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C1 COORDINATE GEOMETRY OF A LINE

Page history last edited by Steph Richards 15 years ago

LINK TO C1 PAST PAPERS

 

There are many sites on the internet that will help you work through the WJEC requirements for this topic (shown in the table below)

 

Try ONLINEMATHLEARNING.com 

 

 WJEC SPECIFICATION

 

2. Equation of a straight line

 

 

including the forms y = mx + c,

y − y1 = m(x − x1)

and ax + by+ c = 0.

Conditions for two straight lines to be parallel or perpendicular to each other.

 

PROPERTIES OF STRAIGHT LINE SEGMENTS

To include finding the gradient, equation, length and mid-point of a line joining two given points.

 

To include finding the equations of lines which are parallel or perpendicular to a given line.

 

Straight lines

 

A straight line has a fixed gradient. The gradient of a line and its y intercept are the two pieces of information that distinguish one line from another line.

 

Gradient of a line

 

The steepness of a line can be measured by its gradient, which is the change in the y direction divided by the change in the x direction. The letter m is used to denote the gradient. The formula to find a gradient is: m=\frac{y_2-y_1}{x_2-x_1}                                          As a side note \tan\theta=m\,.

 

 

Point-Gradient Form

 The equation of a line having the co-ordinate \left(x_1,y_1\right) and having a gradient of m is: y-y_1=m\left(x - x_1\right). Then you simply rearrange the equation into the form y = mx + c

.

Parallel lines

 A pair of lines are parallel (the symbol is\|) if their gradients are equal, m_1=m_2\,. So in order to find to find the equation of a parallel line you need the slope of the original line and one set of co-ordinates on the parallel line. Then you use the Point-Gradient formula to find the equation of the parallel line.

 

 Perpendicular lines

 A pair of lines are perpendicular (the symbol is \perp) if the product of their gradients is m_1 \times m_2=-1, . So if you need the equation of a line perpendicular to another line, all you need to do is replace the gradient m with the negative reciprocal of m.

 So for example if line 1 is y = 2x +3 and you need to find the line perpendicular to this that goes through the point (0,1), then the gradient m = -1/2 (because 2 x -1/2 = -1).

 

 Distance between two points

 Using the co-ordinates of two points, it is possible to calculate the distance between them using Pythagoras' theorem. The distance d between any two points A(x1,y1) and B(x2,y2) is given by: d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

 

 Mid-point of a line

 When the co-ordinate of two points are known, the mid-point is the point halfway between those two points. For any two points A(x1,y1) and B(x2,y2), the co-ordinates of the mid-point of AB can be found by \left(\frac {{x_1} + {x_2}}{2}, \frac {{y_1} + {y_2}}{2}\right)

 

Intersection of lines

Any two straight lines will meet at a point, as long as they are not parallel. You can find the point of intersection simply by solving the two equations simultaneously. This is also true for curves, although non linear curves may intersect at multiple points or not at all and usually require different methods to solve.

SUMMARY OF THE THINGS YOU NEED TO KNOW

 

 WJEC PAST PAPER QUESTIONS LOTS AND LOTS OF EXAM PRACTICE!!! 

 

Video of WJEC PAST PAPER QUESTIONS

 

                        

 

HERE IS A PAST PAPER QUESTION C1 MAY 2008 QUESTION 1

CONTRIBUTED BY j3w@btinternet.com MANY THANKS FOR YOR CONTRIBUTION :-)

 

 

 

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