Equation of a line

The coordinates of a point are the values of $x$ and $y$ enclosed in brackets, which describe the position of the point in a line on the plane. The plane is called the xy-plane, and it has two axes:

1. Horizontal axis known as the x-axis
2. Vertical axis known as the y-axis

Consider the xy-plane below: The coordinates of points A, B, C, D, and E are:

A(2, 3)

B(4, 4)

C(-3, -1)

D(2, -4)

E(1, 0)

The gradient of a line is the rate of change of $y$ with respect to $x$. The general formula for the gradient $m$ of a line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

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

Example 1

Find the gradient of the line joining the points $A(2, 3)$ and $B(4, 4)$:

Using the gradient formula:

$$m = \frac{4 - 3}{4 - 2} = \frac{1}{2}$$

Therefore, the gradient of the line joining the points $A(2, 3)$ and $B(4, 4)$ is $\frac{1}{2}$.

Example 2

Find the value of $k$ if the gradient of the line joining the points $(2, -3)$ and $(k, 5)$ is -2:

We use the gradient formula and the given gradient value:

$$m = \frac{5 - (-3)}{k - 2} = -2$$ $$\frac{8}{k - 2} = -2$$ $$8 = -2(k - 2)$$ $$8 = -2k + 4$$ $$8 - 4 = -2k$$ $$4 = -2k$$ $$k = -2$$

Therefore, the value of $k$ is $-2$.

The equation of a straight line can be determined if one of the following is given:

1. Gradient and y-intercept
2. Gradient and a point on the line
3. Two points on the line

Example 3

Find the equation of the line with the following:

1. Gradient $2$ and y-intercept $3$
2. Gradient $-1$ and passing through the point $(4, 2)$
3. Passing through the points $(1, 2)$ and $(3, 6)$

Solution:

1. For gradient $2$ and y-intercept $3$, the equation of the line is:

$$y = 2x + 3$$

2. For gradient $-1$ and passing through $(4, 2)$, use the point-slope form of the equation:

$$y - 2 = -1(x - 4)$$ $$y - 2 = -x + 4$$ $$y = -x + 6$$

3. For the points $(1, 2)$ and $(3, 6)$, calculate the gradient first:

$$m = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$$

Then use the point-slope form of the equation with point $(1, 2)$:

$$y - 2 = 2(x - 1)$$ $$y - 2 = 2x - 2$$ $$y = 2x$$

The equation of a line can be expressed in two forms:

• Slope-intercept form: $y = mx + c$, where $m$ is the gradient and $c$ is the y-intercept.
• Point-slope form: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the gradient.

Example 4

Find the gradient of the following lines:

• Line 1: $2x + 3y = 6$
• Line 2: $5x - 2y = 10$

Solution:

1. For line 1, rewrite in slope-intercept form:

$$2x + 3y = 6$$ $$3y = -2x + 6$$ $$y = -\frac{2}{3}x + 2$$

The gradient is $-\frac{2}{3}$.

2. For line 2, rewrite in slope-intercept form:

$$5x - 2y = 10$$ $$-2y = -5x + 10$$ $$y = \frac{5}{2}x - 5$$

The gradient is $\frac{5}{2}$.

The x-intercept of a line is the point where the line crosses the x-axis (where $y = 0$). The y-intercept is the point where the line crosses the y-axis (where $x = 0$).

Example 5

Find the y-intercept of the following lines:

• Line 1: $2x - 3y = 6$
• Line 2: $4x + y = 8$

Solution:

1. For line 1, set $x = 0$ to find the y-intercept:

$$2(0) - 3y = 6$$ $$-3y = 6$$ $$y = -2$$

The y-intercept is $-2$.

2. For line 2, set $x = 0$ to find the y-intercept:

$$4(0) + y = 8$$ $$y = 8$$

The y-intercept is $8$.