Coordinate Geometry in the (x,y) plane

Introduction This Chapter focuses on coordinate geometry, mainly involving straight line graphs We will be looking at working out equations of graphs based on various sets of information

Teachings for Exercise 5A

Coordinate Geometry in the (x,y) plane Equation of a straight line The equation of a straight line is usually written in one of 2 forms. One you will have seen before; Where m is the gradient and c is the y-intercept. Or, the general form: Where a, b and c are integers. y-intercept gradient 1 x 5A

Coordinate Geometry in the (x,y) plane Equation of a straight line The equation of a straight line is usually written in one of 2 forms. One you will have seen before; Where m is the gradient and c is the y-intercept. Or, the general form: Where a, b and c are integers. Example 1 Write down the gradient and y-intercept of the following graphs a)  Gradient = -3  y-intercept = (0,2) b) Rearrange to get ‘y’ on one side Divide by 2  Gradient = 2  y-intercept = (0, 5/2) 5A

Coordinate Geometry in the (x,y) plane Equation of a straight line The equation of a straight line is usually written in one of 2 forms. One you will have seen before; Where m is the gradient and c is the y-intercept. Or, the general form: Where a, b and c are integers. Example 2 Write each equation in the form ax + by + c = 0 a) -y Correct form b) +1/2x and -5 x2 (to remove fraction) Correct form 5A

Coordinate Geometry in the (x,y) plane Equation of a straight line The equation of a straight line is usually written in one of 2 forms. One you will have seen before; Where m is the gradient and c is the y-intercept. Or, the general form: Where a, b and c are integers. Example 3 A line is parallel to the line y = 3x + 2 and passes through (0,-1). Write the equation of the line. Parallel so the gradient will be the same Crosses through (0,-1), which is on the y-axis 2 -1 5A

Coordinate Geometry in the (x,y) plane Equation of a straight line The equation of a straight line is usually written in one of 2 forms. One you will have seen before; Where m is the gradient and c is the y-intercept. Or, the general form: Where a, b and c are integers. Example 4 A line is parallel to the line 6x + 3y – 2 = 0 and passes through (0,3). Write the equation of the line. Rearrange to the form y = mx + c Divide by 3 The new line has the same gradient, but intercepts the y-axis at 3 5A

Coordinate Geometry in the (x,y) plane Equation of a straight line The equation of a straight line is usually written in one of 2 forms. One you will have seen before; Where m is the gradient and c is the y-intercept. Or, the general form: Where a, b and c are integers. Example 5 The line y = 4x + 8 crosses the y-axis at 8. It crosses the x-axis at P. Work out the coordinates of P. Crosses the x-axis where y=0 -8 Divide by 4 So the line crosses the x-axis at (-2,0) 5A

Teachings for Exercise 5B

Coordinate Geometry in the (x,y) plane The gradient of a line You can work out the gradient of a line if you know 2 points on it. Let the first point be (x1,y1) and the second be (x2,y2). The following formula gives the gradient: ‘The change in the y values, divided by the change in the x values’ y (x2,y2) (x1,y1) y2 - y1 x2 - x1 x 5B

Coordinate Geometry in the (x,y) plane The gradient of a line You can work out the gradient of a line if you know 2 points on it. Let the first point be (x1,y1) and the second be (x2,y2). The following formula gives the gradient: ‘The change in the y values, divided by the change in the x values’ Example 1 Calculate the gradient of the line which passes through (2,3) and (5,7) (x1, y1) = (2, 3) (x2, y2) = (5, 7) Substitute numbers in   Work out or leave as a fraction 5B

Coordinate Geometry in the (x,y) plane The gradient of a line You can work out the gradient of a line if you know 2 points on it. Let the first point be (x1,y1) and the second be (x2,y2). The following formula gives the gradient: ‘The change in the y values, divided by the change in the x values’ Example 2 Calculate the gradient of the line which passes through (-2,7) and (4,5) (x1, y1) = (-2, 7) (x2, y2) = (4, 5) Substitute numbers in Work out or leave as a fraction Simplify if possible 5B

Coordinate Geometry in the (x,y) plane The gradient of a line You can work out the gradient of a line if you know 2 points on it. Let the first point be (x1,y1) and the second be (x2,y2). The following formula gives the gradient: ‘The change in the y values, divided by the change in the x values’ Example 3 Calculate the gradient of the line which passes through (2d,-5d) and (6d,3d) (x1, y1) = (2d, -5d) (x2, y2) = (6d, 3d) Substitute numbers in Work out or leave as a fraction Simplify if possible (the d’s cancel out) 5B

Coordinate Geometry in the (x,y) plane The gradient of a line You can work out the gradient of a line if you know 2 points on it. Let the first point be (x1,y1) and the second be (x2,y2). The following formula gives the gradient: ‘The change in the y values, divided by the change in the x values’ Example 4 The line joining (2, -5) to (4, a) has a gradient of -1. Calculate the value of a. (x1, y1) = (2, -5) (x2, y2) = (4, a) Substitute numbers in Simplify Multiply by 2 Subtract 5 5B

Teachings for Exercise 5C

Coordinate Geometry in the (x,y) plane Finding the Equation of a line You can find the equation of the line with gradient m, and coordinate (x1, y1) by using the following formula: Example 1 Find the equation of the line with gradient 5 that passes through the point (3,2) (x1, y1) = (3, 2) m = 5 Substitute the numbers in Expand the bracket Add 2 5C

Coordinate Geometry in the (x,y) plane Finding the Equation of a line You can find the equation of the line with gradient m, and coordinate (x1, y1) by using the following formula: Example 2 Find the equation of the line with gradient -1/2 that passes through the point (4,-6) (x1, y1) = (4, -6) m = -1/2 Substitute the numbers in Expand the brackets Subtract 6 5C

Coordinate Geometry in the (x,y) plane Finding the Equation of a line You can find the equation of the line with gradient m, and coordinate (x1, y1) by using the following formula: 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’ Example 3 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. At point A, y = 0 Subtract 9 Divide by 3 A = (3,0) 5C

Coordinate Geometry in the (x,y) plane Finding the Equation of a line You can find the equation of the line with gradient m, and coordinate (x1, y1) by using the following formula: 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’ Example 3 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. (x1, y1) = (3, 0) m = 2/3 Substitute in values Multiply out bracket Subtract y Multiply by 3 5C

Teachings for Exercise 5D

Coordinate Geometry in the (x,y) plane Finding the Equation of a line You can find the equation of a line from 2 points by using the following formula: Example 1 Work out the equation of the line that goes through points (3,-1) and (5, 7). Give your answer in the form y = mx + c. (x1, y1) = (3, -1) (x2, y2) = (5, 7) Substitute in values Work out any sums Multiply the right side by 4 to make fractions the same Multiply by 8 Subtract 1 5D

Coordinate Geometry in the (x,y) plane Finding the Equation of a line You can find the equation of a line from 2 points by using the following formula: Thought Process ‘We need to find point A’ ‘If the equations intersect at A, they have the same value for y (and x)’ If I can write one of the equations in terms of y, I can replace the y in the second equation and solve it’ Example 1 The lines y = 4x – 7 and 2x + 3y – 21 = 0 intersect at point A. Point B has co-ordinates (-2, 8). Find the equation of the line that passes through A and B 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 A = (3,5) 5D

Coordinate Geometry in the (x,y) plane Finding the Equation of a line You can find the equation of a line from 2 points by using the following formula: Thought Process ‘We need to find point A’ ‘If the equations intersect at A, they have the same value for y (and x)’ If I can write one of the equations in terms of y, I can replace the y in the second equation and solve it’ Example 1 A = (3,5) The lines y = 4x – 7 and 2x + 3y – 21 = 0 intersect at point A. Point B has co-ordinates (-2, 8). Find the equation of the line that passes through A and B (x1, y1) = (3, 5) (x2, y2) = (-2, 8) Substitute in values Work out the denominators Multiply all of left by -5 and all of right by 3 (makes denominators equal) Multiply by -15 Rearrange, keeping x positive 5D

Teachings for Exercise 5E

Coordinate Geometry in the (x,y) plane Finding the Perpendicular to a line You need to be able to work out the gradient of a line which is Perpendicular to another.  Perpendicular means ‘intersects at a right angle… If a line has a gradient of m, the line perpendicular has gradient -1/m Two perpendicular lines have gradients that multiply to give -1 Example 1 Work out the gradient of the line that is perpendicular to the lines with the following gradients. Line gradient Perpendicular These lines are perpendicular 3 -1/3 1/2 -2 -2/5 5/2 2x -1/2x 5E

Coordinate Geometry in the (x,y) plane Finding the Perpendicular to a line You need to be able to work out the gradient of a line which is Perpendicular to another.  Perpendicular means ‘intersects at a right angle… If a line has a gradient of m, the line perpendicular has gradient -1/m Two perpendicular lines have gradients that multiply to give -1 Example 2 Is the line y = 3x + 4 perpendicular to the line x + 3y – 3 = 0? Gradient = 3 These lines are perpendicular Gradient = -1/3 The lines are perpendicular since their gradients multiply to give -1 5E

Coordinate Geometry in the (x,y) plane Finding the Perpendicular to a line You need to be able to work out the gradient of a line which is Perpendicular to another.  Perpendicular means ‘intersects at a right angle… If a line has a gradient of m, the line perpendicular has gradient -1/m Two perpendicular lines have gradients that multiply to give -1 Example 3 Find an equation for the line that passes through (3,-1) and is perpendicular to the line y = 2x - 4 Gradient = 2 These lines are perpendicular Gradient of the perpendicular = -1/2 (x1, y1) = (3, -1) m = -1/2 Substitute in values Expand brackets Subtract 1 5E

Summary We have learnt how to write equations of a line in 2 different forms We have done this from varying sets of information We have also looked at the link between parallel and perpendicular lines