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Coordinate Geometry GCSE

10/06/2022

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COORDINATE GEOMETRY
1- Coordinates
Two points (x₁, y₁) and (x₂, 1₂) es (-3.5) (2,8)
1₂ - Y₁ eig.
x₂-x,
1 GRADIENT = AY
Ax
Midpoint is the an

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COORDINATE GEOMETRY
1- Coordinates
Two points (x₁, y₁) and (x₂, 1₂) es (-3.5) (2,8)
1₂ - Y₁ eig.
x₂-x,
1 GRADIENT = AY
Ax
Midpoint is the an

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Join milions of students

Improve your grades

By signing up you accept Terms of Service and Privacy Policy

COORDINATE GEOMETRY
1- Coordinates
Two points (x₁, y₁) and (x₂, 1₂) es (-3.5) (2,8)
1₂ - Y₁ eig.
x₂-x,
1 GRADIENT = AY
Ax
Midpoint is the an

Register

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Access to all documents

Join milions of students

Improve your grades

By signing up you accept Terms of Service and Privacy Policy

COORDINATE GEOMETRY 1- Coordinates Two points (x₁, y₁) and (x₂, 1₂) es (-3.5) (2,8) 1₂ - Y₁ eig. x₂-x, 1 GRADIENT = AY Ax Midpoint is the anwage point (+24317) Distance -342 2 is found using Pyriagoras Theorem 2 √ (x₂-x,₁)² + (Yz - Y₁)" T= mx + c m=4 Y = 4x +C 8-53 2-(-3) If 3 points, A, B and C ke on the same straight line, they have the same gradient. 548 2- Straight Lines - Finding Equations The general equation of a strought line is: y = mx + c L M is the gradient LC is PHC y-intercept To find the equation you need two things Ime gradient, in L any point on the whe 5- 4(-2) + C 5--8-C C (3 ·2·3(-2.5) -> worked example: The live x has an equation Y = 4x + 5. Find the equation of line ĭ parallel tox which passes through the point. (-2,5). $ Y 4x + 13 ✓ 3- Strought Lines Drawing Graphs Y = mx + c Por con me i-axis ·go 1 avoSS m up (repeat) From W • From ax + b1 =C • Put XO to Find Y - intercept • Put Y- O to find x-intercept Worked comple On me axes below ovaw the lines Y 3x - 1 and 5x - SY IS H 6- 5- 21 11 TO Sim 34m N 5 y = 3x - 1 3x +31-15 When 19 917 15. . ۲۰ ۸ است 3x+15 X-5 - 4. Perpendiculas Lines What are perpendicular lines? Perpendicuar tires do meet calm other and what they do the two lines form...

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Alternative transcript:

a right augle-ic. they meet out 90° Gradients m, • Use m₁ = 2 m₂ = -1 2 m₂ = -1/1/201 and m₂ m₂ que popendiculas if: m, xm, -1 Worked Example The live I was equation 1-2x 2 Find an equation of the line perpendicular to which passes twough the point (2,-3). to find a perpendicular gradient Y'. — xrc - 3-LX2 + C 2 C-3-(-1). =-2 2 > Y = -1x -2 ✓