# Parallel and Perpendicular Lines. Gradient-Intercept Form Useful for graphing since m is the gradient and b is the y- intercept Point-Gradient Form Use.

## Presentation on theme: "Parallel and Perpendicular Lines. Gradient-Intercept Form Useful for graphing since m is the gradient and b is the y- intercept Point-Gradient Form Use."— Presentation transcript:

Parallel and Perpendicular Lines

Gradient-Intercept Form Useful for graphing since m is the gradient and b is the y- intercept Point-Gradient Form Use this form when you know a point on the line and the gradient Also can use this version if you have two points on the line because you can first find the gradient using the gradient formula and then use one of the points and the gradient in this equation. General Form Commonly used to write linear equation problems or express answers

The gradient is a number that tells "how steep" the line is and in which direction. So as you can see, parallel lines have the same gradients so if you need the gradient of a line parallel to a given line, simply find the gradient of the given line and the gradient you want for a parallel line will be the same. Perpendicular lines have negative reciprocal gradients so if you need the gradient of a line perpendicular to a given line, simply find the gradient of the given line, take its reciprocal (flip it over) and make it negative.

Let's look at a line and a point not on the line (2, 4) Let's find the equation of a line parallel to y = - x that passes through the point (2, 4) y = - x What is the gradient of the first line, y = - x ? This is in gradient intercept form so y = mx + b which means the gradient is –1. 1 4 Distribute and then solve for y to leave in gradient-intercept form. 2 So we know the gradient is –1 and it passes through (2, 4). Having the point and the gradient, we can use the point-gradient formula to find the equation of the line

What if we wanted perpendicular instead of parallel? (2, 4) Let's find the equation of a line perpendicular to y = - x that passes through the point (2, 4) y = - x The gradient of the first line is still –1. 1 2 4 Distribute and then solve for y to leave in gradient-intercept form. The gradient of a line perpendicular is the negative reciprocal so take –1 and "flip" it over and make it negative. So the gradient of a perpendicular line is 1 and it passes through (2, 4).

Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.www.mathxtc.com Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au