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Content  Page ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
y-intercept (x = 0) x-intercept (y = 0) Compare Line PQ and AB. Which line is steeper? The  gradient   is a value that tells us about the steepness of a line X = midpoint If AB = 10 cm and AX = 5 cm, then X is the  midpoint  of AB A B X P Q y x
Distance  between two points A( x 1 , y 1 ) and B( x 2 , y 2 ) :  AB =  The Distance Formula B ( x 2 , y 2 ) A ( x 1 , y 1 )
Example Find the  length of AB  if A = (- 4,3) and B = (6, 8). Distance formula  = AB
The Midpoint Formula Coordinates of the  midpoint (X)  of two points A( x 1 , y 1 ) and B( x 2 , y 2 ) :  X =
Example Find the coordinates of the  midpoint of AB  if  A = (- 1, 8) and B = (2, - 5). Midpoint Formula Midpoint Of AB
Gradient of a Straight Line The  GRADIENT (m)  is a value that tells us about the steepness of a line The  gradient (m)  of a line AB where A = ( x 1 , y 1 ) and B = ( x 2 , y 2 ) :  m =
Gradient is   positive i.e.   m > 0 Gradient is   negative i.e.   m < 0 I am running down a slope, it’s so easy! LEFT RIGHT LEFT RIGHT I am running up a slope, how tiring!
Gradient is   zero. i.e.   m = 0 Gradient is   infinity  for a vertical line.   i.e.   m =  ∞ LEFT RIGHT This is a level road! It’s impossible to run up this! It’s so steep!
Equation of a Straight Line y =  m x +  c First Method gradient y-intercept
We must find gradient (m) and y-intercept (c) (Substitute into formula y =  m x +  c ) (To find c, we can either substitute coordinates of A or B into above) Example Find the  equation of line AB  if A is (2, 5) and B is (-1, 6).  Hence, equation of AB :
y  –  y 1   =  m  ( x  –  x 1 ) Second Method gradient
Example Find the  equation of line AB  if A is (2, 5) and B is (-1, 6).  After finding the gradient, we can use the formula straight away. Equation of AB :
[object Object],[object Object],[object Object],[object Object],TRY THIS QUESTION NOW…
Parallel Lines Gradient of AB = Gradient of CD m 1  = m 2 A B C D
If ABCD is a rectangle/ square rhombus or //gram, then  X is the  midpoint   of  BD and AC   Rectangle/Square/ Rhombus/Parallelogram A D C B X midpoint
If ABCD is a rhombus/square, then the diagonals will be  perpendicular  to each other.  Rhombus/Square A B C D A D C B
Example The line 3x + 7y = 13 is  parallel  to the line kx +  8 = 3y. Find the value of k. 3x + 7y = 13  and  kx + 8 = 3y  share the  same gradient.
Example The equation of a straight line  l   is 5x + 6y + 30 = 0. K is the point (3, -1).  a) Find the  coordinates  of the point where the line  l  crosses the x-axis ;  When  l   crosses the x-axis, y = 0. Substitute into equation.
b) Find the  coordinates  of the point M, at which  the line  l   intersects the line x = 2 ;  Example The equation of a straight line  l   is 5x + 6y + 30 = 0. K is the point (3, -1).  When  l   crosses x = 2, we can find M by substituting x = 2 into equation. x = 2
c) Find the  equation  of the line passing  through K   and  parallel to  l ;  The line shares the  same gradient  as the line  l . Since the line passes through K, we can substitute K(3, -1) to find the equation. Example The equation of a straight line  l   is 5x + 6y + 30 = 0. K is the point (3, -1).
d) Find the  equation  of the line passing  through K   and  parallel to the line 5y – 10 = 0 ;  The line shares the  same gradient  as the line 5y – 10 =  0. This is a vertical line.  Gradient = infinity Since the required line is parallel to this line, it is a  vertical line  too. Since it passes through K, its equation should be  Example The equation of a straight line  l   is 5x + 6y + 30 = 0. K is the point (3, -1).
Q1a) Find the gradient of the straight line 5x  + y = 14.  b) The point (p, 2p) lies on the straight line x + 4y = 36. Calculate  the value of p. TRY THESE QUESTIONS NOW… ,[object Object],[object Object],[object Object],[object Object],[object Object]
Perpendicular  Lines (Gradient of AB)  x  (Gradient of CD) =  -1 (m 1 )(m 2 ) = -1 A B C D
Perpendicular  Distance Perpendicular Distance  = CX Foot  of the perpendicular from Point C  = X X A B C
Example 1 If the line PQ is  perpendicular  to 3y + 1 = x  and PQ  crosses the x-axis at (2, 0),  find its equation. Therefore gradient of PQ = - 3 y = - 3x + 6 Equation of line :  y –  y 1  =  m  (x –  x 1 ) y – 0 =  - 3 (x – 2) P Q Rearrange:  3y + 1 = x
Example 2 Find the equation of the straight line  passing through A(4,5)  and  perpendicular to the line x + 2y – 4 = 0.   Rearrange:  x + 2y - 4 = 0 Therefore gradient of required line = 2 y = 2x - 3 Equation of line :  y –  y 1  =  m  (x –  x 1 ) y – 5 =  2 (x – 4) A(4,5 )
b) These 2 lines  intersect at F . Find the coordinates of F. Sub (1) into (2) Sub into (1) y = 2(2) – 3 = 1 Therefore, F = (2,1). y = 2x – 3… ……..(1) x + 2y – 4 = 0  …..(2) x + 2(2x - 3) – 4 = 0 5x – 10 = 0 x = 2
Perpendicular Bisector If PQ is the  perpendicular bisector  of MN, then MZ = ZN (Same distance) M N P Z Q
Q1) 2 points A and B have coordinates (-1, -2) and (7, 4) respectively. Given that the perpendicular bisector of the line joining A and B meets the y-axis at C, calculate the coordinates of C.  TRY THESE QUESTIONS NOW… Q2) Find the equation of the perpendicular bisector of the line joining A (-7, 2) and B(-1, 10). This perpendicular bisector meets the x-axis at C. Calculate the length of CM, where M is the midpoint of AB.
Intersection All non-// straight lines will intersect at  1  point.  However, a straight line and a curve  may  intersect at  more than 1 point. A A B A
Solving for intersection points To find coordinates of intersection point/s,  ALWAYS  solve the equations  simultaneously .
Can use  ELIMINATION/ SUBSTITUTION   method Can only use  SUBSTITUTION   method B A A
Area  Of  A  Polygon If we are given  ALL   the coordinates of the vertices of a polygon we can find the  area  easily with the use of a formula.
Area  Of  A  Polygon Area of a n-sided polygon = where (x 1 ,y 1 ), (x 2 ,y 2 )… are thecoordinates of its vertices.
Area Example Find the area of the figure shown below.
‘ Shoelace’ Method
Area Is there a need  to apply the formula to find the area of this triangle?
[object Object],[object Object],[object Object],TRY THESE QUESTIONS NOW… Q2) Calculate the distance of the point A(5, 8) from M, the mid-point of the line joining the points B(-1, 10) and C(3, 2).  Show that AM is perpendicular to BC.  Calculate the area of triangle ABC.
[object Object],[object Object],[object Object],AND TRY THESE TOO … Q4) 3 points have coordinates A(1, 2), B(9, 6) and C(3, 8).  Find the equation of the perpendicular bisector of AB and show that it passes through C.  Hence, or otherwise, find the area of triangle ABC.

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Coordinate geometry

  • 1.  
  • 2.
  • 3. y-intercept (x = 0) x-intercept (y = 0) Compare Line PQ and AB. Which line is steeper? The gradient is a value that tells us about the steepness of a line X = midpoint If AB = 10 cm and AX = 5 cm, then X is the midpoint of AB A B X P Q y x
  • 4. Distance between two points A( x 1 , y 1 ) and B( x 2 , y 2 ) : AB = The Distance Formula B ( x 2 , y 2 ) A ( x 1 , y 1 )
  • 5. Example Find the length of AB if A = (- 4,3) and B = (6, 8). Distance formula = AB
  • 6. The Midpoint Formula Coordinates of the midpoint (X) of two points A( x 1 , y 1 ) and B( x 2 , y 2 ) : X =
  • 7. Example Find the coordinates of the midpoint of AB if A = (- 1, 8) and B = (2, - 5). Midpoint Formula Midpoint Of AB
  • 8. Gradient of a Straight Line The GRADIENT (m) is a value that tells us about the steepness of a line The gradient (m) of a line AB where A = ( x 1 , y 1 ) and B = ( x 2 , y 2 ) : m =
  • 9. Gradient is positive i.e. m > 0 Gradient is negative i.e. m < 0 I am running down a slope, it’s so easy! LEFT RIGHT LEFT RIGHT I am running up a slope, how tiring!
  • 10. Gradient is zero. i.e. m = 0 Gradient is infinity for a vertical line. i.e. m = ∞ LEFT RIGHT This is a level road! It’s impossible to run up this! It’s so steep!
  • 11. Equation of a Straight Line y = m x + c First Method gradient y-intercept
  • 12. We must find gradient (m) and y-intercept (c) (Substitute into formula y = m x + c ) (To find c, we can either substitute coordinates of A or B into above) Example Find the equation of line AB if A is (2, 5) and B is (-1, 6). Hence, equation of AB :
  • 13. y – y 1 = m ( x – x 1 ) Second Method gradient
  • 14. Example Find the equation of line AB if A is (2, 5) and B is (-1, 6). After finding the gradient, we can use the formula straight away. Equation of AB :
  • 15.
  • 16. Parallel Lines Gradient of AB = Gradient of CD m 1 = m 2 A B C D
  • 17. If ABCD is a rectangle/ square rhombus or //gram, then X is the midpoint of BD and AC Rectangle/Square/ Rhombus/Parallelogram A D C B X midpoint
  • 18. If ABCD is a rhombus/square, then the diagonals will be perpendicular to each other. Rhombus/Square A B C D A D C B
  • 19. Example The line 3x + 7y = 13 is parallel to the line kx + 8 = 3y. Find the value of k. 3x + 7y = 13 and kx + 8 = 3y share the same gradient.
  • 20. Example The equation of a straight line l is 5x + 6y + 30 = 0. K is the point (3, -1). a) Find the coordinates of the point where the line l crosses the x-axis ; When l crosses the x-axis, y = 0. Substitute into equation.
  • 21. b) Find the coordinates of the point M, at which the line l intersects the line x = 2 ; Example The equation of a straight line l is 5x + 6y + 30 = 0. K is the point (3, -1). When l crosses x = 2, we can find M by substituting x = 2 into equation. x = 2
  • 22. c) Find the equation of the line passing through K and parallel to l ; The line shares the same gradient as the line l . Since the line passes through K, we can substitute K(3, -1) to find the equation. Example The equation of a straight line l is 5x + 6y + 30 = 0. K is the point (3, -1).
  • 23. d) Find the equation of the line passing through K and parallel to the line 5y – 10 = 0 ; The line shares the same gradient as the line 5y – 10 = 0. This is a vertical line. Gradient = infinity Since the required line is parallel to this line, it is a vertical line too. Since it passes through K, its equation should be Example The equation of a straight line l is 5x + 6y + 30 = 0. K is the point (3, -1).
  • 24.
  • 25. Perpendicular Lines (Gradient of AB) x (Gradient of CD) = -1 (m 1 )(m 2 ) = -1 A B C D
  • 26. Perpendicular Distance Perpendicular Distance = CX Foot of the perpendicular from Point C = X X A B C
  • 27. Example 1 If the line PQ is perpendicular to 3y + 1 = x and PQ crosses the x-axis at (2, 0), find its equation. Therefore gradient of PQ = - 3 y = - 3x + 6 Equation of line : y – y 1 = m (x – x 1 ) y – 0 = - 3 (x – 2) P Q Rearrange: 3y + 1 = x
  • 28. Example 2 Find the equation of the straight line passing through A(4,5) and perpendicular to the line x + 2y – 4 = 0. Rearrange: x + 2y - 4 = 0 Therefore gradient of required line = 2 y = 2x - 3 Equation of line : y – y 1 = m (x – x 1 ) y – 5 = 2 (x – 4) A(4,5 )
  • 29. b) These 2 lines intersect at F . Find the coordinates of F. Sub (1) into (2) Sub into (1) y = 2(2) – 3 = 1 Therefore, F = (2,1). y = 2x – 3… ……..(1) x + 2y – 4 = 0 …..(2) x + 2(2x - 3) – 4 = 0 5x – 10 = 0 x = 2
  • 30. Perpendicular Bisector If PQ is the perpendicular bisector of MN, then MZ = ZN (Same distance) M N P Z Q
  • 31. Q1) 2 points A and B have coordinates (-1, -2) and (7, 4) respectively. Given that the perpendicular bisector of the line joining A and B meets the y-axis at C, calculate the coordinates of C. TRY THESE QUESTIONS NOW… Q2) Find the equation of the perpendicular bisector of the line joining A (-7, 2) and B(-1, 10). This perpendicular bisector meets the x-axis at C. Calculate the length of CM, where M is the midpoint of AB.
  • 32. Intersection All non-// straight lines will intersect at 1 point. However, a straight line and a curve may intersect at more than 1 point. A A B A
  • 33. Solving for intersection points To find coordinates of intersection point/s, ALWAYS solve the equations simultaneously .
  • 34. Can use ELIMINATION/ SUBSTITUTION method Can only use SUBSTITUTION method B A A
  • 35. Area Of A Polygon If we are given ALL the coordinates of the vertices of a polygon we can find the area easily with the use of a formula.
  • 36. Area Of A Polygon Area of a n-sided polygon = where (x 1 ,y 1 ), (x 2 ,y 2 )… are thecoordinates of its vertices.
  • 37. Area Example Find the area of the figure shown below.
  • 39. Area Is there a need to apply the formula to find the area of this triangle?
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