1. Straight Lines and Gradients Objectives: To find linear equations from minimum information. To use linear equations in any form to find the gradient and y-axis intercept.

5. Straight Lines and Gradients Objectives: To find linear equations from minimum information. To use linear equations in any form to find the gradient and y-axis intercept.

6. c is the point where the line meets the y -axis, the y -intercept and y -intercept, c = e.g. has gradient m = The equation of a straight line is m is the gradient of the line gradient = 2 x intercept on y -axis

7. 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 showing that the point ( 4,7 ) lies on the line. e.g. Substituting x = 4 in gives

8. 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.  Finding the equation of a straight line when we know e.g.Find the equation of the line with gradient passing through the point its gradient, m and the coordinates of a point on the line. Solution: So, Using, m is given, so we can find c by substituting for y, m and x. (-1, 3) x

9.  The gradient of the straight line joining the points and is e.g. Find the gradient of the straight line joining the points and To use this formula, we don’t need a diagram! Solution:

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

11. SUMMARY  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

12. Activities Matching line graphs and equations Shooting coordinates

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

14. e.g. 1Find 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

15. 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

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

17. Exercise Solution: So, Solution: So, 1.Find the equation of the line parallel to the line which passes through the point. Parallel line is 2.Find the equation of the line through the point (1, 2), perpendicular to the line So,

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

19. Solution: First find the gradient We could use the 2 nd point, (-1, 3) instead of (2, -3) 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 Now use with

20. Straight Lines and Gradients  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

21. Straight Lines and Gradients Solution: First find the gradient: e.g. Find the equation of the line through the points Now on the line: Equation of line is

22. Straight Lines and Gradients 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