4.1 Straight Lines

In geometry, a straight line is a fundamental concept that represents the shortest distance between two points. Straight lines can be described algebraically, geometrically, and through their various properties.

1. Definition of a Straight Line

A straight line can be defined as a set of points extending infinitely in two directions. In a Cartesian coordinate system, a straight line can be represented using an equation that relates the coordinates of points on the line.

2. Slope-Intercept Form

One of the most common ways to express the equation of a straight line is in the slope-intercept form:

$$y=mx+b$$

Where:

• $y$ is the dependent variable (the output).
• $x$ is the independent variable (the input).
• $m$ is the slope of the line (the rate of change of $y$ with respect to $x$).
• $b$ is the y-intercept (the value of $y$ when $x=0$).

3. Slope of a Line

The slope $m$ measures how steep the line is. It can be calculated using the formula:

m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1}m=x2​−x1​y2​−y1​​

Where (x1,y1)(x_1, y_1)(x1​,y1​) and (x2,y2)(x_2, y_2)(x2​,y2​) are two distinct points on the line.

• If $m>0$: The line rises as it moves from left to right.
• If $m<0$: The line falls as it moves from left to right.
• If $m=0$: The line is horizontal.
• If $m$ is undefined: The line is vertical.

4. General Form of the Line

The general form of a line in two dimensions is given by:

$$Ax+By+C=0$$

Where:

• $A$, $B$, and $C$ are constants.
• $A$ and $B$ cannot both be zero.

To convert from slope-intercept form to general form, you can rearrange the equation:

$$y-mx-b=0\Rightarrow mx-y+b=0$$

5. Point-Slope Form

Another useful form of the equation of a line is the point-slope form:

y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1​=m(x−x1​)

Where:

• (x1,y1)(x_1, y_1)(x1​,y1​) is a specific point on the line.
• $m$ is the slope.

6. Parallel and Perpendicular Lines

• Parallel Lines: Two lines are parallel if they have the same slope (m1=m2m_1 = m_2m1​=m2​). They will never intersect.
• Perpendicular Lines: Two lines are perpendicular if the product of their slopes is $-1$: m1⋅m2=−1m_1 \cdot m_2 = -1m1​⋅m2​=−1

7. Finding the Equation of a Line

To find the equation of a line, you typically need:

1. Two Points: Use the two-point formula or calculate the slope and then use point-slope form.
2. Point and Slope: Use the point-slope form directly.

Example: Find the equation of the line that passes through the points $(1,2)$ and $(3,6)$.

1. Calculate the slope $m$:m=6−23−1=42=2m = \frac{6 – 2}{3 – 1} = \frac{4}{2} = 2m=3−16−2​=24​=2
2. Use point-slope form with point $(1,2)$:$$y-2=2(x-1)$$
3. Simplifying:$$y-2=2x-2\Rightarrow y=2x$$

4.1.1 Equation of straight lines

The equation of a straight line can be expressed in several forms, each useful for different purposes. Here’s a detailed look at the various forms and how to derive and use them.

1. Slope-Intercept Form

The slope-intercept form of a straight line is given by:

$$y=mx+b$$

Where:

• $m$ is the slope of the line, which represents the rate of change of $y$ with respect to $x$.
• $b$ is the y-intercept, the point where the line crosses the y-axis (when $x=0$).

Example: If a line has a slope of $3$ and a y-intercept of $-2$, the equation is:

$$y=3x-2$$

2. Point-Slope Form

The point-slope form is useful when you know a point on the line and its slope:

y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1​=m(x−x1​)

Where:

• (x1,y1)(x_1, y_1)(x1​,y1​) is a point on the line.
• $m$ is the slope.

Example: Given a point $(2,3)$ and a slope of $4$, the equation is:

$$y-3=4(x-2)$$

Simplifying this gives:

$$y-3=4x-8\Rightarrow y=4x-5$$

3. General Form

The general form of a line is represented as:

$$Ax+By+C=0$$

Where:

• $A$, $B$, and $C$ are constants.
• $A$ and $B$ cannot both be zero.

To convert from slope-intercept or point-slope form to general form, rearrange the equation.

Example: Starting from the slope-intercept form $y=2x+3$:

Rearranging gives:

$$-2x+y-3=0\Rightarrow 2x-y+3=0$$

Here, $A=2$, $B=-1$, and $C=3$.

4. Intercept Form

The intercept form of a straight line is:

xa+yb=1\frac{x}{a} + \frac{y}{b} = 1ax​+by​=1

Where:

• $a$ is the x-intercept (the point where the line crosses the x-axis).
• $b$ is the y-intercept.

Example: If a line crosses the x-axis at $4$ and the y-axis at $3$, the equation is:

x4+y3=1\frac{x}{4} + \frac{y}{3} = 14x​+3y​=1

5. Finding the Equation of a Line Given Two Points

To find the equation of a line given two points, follow these steps:

1. Find the Slope ($m$): Using the formula:m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1}m=x2​−x1​y2​−y1​​
2. Use Point-Slope Form: With one of the points and the slope, use:y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1​=m(x−x1​)

Example: Given points $(1,2)$ and $(3,6)$:

1. Calculate the slope:m=6−23−1=42=2m = \frac{6 – 2}{3 – 1} = \frac{4}{2} = 2m=3−16−2​=24​=2
2. Use the point $(1,2)$ in point-slope form:$$y-2=2(x-1)$$Simplifying gives:

   $$y=2x$$

6. Parallel and Perpendicular Lines

• Parallel Lines: Two lines are parallel if they have the same slope. If line L1:y=m1x+b1L_1: y = m_1x + b_1L1​:y=m1​x+b1​ and line L2:y=m2x+b2L_2: y = m_2x + b_2L2​:y=m2​x+b2​ are parallel, then m1=m2m_1 = m_2m1​=m2​.
• Perpendicular Lines: Two lines are perpendicular if the product of their slopes is $-1$. If line L1:y=m1x+b1L_1: y = m_1x + b_1L1​:y=m1​x+b1​ and line L2:y=m2x+b2L_2: y = m_2x + b_2L2​:y=m2​x+b2​ are perpendicular, then:m1⋅m2=−1m_1 \cdot m_2 = -1m1​⋅m2​=−1

Example: If one line has a slope of $2$, a line perpendicular to it will have a slope of −12-\frac{1}{2}−21​.

4.1.2 Perpendicular distance of a line

The perpendicular distance from a point to a line is the shortest distance from the point to the line, measured along a line that is perpendicular to the given line. This concept is crucial in various fields such as geometry, computer graphics, and engineering.

1. Equation of a Line

Consider the line given by the general form:

$$Ax+By+C=0$$

Where $A$, $B$, and $C$ are constants.

2. Distance Formula

To find the perpendicular distance $d$ from a point (x0,y0)(x_0, y_0)(x0​,y0​) to the line $Ax+By+C=0$, we use the formula:

d=∣Ax0+By0+C∣A2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}d=A2+B2​∣Ax0​+By0​+C∣​

• ∣Ax0+By0+C∣|Ax_0 + By_0 + C|∣Ax0​+By0​+C∣ gives the absolute value of the line equation evaluated at the point, which represents the algebraic distance from the point to the line.
• A2+B2\sqrt{A^2 + B^2}A2+B2​ normalizes this distance by dividing by the magnitude of the vector $(A,B)$, which represents the direction of the line.

3. Derivation of the Distance Formula

To derive the distance formula, consider the following steps:

1. The line can be represented as $Ax+By+C=0$.
2. The normal vector to the line is given by $(A,B)$.
3. The distance from the point to the line is the projection of the vector from the point to the line onto this normal vector.

4. Example Calculation

Example: Find the perpendicular distance from the point $(3,4)$ to the line $2x+3y-6=0$.

1. Identify $A=2$, $B=3$, $C=-6$, x0=3x_0 = 3x0​=3, and y0=4y_0 = 4y0​=4.
2. Substitute into the distance formula: d=∣2(3)+3(4)−6∣22+32d = \frac{|2(3) + 3(4) – 6|}{\sqrt{2^2 + 3^2}}d=22+32​∣2(3)+3(4)−6∣​ =∣6+12−6∣4+9=∣12∣13=1213= \frac{|6 + 12 – 6|}{\sqrt{4 + 9}} = \frac{|12|}{\sqrt{13}} = \frac{12}{\sqrt{13}}=4+9​∣6+12−6∣​=13​∣12∣​=13​12​
3. Simplifying gives: $$d\approx 3.32$$

5. Application of Perpendicular Distance

Understanding the perpendicular distance is useful in various applications:

• Geometric Problems: To find the closest point on a line from a given point.
• Optimization: In operations research and linear programming, to determine the optimal path or solution.
• Computer Graphics: For rendering and calculating the intersection of objects and lines.

4.1.3 Angle between lines

The angle between two lines can be determined using the slopes of the lines. The formula for finding the angle $\theta$ between two lines with slopes m1m_1m1​ and m2m_2m2​ is derived from the tangent function.

1. Slopes of Lines

For two lines represented in slope-intercept form:

• Line 1: y=m1x+b1y = m_1x + b_1y=m1​x+b1​
• Line 2: y=m2x+b2y = m_2x + b_2y=m2​x+b2​

Here, m1m_1m1​ and m2m_2m2​ are the slopes of the two lines.

2. Formula for the Angle

The angle $\theta$ between the two lines can be calculated using the following formula:

tan⁡(θ)=∣m1−m21+m1m2∣\tan(\theta) = \left| \frac{m_1 – m_2}{1 + m_1 m_2} \right|tan(θ)=​1+m1​m2​m1​−m2​​​

Where:

• $\theta$ is the angle between the two lines.
• The absolute value ensures that we get a positive angle.

To find $\theta$, you can take the arctangent:

θ=tan⁡−1(∣m1−m21+m1m2∣)\theta = \tan^{-1}\left( \left| \frac{m_1 – m_2}{1 + m_1 m_2} \right| \right)θ=tan−1(​1+m1​m2​m1​−m2​​​)

3. Conditions for Angle Calculation

• If m1m_1m1​ and m2m_2m2​ are equal, the lines are parallel, and the angle θ=0∘\theta = 0^\circθ=0∘.
• If m1⋅m2=−1m_1 \cdot m_2 = -1m1​⋅m2​=−1, the lines are perpendicular, and the angle θ=90∘\theta = 90^\circθ=90∘.

4. Example Calculation

Example: Find the angle between the lines given by the equations:

1. $y=2x+1$ (slope m1=2m_1 = 2m1​=2)
2. y=−12x+3y = -\frac{1}{2}x + 3y=−21​x+3 (slope m2=−12m_2 = -\frac{1}{2}m2​=−21​)

Step 1: Identify the slopes:

• m1=2m_1 = 2m1​=2
• m2=−12m_2 = -\frac{1}{2}m2​=−21​

Step 2: Use the angle formula:

tan⁡(θ)=∣2−(−12)1+2(−12)∣\tan(\theta) = \left| \frac{2 – \left(-\frac{1}{2}\right)}{1 + 2\left(-\frac{1}{2}\right)} \right|tan(θ)=​1+2(−21​)2−(−21​)​​

Calculating the numerator and denominator:

• Numerator:2+12=42+12=522 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}2+21​=24​+21​=25​
• Denominator:$$1-1=0$$

Since the denominator is zero, it indicates that the lines are perpendicular:

tan⁡(θ)=undefined⇒θ=90∘\tan(\theta) = \text{undefined} \Rightarrow \theta = 90^\circtan(θ)=undefined⇒θ=90∘

5. Summary of Angle Relationships

• Parallel Lines: If two lines have the same slope (m1=m2m_1 = m_2m1​=m2​), they do not intersect, and the angle between them is 0∘0^\circ0∘.
• Perpendicular Lines: If the product of their slopes is $-1$ (m1m2=−1m_1 m_2 = -1m1​m2​=−1), the lines are perpendicular, and the angle between them is 90∘90^\circ90∘.