“Teach A Level Maths” Vol. 1: AS Core Modules 1: Straight Lines and Gradients

Finding the Gradient 4 2

is The gradient of the straight line joining the points and To use this formula, we don’t need a diagram! e.g. Find the gradient of the straight line joining the points and Solution:

The gradient of the straight line joining the points and is

Find the gradient of the line joining the points A(3,–2) and B(–5,6) x1= 3 y1= –2 x2 = –5 y2 = 6

Find the gradient of the line joining the points A(3,–2) and B(–5,6) x1 = 3 y1= –2 x2 = –5 y2 = 6

m is the gradient of the line The equation of a straight line is m is the gradient of the line c is the point where the line meets the y-axis, the y-intercept e.g. has gradient m = and y-intercept, c = gradient = 2 x intercept on y-axis

e.g. Substituting x = 4 in gives gradient = 2 x intercept on y-axis ( 4, 7 ) x The coordinates of any point lying on the line satisfy the equation of the line e.g. Substituting x = 4 in gives showing that the point ( 4,7 ) lies on the line.

Finding the equation of a straight line when we know its gradient, m and the coordinates of a point on the line (x1,y1). Using , m is given, so we can find c by substituting for y, m and x. e.g. Find the equation of the line with gradient passing through the point Solution: (-1, 3) x Add 2 to both sides C = 5 So,

Using the formula when we are given two points on the line e.g. Find the equation of the line through the points Solution: First find the gradient Now use with We could use the 2nd point, (-1, 3) instead of (2, -3) Add 4 to both sides -3 = -4 + c

SUMMARY Equation of a straight line where m is the gradient and c is the intercept on the y-axis Gradient of a straight line where and are points on the line

Exercise 1. Find the equation of the line with gradient 2 which passes through the point . Solution: So, 2. Find the equation of the line through the points Solution: So,

We sometimes rearrange the equation of a straight line so that zero is on the right-hand side ( r.h.s. ) e.g. can be written as We must take care with the equation in this form. e.g. Find the gradient of the line with equation Solution: Rearranging to the form : so the gradient is