__6.4 Axes Intercepts and Gradient__**(A) Formula for gradient:**

**2.**Gradient of the line with knowing

**and**

*x*–intercept

*y*–interceptis:

**3.**The

**gradient of the straight line joining**

*P*and

*Q*is equal to the

**tangent of angle**, where

*θ**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(–3

$$\begin{array}{l}\text{Gradient,}m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\ \frac{-3-\left(1-k\right)}{-3k-k}=5\\ \frac{-3-1+k}{-4k}=5\\ -4+k=-20k\\ 21k=4\\ k=\frac{4}{21}\end{array}$$
*k*, –3) is 5. Find the value of*k.*

*Solution:***Example 2:**

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

*Solution:*