[b]uEquation of a straight line[/u]/b

There are 3 forms for the equation of a straight line.

(y=mx+c)

This is the most familiar form, where (m) is the gradient and (c) is the y-intercept. You will usually have to calculate the (y)-intercept once you know the gradient and a point the line goes through.

(y-y_1=m(x-x1))

This is the most useful form, where (m) is the gradient and (x₁₎ and (y_1) are the co-ordinates of a point that you know the line goes through. This is most useful because it does not require calculation of the (y)-intercept.

(ax+by+c=0)

This is the least useful form, where (a), (b) and (c) are integers. However, sometimes a question will ask you to give the equation of a line in this form, and you can do so by finding the equation of the line in any of the other 2 forms, then rearranging the equation to get this form.

[b]uCalculation of the gradient[/u]/b

If you have two points ((x_1,y1)) and ((x_2,y_2)), then the gradient ((m)) of the line that goes through these two points can be calculated by the difference of the (y)-coordinates divided by the difference of the (x)-coordinates:

(m=\dfrac{y_{2-y1}{x2-x1})}

The order in which you do the subtraction does not matter, as long as you are consistent.

[b]uParallel and perpendicular lines[/u]/b

For two lines with gradients (m₁₎ and (m_2),
[ul]liif the lines are parallel, then the gradients are equal, i.e. (m_1=m2);[/li]liif the lines are perpendicular, then the gradients are bnegative reciprocals/b of each other, i.e. (m_{1=\dfrac{-1}{m2})[/li]}/ul
[b]uDistance between two points[/u]/b

If you have two points ((x_1,y1)) and ((x_2,y_2)), then the distance ((d)) between these two points can be calculated by using Pythagoras' Theorem:

(d=\sqrt{(x_{2-x1)^2+(y2-y1)2})}

As the calculation involves squaring the difference of the coordinates, the order in which you subtract does not matter.

[b]uMidpoint of a line segment[/u]/b

If you have a line segment with endpoints ((x_1,y1)) and ((x_2,y_2)), the midpoint (M) can be calculated by finding the average of both the (x) and (y) coordinates:

(M=\Big(\dfrac{x_{1+x2}{2},\dfrac{y1+y2}{2}\Big))}

[b]uPoint of intersection between two lines[/u]/b

The point of intersection between two lines can be found by solving simultaneous equations.

[b]uModelling using straight lines[/u]/b

Straight line graphs can be used for modelling in a variety of contexts, such as the line for converting degrees Celsius to degrees Fahrenheit, distance against time for constant speed, etc.

There are 3 forms for the equation of a straight line:
ul
li(y=mx+c)/li
li(y-y_1=m(x-x1))/li
li(ax+by+c=0)[/li]/ul
The gradient ((m)) of the line that goes through ((x_1,y1)) and ((x_2,y_2)) is:

(m=\dfrac{y_{2-y1}{x2-x1})}

Parallel lines: (m_1=m2)
Perpendicular lines (m_{1=\dfrac{-1}{m2})}

The distance (d) between two points ((x_1,y1)) and ((x_2,y_2)) is:

(d=\sqrt{(x_{2-x1)^2+(y2-y1)2})}

The midpoint (M) of a line segment with endpoints ((x_1,y1)) and ((x_2,y_2)) is:

(M=\Big(\dfrac{x_{1+x2}{2},\dfrac{y1+y2}{2}\Big))}

The point of intersection between two lines can be found by solving simultaneous equations.