Key Concepts:
Slope (Gradient) of a Line:
The slope of a line measures its steepness and is calculated as: \[ m = \frac{y_2 - y_1}{x_2 - x_1}, \] where \((x_1, y_1)\) and \((x_2, y_2)\) are two points on the line.Equation of a Line:
Slope-Intercept Form:
\(y = mx + c\), where \(m\) is the slope and \(c\) is the \(y\)-intercept.Point-Slope Form:
\(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line.Two-Intercept Form:
\(\frac{x}{a} + \frac{y}{b} = 1\), where \(a\) and \(b\) are the \(x\)- and \(y\)-intercepts, respectively.Parallel and Perpendicular Lines:
Midpoint of a Line Segment:
The midpoint of a line segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right). \]Distance Between Two Points:
The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \]Equation of a Circle:
The equation of a circle with center \((h, k)\) and radius \(r\) is: \[ (x - h)^2 + (y - k)^2 = r^2. \]Finding the Equation of a Line Parallel to a Given Line:
A line parallel to \(y = mx + c\) has the same slope \(m\) but a different \(y\)-intercept.Finding the Equation of a Line Perpendicular to a Given Line:
A line perpendicular to \(y = mx + c\) has a slope of \(-\frac{1}{m}\).Examples:
Example 1: Find the slope of the line passing through the points \((2, 3)\) and \((5, 7)\).
Solution:
Use the slope formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 3}{5 - 2} = \frac{4}{3}. \] Therefore, the slope is \(\frac{4}{3}\).Example 2: Write the equation of the line with slope \(2\) passing through the point \((3, 4)\).
Solution:
Use the point-slope form: \[ y - y_1 = m(x - x_1). \] Substitute \(m = 2\), \(x_1 = 3\), and \(y_1 = 4\): \[ y - 4 = 2(x - 3). \] Simplify: \[ y = 2x - 6 + 4 = 2x - 2. \] Therefore, the equation is \(y = 2x - 2\).Example 3: Find the distance between the points \((1, 2)\) and \((4, 6)\).
Solution:
Use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \] Substitute the values: \[ d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5. \] Therefore, the distance is \(5\) units.Example 4: Find the equation of a circle with center \((2, -3)\) and radius \(5\).
Solution:
Use the equation of a circle: \[ (x - h)^2 + (y - k)^2 = r^2. \] Substitute \(h = 2\), \(k = -3\), and \(r = 5\): \[ (x - 2)^2 + (y + 3)^2 = 25. \] Therefore, the equation of the circle is: \[ (x - 2)^2 + (y + 3)^2 = 25. \]Example 5: Find the equation of a line perpendicular to \(y = \frac{1}{2}x + 3\) that passes through the point \((4, 2)\).
Solution:
Step 1: Find the slope of the perpendicular line: \[ m_{\text{perpendicular}} = -\frac{1}{m_{\text{original}}} = -\frac{1}{\frac{1}{2}} = -2. \] Step 2: Use the point-slope form: \[ y - y_1 = m(x - x_1). \] Substitute \(m = -2\), \(x_1 = 4\), and \(y_1 = 2\): \[ y - 2 = -2(x - 4). \] Simplify: \[ y = -2x + 8 + 2 = -2x + 10. \] Therefore, the equation of the line is \(y = -2x + 10\).Example 6: The \(x\)- and \(y\)-intercepts of a straight line are \(-\frac{3}{4}\) and \(\frac{2}{7}\), respectively. Find the equation of the line.
Solution:
Step 1: Use the two-intercept form of a line: \[ \frac{x}{a} + \frac{y}{b} = 1, \] where \(a\) is the \(x\)-intercept and \(b\) is the \(y\)-intercept. Step 2: Substitute \(a = -\frac{3}{4}\) and \(b = \frac{2}{7}\): \[ \frac{x}{-\frac{3}{4}} + \frac{y}{\frac{2}{7}} = 1. \] Step 3: Simplify the fractions: \[ -\frac{4x}{3} + \frac{7y}{2} = 1. \] Step 4: Eliminate the fractions by multiplying through by the least common denominator (LCD), which is \(6\): \[ 6 \times \left(-\frac{4x}{3}\right) + 6 \times \left(\frac{7y}{2}\right) = 6 \cdot 1. \] Simplify: \[ -8x + 21y = 6. \] Therefore, the equation of the line is: \[ -8x + 21y = 6. \] Alternatively, in point-slope form: Using the slope \(m = \frac{\Delta y}{\Delta x} = \frac{\frac{2}{7} - 0}{0 - \left(-\frac{3}{4}\right)} = \frac{\frac{2}{7}}{\frac{3}{4}} = \frac{8}{21}\), and passing through the point \((0, \frac{2}{7})\), the equation becomes: \[ y - \frac{2}{7} = \frac{8}{21}(x - 0). \] Therefore, the gradient-point form is: \[ y = \frac{8}{21}x + \frac{2}{7}. \]