01Flinearequation

= Finding the Equation of a Straight Line =

We know that for a straight line that passes through
 * (x 1, y 1 ) .. and
 * (x 2, y 2 )

The gradient (m) is given by the rule:

If we replace the point (x 2, y 2 ) with the general coordinates of any point on the line (x, y), and reverse the order of the equation, we get:

math . \qquad \dfrac{y - y_1}{x - x_1} = m math

Now multiply both sides by (x – x 1 )

This gives us a new rule for the equation of a straight line:

We can use this rule to find the equation of any straight line, if we know:
 * the gradient (m)
 * the coordinates of one point (x 1, y 1 )


 * Example 1 **

Find the equation of a line with a gradient of 3, that passes through the point (2, –7)


 * Solution:**

We are given
 * m = 3
 * x 1 = 2
 * y 1 = –7

Substitute these values into the rule:

math \\ . \qquad y - y_1 = m \big( x - x_1 \big) \\. \\ . \qquad y - \,^-7 = 2 \big( x - 2 \big) math

Now expand the brackets and simplify

math \\ . \qquad y + 7 = 2x - 4 \\. \\ . \qquad y = 2x - 11 math


 * Finding the equation, given 2 points **

If you are given two points, then
 * Use the gradient rule to find the value of m
 * Use m and one of the two points in the rule to find the equation of the straight line


 * Example 2 **

Find the equation of a straight line that passes through
 * (2, 8) .. and
 * (5, –4)


 * Solution:**

First find the gradient (m)

math \\ . \qquad m = \dfrac{y_2 - y_1}{x_2 - x_1} \\. \\ . \qquad m = \dfrac{^-4 - 8}{5 - 2} \\. \\ . \qquad m = \dfrac{-12}{3} \\. \\ . \qquad m = -4 math

Then find the equation of the line using (2, 8)

math \\ . \qquad y - y_1 = m \big( x - x_1 \big) \\. \\ . \qquad y - 8 = -4 \big( x - 1 \big) \\. \\ . \qquad y - 8 = -4x + 4 \\. \\ . \qquad \quad y = -4x + 12 math


 * Dealing with fractions **

If the gradient or the y-intercept turn out to be a fraction, it is still legitimate to write the equation in the form: .. //y = mx + c //.
 * If the gradient is a fraction, leave it as an improper fraction
 * Don't change it to a mixed number

However some people prefer to rearrange the equation to get rid of the fractions.
 * This can be done by multiplying both sides of the equation by the Lowest Common Denominator of the fractions
 * Then rearranging the equation to get the pronumerals (//x // and //y //) on the same side of the equation


 * Example **

Find the equation of a straight line passing through (–1, 4) and (3, 9)


 * Solution:**

First find the gradient (m)

math \\ . \qquad m = \dfrac{y_2 - y_1}{x_2 - x_1} \\. \\ . \qquad m = \dfrac{9 - 4}{3 - \,^-1} \\. \\ . \qquad m = \dfrac{5}{4} math

Then find the equation of the line using (3, 9)

math \\ . \qquad y - y_1 = m \big( x - x_1 \big) \\. \\ . \qquad y - 9 = \dfrac{5}{4} \big( x - 3 \big) \\. \\ . \qquad y - 9 = \dfrac{5x}{4} - \dfrac{15}{4} \\. \\ . \qquad \quad y = \dfrac{5x}{4} + \dfrac{21}{4} math

This is an acceptable answer, But we could remove the fractions Multiple both sides by the Lowest Common Denominator (which is 4)

math \\ . \qquad \;\; 4y = 5x + 21 \\. \\ . \qquad \;\; 4y - 5x = 21 \\. \\ . \qquad \{ \textit{ or } \} \\. \\ . \qquad \;\; -5x + 4y = 21 \\. \\ . \qquad \;\; 5x - 4y = -21 math

As you can see, this equation looks simpler than the version with fractions. .