03Bfindingrule

toc = Determining Linear Equations =

To find the equation of a straight line, we need or
 * two points
 * one point and the gradient

In either case, the method is essentially the same:
 * 1) Obtain the gradient ** (m) **
 * 2) Obtain the y-intercept ** (c) **
 * 3) Write the equation in the form: ** y = mx + c **

Finding the gradient
Recall, the gradient is given by: where
 * the ** rise ** is how much the y value changes between two points (how far up)
 * the ** run ** is how much the x value changes between two points (how far to the right)

If we are given two points (x 1, y 1 ) and (x 2 , y 2 ) then
 * rise = y 2 – y 1 {the change in y-value}
 * run = x 2 – x 1 {the change in x-value}

So the rule for gradient becomes:

** Example 1 **
Find the equation of the line shown at the right __**Solution:**__

We are given two points
 * x-intercept at –3,
 * so (x 1, y 1 ) = (–3, 0)
 * y-intercept at 5 so
 * so (x 2, y 2 ) = (0, 5)

To find the equation, we follow 3 steps: math \\ . \qquad m = \dfrac{y_2-y_1}{x_2-x_1} \\ \\ . \qquad \quad = \dfrac{5-0}{0- ^-3} \\ \\ . \qquad \quad = \dfrac{5}{3} math
 * 1: Find the gradient (m) **

math . \qquad c = 5 \qquad \textit{from the graph} math
 * 2: Find the y-intercept (c) **

math . \qquad y = \dfrac{5}{3}x + 5 math
 * 3: Write in the form: y = mx + c **

Finding the y-intercept
Given we have the gradient (we should because that was step 1)
 * write the information we know in the form ** y = mx + c **
 * use __one__ of the points we are given and substitute ** x ** and ** y **
 * solve for c

** Example 2 **
Find the equation of a straight line with a gradient of 3 and that passes through (2, 10)

__**Solution:**__

To find the equation, we follow 3 steps: math . \qquad m = 3 \qquad \textit{from question} math
 * 1: Find the gradient (m) **

math \\ . \qquad \textit{write the information we know in the form } y = mx + c \\ \\ . \qquad y = 3x + c math
 * 2: Find the y-intercept (c) **

math \\ . \qquad \textit{given } (2, \;10) \; \textit{ so substitute } x = 2 \textit{ and } y = 10 \\ \\ . \qquad 10 = 3 \times 2 + c \\ \\ . \qquad 10 = 6 + c \qquad \{-6\} \\ \\ . \qquad c = 4 math

math . \qquad y = 3x + 4 math
 * 3: Write in the form: y = mx + c **

** Example 3 **
Find the equation of a straight line that passes through (–2, 6) and (4, 3)

__**Solution:**__ To find the equation, we follow 3 steps:
 * 1: Find the gradient (m) **
 * we are given two points
 * (x 1, y 1 ) = (–2, 6)
 * (x 2, y 2 ) = (4, 3)

math \\ . \qquad m = \dfrac{y_2-y_1}{x_2-x_1} \\ \\ . \qquad \quad = \dfrac{3-6}{4- ^-2} \\ \\ . \qquad \quad = \dfrac{-3}{6} \\ \\ . \qquad \quad = -\dfrac{1}{2} math

math \\ . \qquad \textit{write the information we know in the form } y = mx + c \\ \\ . \qquad y = -\dfrac{1}{2}x + c math
 * 2: Find the y-intercept (c) **

math \\ . \qquad \textit{given } (4, \;3) \; \textit{ so substitute } x = 4 \textit{ and } y = 3 \\ \\ . \qquad 3 = -\dfrac{1}{2} \times 4 + c \\ \\ . \qquad 3 = -2 + c \qquad \{+2\} \\ \\ . \qquad c = 5 math

math . \qquad y = -\dfrac{1}{2}x + 5 math Go to top of page flat .
 * 3: Write in the form: y = mx + c **