SLOPE (m) of a line
\(m = \frac{rise}{run}=\frac{y}{x} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\)
Remarks:
Positive Slopes (m > 0). When x is increasing y is also increasing and vice versa. Typically described the line as sloping upwards to the right or downwards to the left.
Negative Slopes (m < 0). When x is increasing y is decreasing and when x is decreasing y is increasing. Characterized by a line moving downwards to the right or upwards to the left.
Zero Slopes (m = 0). These are lines parallel to x - axis.
Undefined Slopes (m = U). These are lines parallel to y - axis.
Parallel and Perpendicular lines
Slopes of Parallel Lines. m1 = m2. Two lines are parallel if they have the same slope and conversely.
Slopes of Perpendicular Lines. m1 = -1/m2. Two lines are perpendicular if and only if their slopes are negative reciprocals.
Given two lines 1 and 2. Assume that their slopes are negative reciprocal (\( m_1 = \frac{-1}{m_2}\)). Show that the angle of intersection is a right angle. This includes derivation on angles between lines.
Show SolutionAssumption: \(m_1 = \frac{-1}{m_2}\)
Show that: \(\angle C = 90^o\)
Using the triangle formed. \(\angle B +\angle C = \angle A\)
\(\tan B = m_2\) and
\(\tan A = m_1\)
\(\tan A = \tan (B+C)= m_1\)
\( \tan (B+C)= \frac{\tan B + \tan C }{1-\tan B \tan C} = \tan A\)
\( \tan A = \frac{m_2 + \tan C }{1- m_2 \tan C}\)
\( m_1= \frac{m_2 + \tan C }{1- m_2 \tan C}\)
\( \tan C = \frac{m_1-m_2}{1+m_2m_1}\)
But \(m_1 m_2 = -1\) from the assumption
\( \tan C = \frac{m_1-m_2}{0} = U\)
Thus \( \angle C\) must be a right angle.
Therefore, \( \angle C = 90^o\)
Let A, B, C, a, b are constants
1. General Form of a line.\(Ax + By + C = 0\)
2. Slope intercept form of a line \(y = mx + b\)
3. Point - Slope form of a line \((y - {y_1}) = m(x - {x_2})\)
4. Two point form \(\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{x - {x_1}}}\)
5. Intercept form. \(\frac{x}{a} + \frac{y}{b} = 1\)
a = x - intercept. The value of x when y is zero.
b = y - intercept. The value of y when x is zero.