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The equation of any straight line can be written in the form \(y = mx + c\) where \(m\) and \(c\) are numbers and \(x\) and \(y\) represent the \(x\) and \(y\) coordinates.

The equation is a rule connecting the \(x\) and \(y\) coordinates - the rule is true for all points on the line.

For example:
For the line \(y = 2x -3\), for any value of \(x\) , the \(y\) coordinate can be found by doubling that value and subtracting 3.

Example

A line has the equation \(y = 5x + 2\). What are the coordinates of a point, P, on the line where \(x =-2\)?

Answer:

When \(x = -2\), \(y = 5(-2) + 2\)
\(=-8\)

Answer:
P is the point (-2, -8)

When the equation of a straight line is in the form \(y = mx + c\), \(m\) and \(c\) are numbers which give useful information about the graph of the line.

In the equation \(y = mx + c\), \(m\) is the gradient of the line. The gradient is a measure of how steep the line is.

To find the gradient of a line joining two points, divide the difference in the y-coordinates by the difference in the x-coordinates.

This is sometimes given as an equation:

For points \((x_1, y_1)\) and \((x_2, y_2)\)
Gradient = \(\frac{y_2 - y_1}{x_2 - x_1}\)

Solution

Difference in y-coordinates \( 11 - 5 = 6\)
Difference in x-coordinates \( 4 - 10 = -6\)

Gradient =\(\frac{difference~in~y~values}{difference~in~x~values} = \frac{6}{-6} = -1\)

It may be helpful to set out your work like this:

\(\matrix{ & ~x & ~y \cr Coordinates~of~D & ~4 & ~11 \cr Coordinates~of~E &10 & 5 \cr Subtract & \hline ~-6 & ~6}\)

Gradient =\(\frac{difference~in~y~values}{difference~in~x~values} = \frac{6}{-6} = -1\)

Alternatively, using the formula:

Gradient =\(\frac{y_2 - y_1}{x_2-x_1} = \frac{11-5}{4-10} = \frac{6}{-6} = -1\)

If the coordinates of two points on a line are known, the equation of the line can be found by letting the equation be \(y = mx + c\) and then finding values for m and c.

If the value of either m or c is already known, the equation can be found using the coordinates of one point on the line.

If the gradient is known or has been calculated, a formula can be used.For points \((x_1, y_1)\) and \((x_2, y_2)\), this is:

\(y - y_1 = m(x - x_1)\)

Example

The equation of a line is \(2y + 3x = 8\)

1. What is the gradient of the line?
2. At what point does the line cross the y-axis?

Solution:

To determine the values of m and c, the line has to be in the form \(y = mx + c\).

Rearrange the equation

\(2y + 3x = 8\)
\(2y = -3x + 8\)
\(y = -\frac{3}{2}x + 4\)

The values for m and c can now be read from the equation

\(m = -\frac{3}{2}\) and \(c = 4\)

1. The gradient of the line is \(m = -\frac{3}{2}\)
2. The intercept is the point (0, 4)

Solution:

Let the equation of the line be \(y = mx + c\)

The gradient is -2

Put this value for m into the equation

\(y = -2x + c\)

To find the value for c, use the x and y coordinates of the point P, (-3, 2)

Substituting \(x = -3\) and \(y =2\) into the equation \(y = -2x + c\)
\(2 = -2(-3) + c\)
\(2 = 6 + c\)
\(c = -4\)

The equation of the line is y = -2x -4

Alternatively, using the formula:

\(y - y_1 = m(x - x_1)\)

Equation of the line is:

\(y - 2 = -2(x - (-3))\)
\(y - 2 = -2(x + 3)\)
\(y - 2 = -2x - 6\)
\(\mathbf{y = -2x - 4}\)

Solution:

Let the equation be \(y = mx + c\).

Values for m and c are to be found.

m is the gradient

\(\matrix{ & ~x & ~y \cr Coordinates~of~A & ~3 & ~2 \cr Coordinates~of~B & -1 & -0 \cr Subtract & \hline ~4 & -8}\)

Gradient = \(\frac{difference~in~y~values}{difference~in~x~values} = \frac{-8}{4} = -2\)

The gradient is -2
Put this value for m into the equation

\(y = -2x + c\)

To find the value for c, use the x and y coordinates of either of the points A or B.
Using the point A may be easier as both values are positive.

Substituting \(x = 3\) and \(y =2\) into the equation \(y = m-2x + c\)

\(2 = -2(3) + c\)
\(2 = -6 + c\)
\(c = 8\)

The equation of the line is \(\mathbf{y = -2x +8}\)

As mentioned in the introduction, parallel lines have the same gradient.
When written in the\(y = mx + c\) form, parallel lines have the same value of m.

Example

Which of these lines are parallel to the line \(y = 5x - 4\)?

1. \(y = 4x + 5\)
2. \(y - 5x = 3\)
3. \(2y = 10x + 4\)
4. \(y = 5 - 5x\)

Solution:

When written as \(y = mx + c\), m should = 5

1. Not parallel   \(m = 4\)
2. Parallel     can be rearranged to give \(y = 5x + 3\)
3. Parallel     can be rearranged to give \(y = 5x + 2\)
4. Not parallel    can be rearranged to give \(y = -5x + 5\)

Example

A straight line goes through the point (-10, -2) and is parallel to the line \(y - x = 8\).
What is the equation of this line?

Solution:

Let the equation of the unknown line be \(y = mx + c\)

Rearrange the equation \(y - x = 8\) to give\(y = x + 8\)
\(y = (1)x + 8\)
\(m = 1\)

Gradient of the unknown line is also 1 as they are parallel
Equation of line is \(y = x + c\)

To find the value of c, use the point (-10, -2)
\(-2 = 1(-10) + c\)
\(c = 8\)

Equation of the line is \(\mathbf{y = x + 8}\)