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**“Teach A Level Maths” Vol. 1: AS Core Modules**

1: Straight Lines and Gradients © Christine Crisp

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**The gradient of the straight line joining the points**

and is

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**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

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**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.

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**Formula for a Straight Line**

Let ( x, y ) be any point on the line x x Let be a fixed point on the line

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**Finding the equation of a straight line when we know**

its gradient, m and the coordinates of a point on the line (x1,y1). Using , m is given, so we can find c by substituting for y1, m and x1. e.g. Find the equation of the line with gradient passing through the point Solution: (-1, 3) x So,

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**Using the formula when we are given 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)

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SUMMARY Equation of a straight line where m is the gradient and c is the intercept on the y-axis Gradient of a straight line where and are points on the line

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

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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

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**Finding the equation of a straight line when we know**

its gradient, m and the coordinates of a point on the line (x1,y1). 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 Add 2 to both sides C = 5 So,

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**Using the formula when we are given 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) Add 4 to both sides -3 = -4 + c

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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.

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**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

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**Solution: First find the gradient:**

e.g. Find the equation of the line through the points Now on the line: Equation of line is

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**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

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43: Quadratic Trig Equations and Use of Identities © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules.

43: Quadratic Trig Equations and Use of Identities © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules.

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