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Prior Knowledge Check 1) Find the point of intersection for each pair of lines: a) y = 4x + 7 and 5y = 2x 1 b) y = 5x 1 and 3x + 7y = 11 c) 2x 5y = 1 and 5x 7y = 14 2) Simplify each of the following: a) 80 b) 200 c) 125 ( 2, 1) 4 5 10 2 5 5 9 19, 26 19 (7,3) 3) Make y the subject of each equation: a) 6x + 3y 15 = 0 y = 5 2x b) 2x 5y 9 = 0 y = 2 5 x 9 5 c) 3x 7y + 12 = 0 y = 3 7 x + 12 7

You can find the gradient of a straight line joining two points by considering the vertical and horizontal distance between them. It can also be found from the equation of the graph. y (x 2,y 2 ) You can work out the gradient of a line if you know 2 points on it. Let the first point be (x 1,y 1 ) and the second be (x 2,y 2 ). The following formula gives the gradient: y y m x x 2 1 2 1 The change in the y values, divided by the change in the x values (x 1,y 1 ) x 2 - x 1 y 2 - y 1 x 5A/B

You can find the gradient of a straight line joining two points by considering the vertical and horizontal distance between them. It can also be found from the equation of the graph. You can work out the gradient of a line if you know 2 points on it. Let the first point be (x 1,y 1 ) and the second be (x 2,y 2 ). The following formula gives the gradient: y y m x x 2 1 2 1 The change in the y values, divided by the change in the x values Calculate the gradient of the line which passes through (-2,7) and (4,5) (x 1, y 1 ) (x 2, y 2 ) y y m x x m = (-2, 7) = (4, 5) m 2 1 2 1 5 7 4 ( 2) 2 6 1 m 3 Substitute numbers in Work out or leave as a fraction Simplify if possible 5A/B

You can find the gradient of a straight line joining two points by considering the vertical and horizontal distance between them. It can also be found from the equation of the graph. You can work out the gradient of a line if you know 2 points on it. Let the first point be (x 1,y 1 ) and the second be (x 2,y 2 ). The following formula gives the gradient: y y m x x 2 1 2 1 The change in the y values, divided by the change in the x values The line joining (2, -5) to (4, a) has a gradient of -1. Calculate the value of a. (x 1, y 1 ) (x 2, y 2 ) = (2, -5) = (4, a) y y m x x 2 1 2 1 a ( 5) 1 4 2 a 5 1 2 2 a 5 7 a Substitute numbers in Simplify Multiply by 2 Subtract 5 5A/B

You can find the gradient of a straight line joining two points by considering the vertical and horizontal distance between them. It can also be found from the equation of the graph. y-intercept y gradient The equation of a straight line is usually written in one of 2 forms. One you will have seen before; y mx c Where m is the gradient and c is the y-intercept. 1 x Or, the general form: ax by c Where a, b and c are integers. 0 5A/B

You can find the gradient of a straight line joining two points by considering the vertical and horizontal distance between them. It can also be found from the equation of the graph. The equation of a straight line is usually written in one of 2 forms. One you will have seen before; Write down the gradient and y- intercept of the following graphs a) y 3x 2 Gradient = -3 y-intercept = (0,2) y mx c Where m is the gradient and c is the y-intercept. Or, the general form: ax by c Where a, b and c are integers. 0 b) 4x 2y 5 0 4x 5 2y 5 2x y 2 Gradient = 2 y-intercept = (0, 5 / 2 ) Rearrange to get y on one side Divide by 2 5A/B

You can find the gradient of a straight line joining two points by considering the vertical and horizontal distance between them. It can also be found from the equation of the graph. The equation of a straight line is usually written in one of 2 forms. One you will have seen before; y mx c Where m is the gradient and c is the y-intercept. Or, the general form: ax by c Where a, b and c are integers. 0 b) Write each equation in the form ax + by + c = 0 a) y 4x 3 0 4x 3 y 4x y 3 0 1 y x 5 2 1 y x 5 0 2 2y x 10 0 x 2y 10 0 Subtract y Correct form + 1 / 2 x and -5 x2 (to remove fraction) Correct form 5A/B

You can find the gradient of a straight line joining two points by considering the vertical and horizontal distance between them. It can also be found from the equation of the graph. The equation of a straight line is usually written in one of 2 forms. One you will have seen before; y mx c Where m is the gradient and c is the y-intercept. Or, the general form: ax by c 0 The line y = 4x + 8 crosses the x-axis at P. Work out the coordinates of P. y 4x 8 0 4x 8 8 4x 2 x Crosses the x- axis where y=0-8 Divide by 4 So the line crosses the x-axis at (-2,0) Where a, b and c are integers. 5A/B

The equation of a straight line can be found if you know two points on the line, or you know its gradient and a single point. Starting with the definition of the gradient y (x,y) (x 1,y 1 ) x - x 1 y - y 1 x m = y y 1 x x 1 m x x 1 = y y 1 y y 1 = m x x 1 Multiply by the denominator Usually written the other way round y y 1 = m x x 1 5C/D

The equation of a straight line can be found if you know two points on the line, or you know its gradient and a single point. m = y 2 y 1 x 2 x 1 y y 1 = m x x 1 Find the equation of the line with gradient 5 that passes through the point (3,2) (x 1, y 1 ) = (3, 2) m = 5 y y m( x x ) 1 1 y 2 5( x 3) y 2 5x 15 y 5x 13 Substitute the numbers in Expand the bracket Add 2 5C/D

The equation of a straight line can be found if you know two points on the line, or you know its gradient and a single point. Find the equation of the line which passes through (5,7) and (3,-1) Start by finding the gradient m = y 2 y 1 x 2 x 1 m = y 2 y 1 x 2 x 1 y y 1 = m x x 1 m = 1 7 3 5 m = 8 2 m = 4 Sub in values carefully Calculate Calculate again 5C/D

The equation of a straight line can be found if you know two points on the line, or you know its gradient and a single point. m = y 2 y 1 x 2 x 1 y y 1 = m x x 1 Find the equation of the line which passes through (5,7) and (3,-1) Now use the gradient with either coordinate, in the second formula y y 1 = m x x 1 y 7 = 4 x 5 y 7 = 4x 20 y = 4x 13 m = 4 Sub in values Expand bracket Rearrange 5C/D

The equation of a straight line can be found if you know two points on the line, or you know its gradient and a single point. y y 1 = m x x 1 The line y = 3x 9 crosses the x-axis at coordinate A. Find the equation of the line with gradient 2 / 3 that passes through A. Give your answer in the form ax + by + c = 0 where a, b and c are integers. m = y 2 y 1 x 2 x 1 y 3x 9 At point A, y = 0 Thought Process To find the equation of the line, I need point A Point A is on the x-axis, so will have a y- coordinate of 0 I can put y=0 into the first equation to find out the x value at A 0 3x 9 9 3x 3 x Subtract 9 Divide by 3 A = (3,0) 5C/D

The equation of a straight line can be found if you know two points on the line, or you know its gradient and a single point. m = y 2 y 1 x 2 x 1 y y 1 = m x x 1 Thought Process To find the equation of the line, I need point A Point A is on the x-axis, so will have a y- coordinate of 0 As the equation I have already, crosses A as well, I can put y=0 into it to find out the x value at A A = (3,0) The line y = 3x 9 crosses the x-axis at coordinate A. Find the equation of the line with gradient 2 / 3 that passes through A. Give your answer in the form ax + by + c = 0 where a, b and c are integers. (x 1, y 1 ) = (3, 0) m = 2 / 3 y y1 m( x x1) 2 y 0 ( x 3) 3 2 y x 2 3 2 0 2 3 x y 0 2x 3y 6 Substitute in values Multiply out bracket Subtract y Multiply by 3 5C/D

The equation of a straight line can be found if you know two points on the line, or you know its gradient and a single point. m = y 2 y 1 x 2 x 1 y y 1 = m x x 1 Thought Process We need to find point A If the equations intersect at A, they have the same value for y (and x) I could try to solve these as simultaneous equations The lines y = 4x 7 and 2x + 3y 21 = 0 intersect at point A. Point B has coordinates (-2, 8). Find the equation of the line that passes through A and B y 4x 7 2x 3y 21 0 2x 3y 21 0 2x 3( 4x 7) 21 0 2x 12x 21 21 0 A = (3,5) 14x 42 x 3 y 5 Replace y with 4x - 7 Expand the bracket Group x s and add 42 Divide by 14 Sub x into one of the first equations to get y 5C/D

A = (3,5) The equation of a straight line can be found if you know two points on the line, or you know its gradient and a single point. m = y 2 y 1 x 2 x 1 y y 1 = m x x 1 Thought Process We need to find point A If the equations intersect at A, they have the same value for y (and x) The lines y = 4x 7 and 2x + 3y 21 = 0 intersect at point A. Point B has coordinates (-2, 8). Find the equation of the line that passes through A and B (x 1, y 1 ) = (3, 5) (x 2, y 2 ) = (-2, 8) m = y 2 y 1 x 2 x 1 m = 8 5 2 3 m = 3 5 or 3 5 Sub in values Calculate I could try to solve these as simultaneous equations 5C/D

A = (3,5) The equation of a straight line can be found if you know two points on the line, or you know its gradient and a single point. m = y 2 y 1 x 2 x 1 y y 1 = m x x 1 Thought Process We need to find point A If the equations intersect at A, they have the same value for y (and x) I could try to solve these as simultaneous equations The lines y = 4x 7 and 2x + 3y 21 = 0 intersect at point A. Point B has coordinates (-2, 8). Find the equation of the line that passes through A and B (x 1, y 1 ) = (3, 5) (x 2, y 2 ) = (-2, 8) y y 1 = m x x 1 y 5 = 3 5 x 3 5y 25 = 3 x 3 5y 25 = 3x + 9 5y + 3x 34 = 0 Sub in values Multiply by 5 Multiply by 5 Add 3x, Subtract 9 m = 3 5 5C/D

y y 1 = m x x 1 m = y 2 y 1 x 2 x 1 You need to be able to solve questions involving parallel lines. Find the gradient of the line we are given If you are told that lines are parallel, or are asked to find a parallel line, then they must have the same gradient 6x + 3y 2 = 0 3y = 2 6x y = 2 3 2x Add 2, Subtract 6x Divide by 3 A line is parallel to the line 6x + 3y 2 = 0 and passes through the coordinate (3,5). Find the equation of the line. So the gradient needs to be -2 m = -2 (x 1,y 1 ) = (3,5) 5E

y y 1 = m x x 1 m = y 2 y 1 x 2 x 1 You need to be able to solve questions involving parallel lines. If you are told that lines are parallel, or are asked to find a parallel line, then they must have the same gradient y y 1 = m x x 1 y 5 = 2 x 3 y 5 = 2x + 6 y = 2x + 11 Sub in values Expand the bracket Add 5 A line is parallel to the line 6x + 3y 2 = 0 and passes through the coordinate (3,5). Find the equation of the line. m = -2 (x 1,y 1 ) = (3,5) 5E

You need to be able to solve questions involving perpendicular lines. Perpendicular lines are at right-angles to each other. If you know the gradient of one, you can find the gradient of the other. 2 2 If a line has gradient m, a line perpendicular will have gradient 1 m 1 2 1 If two lines are perpendicular, their gradients have a product of -1 1 2 These rules mean the same thing and either can be used 5F

If a line has gradient m, a line perpendicular will have gradient 1 m You need to be able to solve questions involving perpendicular lines. Are the following lines perpendicular? 3x y 2 = 0 3x 2 = y Gradient = 3 Add y 3x y 2 = 0 x + 3y 6 = 0 x + 3y 6 = 0 3y = 6 x y = 2 1 3 x Add 6, subtract x Divide by 3 Gradient = - 1 / 3 So these lines are perpendicular! 5F

If a line has gradient m, a line perpendicular will have gradient 1 m You need to be able to solve questions involving perpendicular lines. Are the following lines perpendicular? 2x y + 4 = 0 2x + 4 = y Gradient = 2 Add y y = 1 2 x Gradient = 1 / 2 2x y + 4 = 0 So these lines are not perpendicular! 5F

If a line has gradient m, a line perpendicular will have gradient 1 m You need to be able to solve questions involving perpendicular lines. A line is perpendicular to the line 2y x 8 = 0, and passes through the coordinate (5,-7). Find the equation of the line. 2y x 8 = 0 2y = x + 8 y = 1 2 x + 4 Gradient = 1 / 2 Add x, add 8 Divide by 2 We already have a coordinate on the line Gradient of the perpendicular = -2 We just need to find its gradient by using what we already know m = -2 (x 1,y 1 ) = (5,-7) 5F

If a line has gradient m, a line perpendicular will have gradient 1 m You need to be able to solve questions involving perpendicular lines. A line is perpendicular to the line 2y x 8 = 0, and passes through the coordinate (5,-7). Find the equation of the line. We already have a coordinate on the line y y 1 = m x x 1 y ( 7) = 2 x 5 y + 7 = 2x + 10 y = 2x + 3 Sub in values Expand the bracket Subtract 7 We just need to find its gradient by using what we already know m = -2 (x 1,y 1 ) = (5,-7) 5F

You can find the distance between two points by using Pythagoras Theorem. This can also be helpful in finding areas. y (x 2, y 2 ) If we have 2 coordinates, (x 1, y 1 ) and (x 2, y 2 ), then we can create a formula for the distance between them using Pythagoras Theorem. y 2 y 1 d = y 2 y 1 2 + x 2 x 1 2 (x 1, y 1 ) x 2 x 1 x 5G

d = y 2 y 1 2 + x 2 x 1 2 You can find the distance between two points by using Pythagoras Theorem. This can also be helpful in finding areas. Find the distance between the coordinates (2,3) and (5,7) d = y 2 y 1 2 + x 2 x 1 2 d = 7 3 2 + 5 2 2 d = 25 d = 5 units Square root Sub in values Simplify You can substitute the coordinates either way round and it will work Be careful though. If you get a negative to substitute, ensure to write (-4) 2 instead of -4 2. Your calculator will always do indices before subtractions, which is why the bracket is needed 5G

d = y 2 y 1 2 + x 2 x 1 2 You can find the distance between two points by using Pythagoras Theorem. This can also be helpful in finding areas. The straight line l 1 with equation 4x y = 0 and the straight line l 2 2x + 3y 21 = 0 intersect at point A. a) Work out the coordinates of A. b) Work out the area of triangle AOB, where O is the origin and B is the point where l 2 meets the x-axis. You can solve them as simultaneous equations to find the coordinates of A 4x y = 0 2x + 3y 21 = 0 4x = y Rearrange 2x + 3y 21 = 0 2x + 3(4x) 21 = 0 14x 21 = 0 14x = 21 x = 3 2 y = 6 Sub in y = 4x Simplify Add 21 Divide by 14 5G

d = y 2 y 1 2 + x 2 x 1 2 You can find the distance between two points by using Pythagoras Theorem. This can also be helpful in finding areas. For this type of problem, a sketch is extremely helpful! Label key points such as where they lines cross the axes y 4x y = 0 The straight line l 1 with equation 4x y = 0 and the straight line l 2 2x + 3y 21 = 0 intersect at point A. 0, 7 A 3 2, 6 a) Work out the coordinates of A. 3 2, 6 b) Work out the area of triangle AOB, where O is the origin and B is the point where l 2 meets the x-axis. O 0, 0 B 21 2, 0 x 2x + 3y 21 = 0 5G

d = y 2 y 1 2 + x 2 x 1 2 You can find the distance between two points by using Pythagoras Theorem. This can also be helpful in finding areas. y 0, 7 4x y = 0 A 3 2, 6 The straight line l 1 with equation 4x y = 0 and the straight line l 2 2x + 3y 21 = 0 intersect at point A. O 0, 0 B 21 2, 0 x a) Work out the coordinates of A. 3 2, 6 b) Work out the area of triangle AOB, where O is the origin and B is the point where l 2 meets the x-axis. 2x + 3y 21 = 0 The area we need is a triangle with a width of 21 2 and a height of 6. 21 2 6 = 31.5 square units 2 5G

Extension (cm) E Two quantities are in direct proportion when they increase at the same rate. The graph of these quantities is a straight line through the origin. 20 The graph shows the extension, E, of a spring where different masses, m, are attached to the end of the spring. a) Calculate the gradient, k, of the line 1 20 b) Write an equation linking E and m c) Explain what the value of k represents in this context 10 0 100 200 300 400 Mass on spring (grams) Gradient = rise run Gradient = 5 100 Gradient = 1 20 100 Sub in values Simplify 5 m 5H

Extension (cm) E Two quantities are in direct proportion when they increase at the same rate. The graph of these quantities is a straight line through the origin. 20 The graph shows the extension, E, of a spring where different masses, m, are attached to the end of the spring. a) Calculate the gradient, k, of the line 1 20 b) Write an equation linking E and m E = 1 20 m c) Explain what the value of k represents in this context 10 0 y = mx + c E = 1 20 m 100 200 300 400 Mass on spring (grams) Replace y and x with E and m. We also know the gradient and y intercept m 5H

Extension (cm) E Two quantities are in direct proportion when they increase at the same rate. The graph of these quantities is a straight line through the origin. 20 The graph shows the extension, E, of a spring where different masses, m, are attached to the end of the spring. a) Calculate the gradient, k, of the line 1 20 b) Write an equation linking E and m E = 1 20 m c) Explain what the value of k represents in this context 10 0 100 200 300 400 Mass on spring (grams) The gradient indicates the increase in the vertical axis for an increase of 1 on the horizontal axis Therefore, k indicates that the extension of the spring increase by 1 cm for every 1g increase in mass 20 m 5H

Depth (cm) Two quantities are in direct proportion when they increase at the same rate. The graph of these quantities is a straight line through the origin. A container was filled with water. A hole was then made at the bottom of the container. The depth of the water was recorded at various time intervals, and the table shows the results. Time, t seconds 0 10 30 60 100 120 Depth, d cm 19.1 17.8 15.2 11.3 6.1 3.5 d 20 15 10 a) Determine whether a linear model is appropriate, by drawing a graph b) Deduce an equation in the form d = at + b c) Interpret the meaning of the coefficients a and b d) Use the model to estimate when the container will be empty 5 0 20 40 60 80 100 120 Time (s) As the points form a straight (or at least very close to straight) line, a linear model is appropriate t 5H

Two quantities are in direct proportion when they increase at the same rate. The graph of these quantities is a straight line through the origin. Time, t seconds 0 10 30 60 100 120 Depth, d cm 19.1 17.8 15.2 11.3 6.1 3.5 A container was filled with water. A hole was then made at the bottom of the container. The depth of the water was recorded at various time intervals, and the table shows the results. m = 0. 13 a) Determine whether a linear model is appropriate, by drawing a graph b) Deduce an equation in the form d = at + b c) Interpret the meaning of the coefficients a and b d) Use the model to estimate when the container will be empty You can find the equation of the line from the data in the table Choose 2 pairs of values and calculate rise/run m = 17.8 19.1 10 0 m = 0.13 5H

Two quantities are in direct proportion when they increase at the same rate. The graph of these quantities is a straight line through the origin. A container was filled with water. A hole was then made at the bottom of the container. The depth of the water was recorded at various time intervals, and the table shows the results. Time, t seconds 0 10 30 60 100 120 Depth, d cm 19.1 17.8 15.2 11.3 6.1 3.5 c = 19. 1 m = 0. 13 a) Determine whether a linear model is appropriate, by drawing a graph b) Deduce an equation in the form d = 0. 13t + 19. 1 d = at + b c) Interpret the meaning of the coefficients a and b d) Use the model to estimate when the container will be empty The y-intercept is also in the table Therefore, the equation is: d = 0.13t + 19.1 5H

Two quantities are in direct proportion when they increase at the same rate. The graph of these quantities is a straight line through the origin. Time, t seconds 0 10 30 60 100 120 Depth, d cm 19.1 17.8 15.2 11.3 6.1 3.5 A container was filled with water. A hole was then made at the bottom of the container. The depth of the water was recorded at various time intervals, and the table shows the results. a) Determine whether a linear model is appropriate, by drawing a graph b) Deduce an equation in the form d = 0. 13t + 19. 1 d = at + b c) Interpret the meaning of the coefficients a and b d) Use the model to estimate when the container will be empty d = 0.13t + 19.1 This is the change of depth per second. So every second the depth decreases by 0.13cm This is the depth when t = 0. It is the starting depth of the water in the tank. 5H

Two quantities are in direct proportion when they increase at the same rate. The graph of these quantities is a straight line through the origin. Time, t seconds 0 10 30 60 100 120 Depth, d cm 19.1 17.8 15.2 11.3 6.1 3.5 A container was filled with water. A hole was then made at the bottom of the container. The depth of the water was recorded at various time intervals, and the table shows the results. a) Determine whether a linear model is appropriate, by drawing a graph b) Deduce an equation in the form d = 0. 13t + 19. 1 d = at + b c) Interpret the meaning of the coefficients a and b d) Use the model to estimate when the container will be empty d = 0.13t + 19.1 0 = 0.13t + 19.1 0.13t = 19.1 t = 146.9 The container will be empty when the depth is 0 Add 0.13t Divide by 0.13 5H

Two quantities are in direct proportion when they increase at the same rate. The graph of these quantities is a straight line through the origin. There are 18,500 people to start with It will increase by 350 people for every year p = 18500 + 350t In 1991 there were 18,500 people living in Bradley Stoke. Planners project that the number of people living in Bradley Stoke would increase by 350 each year. Population does not increase at a linear rate (more people = faster growth), so this model may not be realistic a) Write down a linear model for the population p of Bradley Stoke t years after 1991 b) Write down one reason why this may not be a realistic model 5H