6.4 Equation of Straight Lines (Part 1)

Posted on April 21, 2020 by user

6.4.1 Axes Intercepts and Gradient

(A) Formula for gradient:

1. Gradient of the line joining (x₁, y_l) and (x₂, y₂) is:

2. Gradient of the line with knowing x–intercept and y–intercept is:

3. The gradient of the straight line joining P and Q is equal to the tangent of angle θ, where θ is the angle made by the straight line PQ and the positive direction of the x-axis.

(B) Collinear points

The gradient of a straight line is always constant i.e.the gradient of AB is equal to the gradient of BC.

Example 1:

The gradient of the line passing through point (k, 1 – k) and point (–3k, –3) is 5. Find the value of k.

Solution:

$\begin{array}{l} Gradient, m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ \frac{- 3 - (1 - k)}{- 3 k - k} = 5 \\ \frac{- 3 - 1 + k}{- 4 k} = 5 \\ - 4 + k = - 20 k \\ 21 k = 4 \\ k = \frac{4}{21} \end{array}$

Example 2:

Based on the diagram below, find the gradient of the line.

Solution:

$Gradient, m = - (\frac{y intercept}{x intercept}) = - (\frac{- 5}{10}) = \frac{1}{2}$