Starter: state the gradient and y intercept of each line Coordinate Geometry KUS objectives BAT find distances between points BAT explore equations of lines, know the general equation of a line BAT use a new formula to find equations of lines Starter: state the gradient and y intercept of each line      

A (3, 7) and B (9, 19) 12 6 Find the distance between these points: Notes 1 Find the distance between these points: A (3, 7) and B (9, 19) 12 6

A (-1, 7) and B (19, 22) 15 20 Find the gradient between these points: Notes 2 Find the gradient between these points: A (-1, 7) and B (19, 22) 15 20

A (x1, y1) and B (x2, y2) 𝒅 y1 – y2 x1 – x2 𝑑= 𝑥− 𝑥 1 2 +( 𝑦 2 − 𝑦 1 2 Notes 3 Formula Find the distance between these points: A (x1, y1) and B (x2, y2) 𝒅 y1 – y2 x1 – x2 𝑑= 𝑥− 𝑥 1 2 +( 𝑦 2 − 𝑦 1 2 Generally: Can you establish this result in a proof with clear steps?

Find the distance between these points Practice 1 Find the distance between these points Find the gradient between these points: A (4, 9) and B (7, 14) A (-2, -5) and B (-7, 7) A (8, -1) and B (7, -11) A (14, 3) and B (22, 13) A (3, 18) and B (9, 2) A (-7, -6) and B (14, -6)

Objectives BAT find distances between points BAT explore equations of lines, know the general equation of a line BAT use a new formula to find equations of lines

Find the gradient between these points Notes 4 What is gradient? (x2, y2) Change in y Change in x y2- y1 y2-y1 x2-x1 x2- x1 (x1, y1) Find the gradient between these points (4, 9) and (7, 12)

We can generalise y - y1 y- y1 x - x1 Rearrange to: x- x1 notes We can generalise (x, y) y - y1 x - x1 y- y1 Rearrange to: x- x1 (x1, y1) How can we use this?

Find the line that joins these points WB5 Find the line that joins these points (-2, 8) and (3,-7) Use this point (3, -7) x1 is 3 and y1 is -7 y – (-7) = -3 (x – 3) y + 7 = -3x + 9 y = -3 x + 2

Find the equation of the line that: Joins (3, 7) and (9, 19) Practice 2 Find the equation of the line that: Joins (3, 7) and (9, 19) Joins (4, 2) and (7, -4) Goes through (3, 7) and has m = 3 Goes through (2, -3) and has m = -4 Joins (-1, -1) and (4, 14) Joins (8, 1) and (18, 6) Goes through (7, 3) and parallel to 3y = x+7 Goes through (5, -5) and parallel to x+ 2y =11

y x (-6, 1) (2, 5) (-3, -5) Find the equations of the lines that WB6 Find the equations of the lines that join these points (-6, 1) (2, 5) (-3, -5) activity: Geogebra three points

y x Choose three points of your own Work out the general Challenge Choose three points of your own Work out the general equations of each line

Objectives 1 find distances between points 2 explore equations of lines, know the general equation of a line 3 use a new formula to find equations of lines

The general equation of a line: ax + by + c = 0 Notes The general equation of a line: ax + by + c = 0 y = 4x + 6  4x – y + 6 = 0 y = ½ x + 2  x – 2y + 4 = 0 y = - x - 3  x + y + 3 = 0 y = -5 x + 1  5x + y – 1 = 0

y – 9 = ¾ (x – 4) y – 9 = ¾ x – 3 y = ¾ x + 6 or 3x - 4y + 24 = 0 WB7 Find the line that joins points (4, 9) and (8, 12) in the form ax + by + c = 0 Use this point (4, 9) x1 is 4 and y1 is 9 y – 9 = ¾ (x – 4) y – 9 = ¾ x – 3 y = ¾ x + 6 or 3x - 4y + 24 = 0

y – (-7) = -3 (x – 3) y + 7 = -3x + 9 y = -3 x + 2 or 3x + y - 2 = 0 WB8 Find the line that joins points (-2, 8) and (3,-7) in the form ax + by + c = 0 Use this point (3, -7) x1 is 3 and y1 is -7 y – (-7) = -3 (x – 3) y + 7 = -3x + 9 y = -3 x + 2 or 3x + y - 2 = 0

y – 7 = -½ (x – 3) y - 7 = -½ x + 3/2 y = -½x + 17/2 or WB9 Find the general equation of the line through (3, 7) that is perpendicular to y = 2x + 8 Use gradient Use the point (3, 7) x1 is 3 and y1 is 7 y – 7 = -½ (x – 3) y - 7 = -½ x + 3/2 y = -½x + 17/2 or x + 2y - 17 = 0

y – 7 = -½ (x – 3) y - 7 = -½ x + 3/2 y = -½x + 17/2 or x + 2y = 17 WB10 Find the equation of the line that goes through (3, 7) and is perpendicular to y = 2x + 8 Giving your answer in the form 𝑎𝑥+𝑏𝑦=𝑐 Use gradient Use the point (3, 7) x1 is 3 and y1 is 7 y – 7 = -½ (x – 3) y - 7 = -½ x + 3/2 y = -½x + 17/2 or x + 2y = 17

y – 3 = -3 (x – (-2)) l1 is y = -3 x - 3 y – 3 = 1/3(x – (-2)) WB11 The line l1 has gradient -3 goes through (-2, 3) Line l2 is perpendicular to l1 and goes through (-2, 3) Find the equations of lines l1 and L2 y – 3 = -3 (x – (-2)) l1 is y = -3 x - 3 y – 3 = 1/3(x – (-2)) l2 is 3y = x + 11

Find the equation of the line in the form ax + by +c = 0 Practice Find the equation of the line in the form ax + by +c = 0 Perpendicular to y = 3x + 2 And goes through (6, 12) Perpendicular to y = -1/2x + 1 And goes through (-3, 8) Find the equation of the line perpendicular to: The midpoint of the line that joins (4, 9) and (10, 21) The midpoint of the line that joins (-3,-6) and (9, 2)

𝑦 = −2𝑥+ 4 3 𝑦−5= −2 𝑥−5 𝑦 =−2𝑥+15 2𝑥+𝑦 = 15 Gradient is -2 WB12   Giving your answer in the form 𝑎𝑥+𝑏𝑦=𝑐 𝑦 = −2𝑥+ 4 3 Gradient is -2 𝑦−5= −2 𝑥−5 𝑦 =−2𝑥+15 2𝑥+𝑦 = 15

WB13   𝑦−11=3 𝑥−3 y=3𝑥+2

Summary You should be able to: Rearrange equations of lines Use to find the equation of a line from two points or gradient plus a point Use to find the equation of a perpendicular line given enough information

One thing to improve is – KUS objectives BAT solve linear geometry problems self-assess One thing learned is – One thing to improve is –

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