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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?
• 40.
• 41.
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