The equation is \(y = x + 3\). Setting \(2x - 1 = x + 3\), we get \(x = 4\). Substituting \(x = 4\) into \(y = x + 3\), we get \(y = 7\). The intersection is \((4, 7)\).
- Calculating Intersection Points (Rows 3 & 4)
Row 3: Line 1 passes through \((1, 0)\) and \((0, -1)\). Slope \(m_1 = \frac{-1 - 0}{0 - 1} = 1\). The equation is \(y = x - 1\). Line 2 passes through \((0, 3)\) and \((-2, -1)\). Slope \(m_2 = \frac{-1 - 3}{-2 - 0} = \frac{-4}{-2} = 2\). The equation is \(y = 2x + 3\). Setting \(x - 1 = 2x + 3\), we get \(x = -4\). Substituting \(x = -4\) into \(y = x - 1\), we get \(y = -5\). The intersection is \((-4, -5)\).
Row 4: Line 1 passes through \((2, 0)\) and \((0, -2)\). Slope \(m_1 = \frac{-2 - 0}{0 - 2} = 1\). The equation is \(y = x - 2\). Line 2 passes through \((4, 5)\) and \((3, 3)\). Slope \(m_2 = \frac{3 - 5}{3 - 4} = 2\). The equation is \(y - 5 = 2(x - 4) \Rightarrow y = 2x - 3\). Setting \(x - 2 = 2x - 3\), we get \(x = 1\). Substituting \(x = 1\) into \(y = x - 2\), we get \(y = -1\). The intersection is \((1, -1)\).
- Final Results and Conclusion
The intersection points are determined as:
- Row 1: \((-4, -7)\)
- Row 2: \((4, 7)\)
- Row 3: \((-4, -5)\)
- Row 4: \((1, -1)\)
These points confirm the matching given in the table. The intersections are \((-4, -7)\), \((4, 7)\), \((-4, -5)\), and \((1, -1)\).
Examples
In urban planning, streets are often modeled as lines on a coordinate plane. If two streets, one running through coordinates \((2, 5)\) and \((-3, -5)\) and another through \((3, 0)\) and \((0, -3)\), need a traffic signal at their crossing point, finding the intersection point using the method above tells city engineers exactly where to place the signal. This ensures traffic flow and safety by pinpointing the precise location where these two paths meet.