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Published byViolet Stevens Modified over 9 years ago
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The Distance Formula (and mid point)
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What is to be learned? How to calculate the distance between two points
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x y A (3, 1) B (7, 1) AB = 4 units
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x y A (3, 1) C (7, 4) ? 4 units
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x y A (3, 1) C (7, 4) ? 4 units (7 – 3) (x 1, y 1 ) (x 2, y 2 ) (x 2 – x 1 ) 3 units (4 – 1) (y 2 – y 1 ) Pythagoras AC 2 = 4 2 + 3 2 = (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2
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The Distance Formula AB 2 = (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 For Points A(x 1, y 1 ) and B(x 2, y 2 ) orAB =√ (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 Distance between A (2, 8) and B (7, 5) (x 1, y 1 ) (x 2, y 2 ) AB =√ (7 – 2) 2 + (5 – 8) 2 AB =√ 5 2 + (-3) 2 = √34
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The Distance Formula AB 2 = (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 For Points A(x 1, y 1 ) and B(x 2, y 2 ) orAB =√ (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2
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Ex Distance between C (2, -5) and D (4, 5) (x 1, y 1 ) (x 2, y 2 ) CD =√ (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 CD =√ (4 – 2) 2 + (5 – (-5)) 2 Careful! CD =√ 2 2 + 10 2 = √104
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What is to be learned? How to calculate the distance between two points How to use the distance formula for problem solving.
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Distance Formula Applications Circle centre A (2, 3) has radius 5 units Is B (4, 7) inside outside or on circle? A (2, 3) B (4, 7) √20 5 Inside
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Key Question A(1, 2) B(3, 0) C (-1, -2) are vertices of triangle ABC. Prove that it is isosceles. AC = √20 and BC = √20 Therefore AB = BC, so triangle is isosceles.
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Mid Points x y A (3, 1) B (9, 5) (, ) Midway between 3 and 9 6 Midway between 1 and 5 3 3 + 9 2 1 + 5 2
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Mid Point of a Line For Points A(x 1, y 1 ) and B(x 2, y 2 ) Mid Point = x 1 + x 2, y 1 + y 2 2 ( ) or use common sense!
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Ex Mid Point EF E(2, -5) and F (-6, 8) (x 1, y 1 ) (x 2, y 2 ) Mid Point = 2 + (-6), -5 + 8 ( ) 2 = -4, 3 ( ) 2 = (-2, 1½ )
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Is this right angled? 7 9 12 7 2 + 9 2 = 12 2 49 + 81 = 144 130 = 144 no
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Is this right angled? 12 5 13 If right angled triangle (if rt. L ) 13 2 = 12 2 + 5 2 LHS13 2 = 169 RHS12 2 + 5 2 = 169 LHS = RHS → rt. L
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