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Summary of Chapter 2 (so far). Parallel lines  y = mx + c y = mx + d Perpendicular lines  y = mx + cy = nx + dm x n = -1 Length of a line using Pythagoras’

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Presentation on theme: "Summary of Chapter 2 (so far). Parallel lines  y = mx + c y = mx + d Perpendicular lines  y = mx + cy = nx + dm x n = -1 Length of a line using Pythagoras’"— Presentation transcript:

1 Summary of Chapter 2 (so far)

2 Parallel lines  y = mx + c y = mx + d Perpendicular lines  y = mx + cy = nx + dm x n = -1 Length of a line using Pythagoras’ theorem Finding equation of a line when given coordinates and / or another line. Intersection of 2 lines by solving equations simultaneously Proofs about geometric properties Circles … intersection lines & curves or 2 curves

3 Centre (4,3) and radius of r – Equation of circle Choose any point on circle (x, y) Using Pythagoras (x – 4) 2 + (y – 3) 2 = r 2 From the diagram we can see that x = 6 and y = 6.4 (ish) (6 – 4) 2 + (6.4 – 3) 2 = r 2 r = 3.945

4 Centre (-2, 4) and radius of r. Passes through the point (1,8) Find the equation of the circle.

5 Question 5 … HINT! Show that x 2 + y 2 + 2x – 4y + 1 =0 Can be written as (x + 1) 2 + (y – 2) 2 = r 2 Exercise 2E – pg 67

6 Using geometric info (so far) If a circle has an equation of the form (x - h) 2 + (y - k) 2 = r 2 How would you find the equation of the tangent that meets the circle at (x, y)


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