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1: Straight Lines and Gradients © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules.

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Presentation on theme: "1: Straight Lines and Gradients © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules."— Presentation transcript:

1 1: Straight Lines and Gradients © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules

2 Equation of a Straight Line  The gradient of the straight line joining the points and is

3 Equation of a Straight Line 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

4 Equation of a Straight Line 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

5 Equation of a Straight Line Formula for a Straight Line Let ( x, y ) be any point on the line Let be a fixed point on the line x x

6 Equation of a Straight Line  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 (x 1, y 1 ). Solution: So, Using, m is given, so we can find c by substituting for y 1, m and x 1. (-1, 3) x

7 Equation of a Straight Line Solution: First find the gradient We could use the 2 nd point, (-1, 3) instead of (2, -3) Using the formula when we are given two points on the line e.g. Find the equation of the line through the points Now use with

8 Equation of a Straight Line 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

9 Equation of a Straight Line 2. Find the equation of the line through the points Exercise 1. Find the equation of the line with gradient 2 which passes through the point. Solution: So, Solution: So,

10 Equation of a Straight Line We sometimes rearrange the equation of a straight line so that zero is on the right-hand side ( r.h.s. ) We must take care with the equation in this form. e.g. can be written as e.g. Find the gradient of the line with equation Solution: Rearranging to the form : so the gradient is

11 Equation of a Straight Line  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 (x 1, y 1 ). Solution: So, Using, m is given, so we can find c by substituting for y, m and x. (-1, 3) x Add 2 to both sides C = 5

12 Equation of a Straight Line Solution: First find the gradient We could use the 2 nd point, (-1, 3) instead of (2, -3) Using the formula when we are given two points on the line e.g. Find the equation of the line through the points Now use with -3 = -4 + c Add 4 to both sides

13 Equation of a Straight Line

14 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.

15 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

16 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

17 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


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