Introduction to Coordinate Geometry and Slope of a Line

Basic Concepts (Recap)

Coordinate geometry provides a powerful connection between algebra and geometry, allowing us to represent geometric figures as algebraic equations.

• Distance Formula: The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is found using the Pythagorean theorem.

$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

• Section Formula: The coordinates of a point dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m:n$ are given by a weighted average.

$$(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$$

Slope of a Line

The slope or gradient (denoted by 'm') of a line is a measure of its steepness and direction. It represents the rate of change in the vertical direction (rise) for each unit of change in the horizontal direction (run).

• Definition: If $\theta$ is the angle a line makes with the positive x-axis (its inclination), then its slope is:

$$m = \tan\theta$$

• Slope from Two Points: The slope of a non-vertical line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$$m = \frac{\text{Rise}}{\text{Run}} = \frac{y_2 - y_1}{x_2 - x_1}$$

• A horizontal line has an inclination of $0^\circ$, so its slope is $m=\tan(0^\circ)=0$.

• A vertical line has an inclination of $90^\circ$, and its slope is $m=\tan(90^\circ)$, which is undefined.

Conditions on Slopes

• Parallel Lines: Two lines are parallel if and only if they have the same steepness and direction. Thus, their slopes must be equal.

$m_1 = m_2$

• Perpendicular Lines: Two lines are perpendicular if and only if their slopes are negative reciprocals of each other. This means their product is -1.

$m_1 m_2 = -1$

Angle Between Two Lines

If $\theta$ is the acute angle between two intersecting lines with slopes $m_1$ and $m_2$, it can be found using the formula:

$$\tan\theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$$

Example 1: Slope and Inclination

Question: Find the slope of a line which passes through the origin and the mid-point of the line segment joining the points P(0, -4) and B(8, 0).

Solution:

Step 1: Find the mid-point of the segment PB.

Mid-point M = $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (\frac{0+8}{2}, \frac{-4+0}{2}) = (4, -2)$.

Step 2: Find the slope of the line passing through the origin O(0,0) and M(4, -2).

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 0}{4 - 0} = -\frac{2}{4} = -\frac{1}{2}$$

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Example 2: Deriving the Point-Slope Form

Question: A line passes through $(x_1, y_1)$ and $(h, k)$. If the slope of the line is m, show that $k-y_1 = m(h-x_1)$.

Solution:

This demonstrates the derivation of the point-slope form, a fundamental equation for a line.

By definition, the slope 'm' of the line passing through the two points is:

$$m = \frac{\text{change in y}}{\text{change in x}} = \frac{k - y_1}{h - x_1}$$

To remove the fraction, we multiply both sides by the denominator $(h-x_1)$:

$$m(h - x_1) = k - y_1$$

This is the point-slope relationship, which is often written as $y-y_1=m(x-x_1)$.

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Example 3: Collinearity of Points

Question: Using the concept of slope, show that the points A(4, 4), B(3, 5), and C(1, 7) are collinear.

Solution:

Three points are collinear (lie on the same straight line) if the slope between any two pairs of the points is the same.

• Slope of AB: $m_{AB} = \frac{5-4}{3-4} = \frac{1}{-1} = -1$.

• Slope of BC: $m_{BC} = \frac{7-5}{1-3} = \frac{2}{-2} = -1$.

Since $m_{AB} = m_{BC}$ and they share a common point B, the points A, B, and C must lie on the same line. Thus, they are collinear.

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Example 4: Angle Between Two Lines

Question: Find the acute angle between two lines whose slopes are $\sqrt{3}$ and $1/\sqrt{3}$.

Solution:

Let $m_1 = \sqrt{3}$ and $m_2 = 1/\sqrt{3}$. We use the formula for the angle between two lines:

$$\tan\theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| = \left| \frac{1/\sqrt{3} - \sqrt{3}}{1 + (\sqrt{3})(1/\sqrt{3})} \right|$$

$$= \left| \frac{(1-3)/\sqrt{3}}{1+1} \right| = \left| \frac{-2/\sqrt{3}}{2} \right| = \left|-\frac{1}{\sqrt{3}}\right| = \frac{1}{\sqrt{3}}$$

Since $\tan\theta = 1/\sqrt{3}$, the acute angle $\theta$ is $30^\circ$ or $\pi/6$ radians.