5.1 Coordinates 1. A (1, 2) B ( 4, 0) C ( 2, 3) D (3, 2) E (1, 4) F (0, 2) G ( 2, 3) H (0, 5)

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1 MEP: Demonstration Project Teacher Support YA, P Practice Book UNIT Linear Graphs and Equations. Coordinates. A (, ) B (, ) C (, ) D (, ) E (, ) F (, ) G (, ) H (, ). C A B G D F E H The points A, B and E all lie on a straight line through the origin.. The shape is an isosceles triangle.. (, ) (, ) Remaining corner (, ). (, ) (, ) The Gatsb Charitable Foundation

2 MEP: Demonstration Project Teacher Support YA, P.. (, ) (, ) Remaining corner (, ) (, ) (, ). (a) units (b) (, ), (, ) and (, ), (, ). (a) (, ) (b) (, ) because both coordinates are equal to double the tile number (c) Daniel is wrong because is an odd number and all the corners with a have even numbers. (d) Tile Number Coordinates of the Corner with a (e) Tile number has a cross in the corner at (, ). (f) Tile number has a cross in the corner at (, ).. (a) and (b) (, ) (, ) (, ) (, ) (a) (c) st Step S W W nd Step W S W rd Step W W S The Gatsb Charitable Foundation

3 MEP: Demonstration Project Teacher Support YA, P.. Straight Line Graphs. (a) (b) =. (a) (b) = + = (c) = (d) = + The Gatsb Charitable Foundation

4 MEP: Demonstration Project Teacher Support YA, P. (e) = (f) =. (a) (b) (c) (d) (e) (f) (g). (a) = + (b) = (c) = + (d) =. Equation Gradient Intercept = + = = + = = + = + = =. (a) C (, ) B (, ) A (, ) (b) Gradient A B = A C = B C = The Gatsb Charitable Foundation

5 MEP: Demonstration Project Teacher Support YA, P.. (a) = (b) = (c) = + (d) = (e) = (f) =. (a) (b) =. (a) = + = (b) The lines cross at (, ).. (a) C (, ) B (, ) A (, ) (b) A B has equation = B C has equation = A C has equation = The Gatsb Charitable Foundation

6 MEP: Demonstration Project Teacher Support YA, P.. (a) = (b) = + (c) =. (a) squares with pins in each square squares with pins in each square (b) The gradient tells ou how steep the line p = s + is. (c) An three points on the line p = s +, e.g. (, ), (, ) and (,) (d) p = s +. Linear Equations. (a) = (b) = (c) = (d) = (e) = (f) = (g) = (h) = (i) = (j) = (k) = (l) =. (a) = (b) = (c) = (d) = (e) = (f) = (g) = (h) = (i) = (j) = (k) = (l) =. (a) = (b) = (c) = (d) = (e) = (f) = (g) = (h) =. (a) = (b) = (c) =. = = The solution is =. The Gatsb Charitable Foundation

7 MEP: Demonstration Project Teacher Support YA, P.. = The solution is =. =. (a) = + (b) The solution is =. = =. (a) = (see graph opposite) = + The Gatsb Charitable Foundation

8 MEP: Demonstration Project Teacher Support YA, P.. (b) = (see graph opposite) = =. (a) = (b) = (c) =. = = (a) = (b) = (c) = = +. (a) = (b) = The Gatsb Charitable Foundation

9 MEP: Demonstration Project Teacher Support YA, P.. Parallel and Perpendicular Lines. (a) =+ =+ = = = (b) An two lines of the form = + c, with c not equal to, or. In this case, the diagram shows the lines = and =.. (a) and (b) = (c) The second line has equation = +. The Gatsb Charitable Foundation

10 MEP: Demonstration Project Teacher Support YA, P.. (a) E (b) D (c) There are no lines parallel to B because B has gradient whilst A and E have gradient, and C and D have gradient.. (a) (b) = (c) =. (a) = + (b) = (c) =. (a) (b) = + (c) = +. (a) (b) (c) The lines are perpendicular because = and the product of the gradient is for perpendicular lines.. (a) D (b) E (c) C. (a) Gradient of A = B = C = (b) Lines A and B are perpendicular.. (a) An two lines of the form = + c, with c, e.g. = + and = + (b) An two lines of the form = + c, e.g. = + and =. (a) and (c) = = (b) An equation of the form = m where m or. (d) = The Gatsb Charitable Foundation

11 MEP: Demonstration Project Teacher Support YA, P.. (a) That the are parallel, with gradient =. (b) The constant term gives the intecept, i.e the value where the line crosses the -ais. (c) (, ) (d) An line of the form = + c, where c,, or.. Simultaneous Equations. (a) = + = (b) Intersection (, ) (c) =, =. (a) = = (b) Intersection (, ) (c) =, =. = = ( ) The solution is =, =. The Gatsb Charitable Foundation

12 MEP: Demonstration Project Teacher Support YA, P.. = ( ) = ( ) The solution is =, =.. (a) + = = (b) = = The solutions to the simultaneous equations are =, =, so the numbers are and.. + = and + =. (a) Because it eliminates the unknown, leaving an equation in onl. (b) = =, =. (a) =, = (b) =, = (c) =, = (d) =, = (e) =, = (f) =, =. (a) Because it eliminates, leaving an equation in onl. (b) =, =. (a) =, = (b) =, = (c) =, = (d) =, = (e) =, = (f) =, = The Gatsb Charitable Foundation

13 MEP: Demonstration Project Teacher Support YA, P.. (a) Line A has gradient (for eample) = +, i.e. + =. (b) = or + = = and it has intercept, so the equation is (c) = + C (d) An suitable method leading to = + =, =.. (a) = (b)... the line through A and B. (c) C B = D A A B E H F G (d) One of: = = = (e) An suitable method to obtain =, =. (f)... G and H at (, ).. Equations in Contet. (a) n (b) n =. (a) c =. m + (b) c =. m + (c). =. m + m = miles. + ( + ) =, i.e. + =, giving =. + m = Distance travelled was. miles.. (a). (b). per gallon (c) pence per litre The Gatsb Charitable Foundation

14 MEP: Demonstration Project Teacher Support YA, P.. (a) Area = m (b) =, giving =. ( ) ( n ) (c) Perimeter = + m (d) + =, giving =. m. (a) Cost = + (b) n + =., giving n =. (a) Number of Fr = ( ) hours, i.e. the repair took hour minutes (b) ( )=, giving =, i.e. ou need to get Fr.. (a) Perimeter = (b) = m (c) Area = (d) =. m. (a) Perimeter = ( + ) m (b) =. m. (a) km per hour (b) Because the third section (Brussels to Madrid) has a steeper gradient than the first section (London to Brussels). (c) Madrid Distance from London (km) (d) Brussels London Time (hours) GMT The Gatsb Charitable Foundation