10.1 Gradient of a Straight Line - Mathematics

10.1.1 Gradient of a Straight Line

The gradient of a straight line is the ratio of the vertical distance to the horizontal distance between any two given points on the straight line.

Gradient, m = \frac{Vertical distance}{Horizontal distance}

$Gradient, m = \frac{Vertical distance}{Horizontal distance}$

Example:

Find the gradient of the straight line above.

Solution:

\begin{matrix} Gradient, m = \frac{Vertical distance}{Horizontal distance} = \frac{4 units}{6 units} = \frac{2}{3} \end{matrix}

$\begin{array}{l} Gradient, m = \frac{Vertical distance}{Horizontal distance} \\ = \frac{4 units}{6 units} \\ = \frac{2}{3} \end{array}$

10.1.2 Finding the Gradient of a Straight Line

The gradient, m, of a straight line which passes through P (x₁, y₁) and Q (x₂, y₂) is given by,

m_PQ =

\frac{y_{2} - y_{1}}{x_{2} - x_{1}}

$\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Example 1:

Find the gradient of the straight line joining two points P and Q in the above diagram.

Solution:

P = (x₁, y₁) = (4, 3), Q = (x₂, y₂) = (10, 5)

Gradient of the straight line PQ

\begin{matrix} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{5 - 3}{10 - 4} = \frac{2}{6} = \frac{1}{3} \end{matrix}

$\begin{array}{l} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{5 - 3}{10 - 4} \\ = \frac{2}{6} = \frac{1}{3} \end{array}$

Example 2:

Calculate the gradient of a straight line which passes through point A (7, -3) and point B (-3, 6).

Solution:

A = (x₁, y₁) = (7, -3), B = (x₂, y₂) = (-3, 6)

Gradient of the straight line AB

\begin{matrix} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{6 - (- 3)}{- 3 - 7} = - \frac{9}{10} \end{matrix}

$\begin{array}{l} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{6 - (- 3)}{- 3 - 7} \\ = - \frac{9}{10} \end{array}$

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