Coordinate geometry © Christine Crisp

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. 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 Notice that to find c, the equation has been solved from right to left. This takes a bit of practice but reduces the chance of errors. So,

To find the equation of a straight line given 2 points on the line. e.g. Find the equation of the line through the points Solution: First find the gradient: Now on the line: Equation of line is

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

Parallel and Perpendicular Lines If 2 lines have gradients and , then: They are parallel if They are perpendicular if

e.g. 1 Find the equation of the line parallel to which passes through the point Solution: The given line has gradient 2. Let For parallel lines, is the equation of any line parallel to Using on the line

e.g. Find the equation of the line perpendicular to passing through the point . Solution: The given line has gradient 2. Let Perpendicular lines: Equation of a straight line: on the line We don’t usually leave fractions ( or decimals ) in equations. So, multiplying by 2:

SUMMARY Method of finding the equation of a straight line: If the gradient isn’t given, find the gradient using either parallel lines: or perpendicular lines: or 2 points on the line: Substitute for y, m and x in into to find c.

A Second Formula for a Straight Line ( optional ) Let ( x, y ) be any point on the line x x Let be a fixed point on the line

To use the formula we need to be given either: one point on the line and the gradient or: 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)

The mid-point is the average of the end points: M X The mid-point is the average of the end points:

A formula for the Mid-Point of AB M X The mid-point is the average of the end points: or

Exercise Find the mid-point, M of the line joining to Solution:

Using Pythagoras’ theorem:

A formula for the length of the line joining A to B Using Pythagoras’ theorem:

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

If 2 lines have gradients and , then: They are parallel if They are perpendicular if If 2 lines have gradients and , then: Equation of a straight line Gradient of a straight line where and are points on the line where m is the gradient and c is the intercept on the y-axis SUMMARY

Solution: First find the gradient: e.g. Find the equation of the line through the points Now on the line: Equation of line is

We don’t usually leave fractions ( or decimals ) in equations We don’t usually leave fractions ( or decimals ) in equations. So, multiplying by 2: e.g. Find the equation of the line perpendicular to passing through the point . Solution: The given line has gradient 2. Let Perpendicular lines: Equation of a straight line: on the line