Linear equations, a cornerstone in algebra, offer a gateway to understanding complex mathematical concepts. This comprehensive guide explores the various forms of linear equations, their practical applications, and techniques for interconversion.

**1. Slope-Intercept Form **$(y = mx + c)$

The slope-intercept form, represented as $y = mx + c$, is invaluable for quickly identifying the slope $(m)$ and y-intercept $(c)$ of a line. It's particularly useful in graphing linear equations and understanding their behavior.

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**Example:**

Suppose we want to find the equation of a line with a slope of 4 and passing through the point (1, 3).

**Solution**:

**Given:**

- Slope (m): $4$
- Point: $(1, 3)$

**1. Use the Slope-Intercept Form**

The slope-intercept form of a line is: $y = mx + c$

Where $m$ is the slope and $c$ is the y-intercept.

**2. Substitute the Given Point**

Substitute $x = 1$ and $y = 3$ into the equation to find $c$:

$3 = 4 \times 1 + c$

**3: Solve for c**

Solve the equation for $c$:

$3 = 4 + c$

$c = 3 - 4$

$c = -1$

With $m = 4$ and $c = -1$, the equation of the line is: $y = 4x - 1$

This is the equation of the line with a slope of $4$ passing through the point $(1, 3)$. It shows the relationship between $x$ and $y$ for every point on this line.

**2. Point-Slope Form **$(y - y_1 = m(x - x_1))$

The point-slope form, $y - y_1 = m(x - x_1)$, is essential when the slope and a specific point on the line are known. It's particularly useful in problems involving tangents to curves or constructing lines parallel or perpendicular to a given line.

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**Example:**

Find the equation of a line that is perpendicular to $y = 2x - 5$ and passes through (3, 4).

**Solution**:

**1. Determine the Slope of the Perpendicular Line**

- The given line is $y = 2x - 5$.
- The slope of this line is the coefficient of $x$, which is $2$.
- The slope of a line perpendicular to this would be the negative reciprocal of $2$. The negative reciprocal of $m$ is $-\frac{1}{m}$.
- Therefore, the slope of the perpendicular line is $-\frac{1}{2}$.

**2. Use the Point-Slope Formula**

- The point-slope formula is $y - y_1 = m(x - x_1)$, where ( m ) is the slope and $(x_1, y_1)$ is the point through which the line passes.
- In this case, $m = -\frac{1}{2}$ and the point is $(3, 4)$.
- Substituting these values, the result is: $y - 4 = -\frac{1}{2}(x - 3)$

**3. Simplify the Equation**

$y = mx + b$

$=y - 4 = -\frac{1}{2}x + \frac{3}{2}$

$=y = -\frac{1}{2}x + \frac{3}{2} + 4$

$=y = -\frac{1}{2}x + \frac{11}{2}$

The equation of the line that is perpendicular to $y = 2x - 5$ and passes through the point $(3, 4)$ is: $y = -\frac{1}{2}x + \frac{11}{2}$

**3. General Form **$(ax + by + c = 0)$

The general form, $ax + by + c = 0$, is a versatile representation used in various mathematical disciplines. It's particularly useful in systems of linear equations and in computational geometry.

**Example:**

Convert the equation $3y = 6x - 9$ into the general form.

**Solution**:

**1. Start with the Given Equation**

The given equation is: $3y = 6x - 9$

**2. Rearrange the Equation**

Rearrange this equation to get it into the form $ax + by + c = 0$.

$-6x + 3y = -9$**3. Adjust the Equation to Match the General Form**

Add ( 9 ) to both sides:

$-6x + 3y + 9 = 0$The equation $3y = 6x - 9$ in the general form is: $-6x + 3y + 9 = 0$

**Intersection Points**

Finding intersection points of lines is crucial in understanding their relationships and is a key concept in algebra and geometry.

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**Example:**

Determine the intersection point of the lines $y = -x + 6$ and $y = 2x + 4$.

**Solution**:

**1. Set the Equations Equal to Each Other**

**2. Solve for x**

Solve this equation for $-x + 6 = 2x + 4$:

$-x - 2x = 4 - 6$$-3x = -2$$x = \frac{-2}{-3}$$x = \frac{2}{3}$**3. Solve for y**

Substitute $x = \frac{2}{3}$ to find $y$.

Choose one between the two original equations: $y = -x + 6$

$y = -\left(\frac{2}{3}\right) + 6$$y = -\frac{2}{3} + \frac{18}{3}$$y = \frac{16}{3}$The intersection point of the lines $y = -x + 6$ and $y = 2x + 4$ is $\left(\frac{2}{3}, \frac{16}{3}\right).$

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.