Co-ordinate Geometry in the (x, y) Plane.

Co-ordinate Geometry in the (x, y) Plane

Introduction This chapter reminds us of how to calculate midpoints and distances between co-ordinates We will see lots of Algebraic versions as well We will also learn to solve problems involving the equation of a circle You will need to remember how to calculate the equation of a line as well know the gradient of the perpendicular to it

Teachings for Exercise 4A

Co-ordinate Geometry in the (x, y) Plane
You need to be able to find the mid-point of a line joining (x1,y1) and (x2,y2) You can find the mid-point of a line by using the following formula: Where (x1,y1) and (x2,y2) are the ends of the line segment Find the midpoint of this pair of points: Let the first coordinate be (x1,y1) and the second be (x2,y2) Calculate

You need to be able to find the mid-point of a line joining (x1,y1) and (x2,y2) You can find the mid-point of a line by using the following formula: Where (x1,y1) and (x2,y2) are the ends of the line segment Find the midpoint of this pair of points: Let the first coordinate be (x1,y1) and the second be (x2,y2) Calculate (leaving fractions in is fine!)

You need to be able to find the mid-point of a line joining (x1,y1) and (x2,y2) You can find the mid-point of a line by using the following formula: Where (x1,y1) and (x2,y2) are the ends of the line segment Find the midpoint of this pair of points: Let the first coordinate be (x1,y1) and the second be (x2,y2) Calculate

AB is a diameter of a circle, where A and B are the coordinates (-3,8) and (5,4) respectively. Find the coordinates of the centre of the circle. You need to be able to find the mid-point of a line joining (x1,y1) and (x2,y2) You can find the mid-point of a line by using the following formula: Where (x1,y1) and (x2,y2) are the ends of the line segment (-3,8) A (1,6) (5,4) B The centre of the circle is at the midpoint of its diameter… Calculate the coordinates

PQ is a diameter of a circle, centre (2,-2). Given that P is (8,-5), find the coordinates of Q. You need to be able to find the mid-point of a line joining (x1,y1) and (x2,y2) You can find the mid-point of a line by using the following formula: Where (x1,y1) and (x2,y2) are the ends of the line segment (x,y) (-4,1) Q (2,-2) (8,-5) P Write the unknown coordinate in terms of x and y, and fill in the formula for the mid-point as before… 𝑥+8 2 , 𝑦−5 2 Set the x-part equal to 2 and the y-part equal to -2…

Teachings for Exercise 4B

Draw a sketch! (-1,4) You need to be able to include topics covered in C1 in answering questions The line AB is the diameter of the circle with centre C, where A and B are (-1, 4) and (5, 2) respectively. The line l passes through C and is perpendicular to AB. Find the equation of l. We need to: a) Find the gradient of the line AB b) Then work out the gradient perpendicular to that c) We also need to find the co-ordinates of the centre d) We can then find the equation of l A C (5,2) B = -1/3 = 3 4B

Draw a sketch! (-1,4) You need to be able to include topics covered in C1 in answering questions The line AB is the diameter of the circle with centre C, where A and B are (-1, 4) and (5, 2) respectively. The line l passes through C and is perpendicular to AB. Find the equation of l. We need to: a) Find the gradient of the line AB b) Then work out the gradient perpendicular to that c) We also need to find the co-ordinates of the centre d) We can then find the equation of l A C (5,2) B = -1/3 = 3 = (2,3) 4B

Draw a sketch! (1,12) Q You need to be able to include topics covered in C1 in answering questions The line PQ is the Chord of the circle centre (-3,5), where P and Q are (5,4) and (1,12) respectively. The line l is perpendicular to PQ and bisects it. Show that it passes through the centre of the circle. We need to: a) Find the midpoint of PQ b) Find the gradient of PQ, and then the perpendicular c) We can then find the equation of line l and substitute (-3,5) into it l C P (-3,5) (5,4) = (3,8) 4B

Draw a sketch! (1,12) Q You need to be able to include topics covered in C1 in answering questions The line PQ is the Chord of the circle centre (-3,5), where P and Q are (5,4) and (1,12) respectively. The line l is perpendicular to PQ and bisects it. Show that it passes through the centre of the circle. We need to: a) Find the midpoint of PQ b) Find the gradient of PQ, and then the perpendicular c) We can then find the equation of line l and substitute (-3,5) into it l C P (-3,5) (5,4) = (3,8) = -2 so 1/2 4B

y = 3x - 11 Draw a sketch! You need to be able to include topics covered in C1 in answering questions The lines AB and CD are chords of a circle. The line y = 3x – 11 is the perpendicular bisector of AB. The line y = -x – 1 is the perpendicular bisector of CD. Find the coordinates of the circle’s centre. THE PERPENDICULAR BISECTOR OF A CHORD GOES THROUGH THE CENTRE! We need to: 1) Set the bisectors equal to each other and solve the equation for x and y. D C A B y = -x - 1 4B

Teachings for Exercise 4C

You can find the distance between 2 points on a line using Pythagoras’ Theorem Find the distance between the co-ordinates (2,3) and (5,7) (y2-y1) (x2-x1) (x1,y1) x (x1,y1) = (2,3) (x2,y2) = (5,7) 4C

You can find the distance between 2 points on a line using Pythagoras’ Theorem Find the distance between the co-ordinates (4a,a) and (-3a,2a) (y2-y1) (x1,y1) = (4a,a) (x2,y2) = (-3a,2a) (x2-x1) (x1,y1) x Break down each part 4C

You can find the distance between 2 points on a line using Pythagoras’ Theorem PQ is the diameter of a circle, where P and Q are (-1,3) and (6,-3) respectively. Find the radius of the circle (x1,y1) = (-1,3) (x2,y2) = (6,-3) Radius is half the diameter 4C

You can find the distance between 2 points on a line using Pythagoras’ Theorem The line AB is the diameter of the circle, where A and B are (-3,21) and (7,-3) respectively. The point C (14,4) lies on the circumference of the circle. Find the values of AB2, AC2 and BC2 and hence show that angle ACB is 90° (-3,21) A Since Pythagoras’ Theorem works, ACB must be right-angled! (14,4) C B (7,-3) 4C

Teachings for Exercise 4D

You can write the equation of a circle in the following form In your books, plot the graph of the equation: x2 + y2 = 25 for values of x from -5 to 5 4D

You can write the equation of a circle in the following form (x – a)2 + (y – b)2 = r2 Where (a,b) is the centre of the circle and r is its radius. Write down the equation of the circle with centre (5,7) and radius 4 (5,7) This is a translation 5 units to the left This is a translation 7 units up A circle with this equation would have centre at (0,0) 4D

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You can write the equation of a circle in the following form Find the coordinates of the centre, and the radius of, the circle with the following equation: 𝑥 𝑦−1 2 = 4 2 Remember the coordinates have the opposite sign to what is in the brackets! 𝐶𝑒𝑛𝑡𝑟𝑒=(−3,1) 𝑅𝑎𝑑𝑖𝑢𝑠=4 If the equation is written as above, the radius is obvious! 4D

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You can write the equation of a circle in the following form Find the coordinates of the centre, and the radius of, the circle with the following equation: 𝑥− 𝑦+4 2 =32 Remember the coordinates have the opposite sign to what is in the brackets! 𝐶𝑒𝑛𝑡𝑟𝑒= 5 2 ,−4 𝑅𝑎𝑑𝑖𝑢𝑠= 32 If the equation is written as above, square root the number =4 2 4D

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You can write the equation of a circle in the following form Show that the circle: Passes through (5,-8) 𝑥− 𝑦+4 2 =20 Sub in values 5− − =20 Calculate inside the brackets 𝑥− 𝑦+4 2 =20 −4 2 =20 Square both 4+16=20 As the statement is true, the circle curve will pass through (5,-8) 4D

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You can write the equation of a circle in the following form The line AB is the diameter of a circle, where A and B are (4,7) and (-8,3) respectively. Find the equation of the circle.  To do this we need to find the centre of the circle, and its radius… Finding the midpoint 𝑥 1 + 𝑥 2 2 , 𝑦 1 + 𝑦 2 2 Sub in the coordinate values −8+4 2 , 3+7 2 Calculate −2,5 (4,7) (-2,5) (-8,3) 𝐶𝑒𝑛𝑡𝑟𝑒= −2,5 4D

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You can write the equation of a circle in the following form The line AB is the diameter of a circle, where A and B are (4,7) and (-8,3) respectively. Find the equation of the circle.  To do this we need to find the centre of the circle, and its radius… Finding the radius The radius will be the distance from the centre to the edge, so you can use 2 coordinates and Pythagoras’ Theorem for this! Using (-2,5) and (4,7) 𝑥 2 − 𝑥 𝑦 2 − 𝑦 1 2 Sub in values… 4−(−2) −5 2 Careful with negatives! (4,7) (-2,5) Calculate (-8,3) 40 Simplify =2 10 𝐶𝑒𝑛𝑡𝑟𝑒= −2,5 𝑅𝑎𝑑𝑖𝑢𝑠=2 10 4D

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane 𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 You can write the equation of a circle in the following form The line AB is the diameter of a circle, where A and B are (4,7) and (-8,3) respectively. Find the equation of the circle.  To do this we need to find the centre of the circle, and its radius… Replace values 𝑥 𝑦−5 2 = (2 10 ) 2 Other form 𝑥 𝑦−5 2 =40 (4,7) (-2,5) (-8,3) 𝐶𝑒𝑛𝑡𝑟𝑒= −2,5 𝑅𝑎𝑑𝑖𝑢𝑠=2 10 4D

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You can write the equation of a circle in the following form The line 4x – 3y – 40 = 0 touches the circle (x – 2)2 + (y – 6)2 = 100 at P = (10,0). Show that the radius at P is perpendicular to this line. To solve this problem, you need to find the gradient of the straight line and compare it to the gradient of the radius at (10,0)  Find the gradient of the straight line by writing it in terms of y 4𝑥−3𝑦−40=0 Add 3y 4𝑥−40=3𝑦 (x – 2)2 + (y – 6)2 = 100 4x – 3y – 40 = 0 Divide by 3 4 3 𝑥− 40 3 =𝑦 (2,6) So the gradient of the straight line is 4/3 (10,0) 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒= 4 3 4D

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You can write the equation of a circle in the following form The line 4x – 3y – 40 = 0 touches the circle (x – 2)2 + (y – 6)2 = 100 at P = (10,0). Show that the radius at P is perpendicular to this line. Now find the gradient of the radius at (10,0) You can do this by finding the gradient of the line segment shown in green… You can use a formula from C1! 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 Sub in values (make sure you put them the same way round!) 𝑚= 0 −(6) (10)−(2) (x – 2)2 + (y – 6)2 = 100 4x – 3y – 40 = 0 Calculate 𝑚= −6 8 (2,6) Simplify 𝑚=− 3 4 (10,0) 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒= 4 3 So this is the gradient of the radius at (10,0) 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑟𝑎𝑑𝑖𝑢𝑠=− 3 4 4D

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You can write the equation of a circle in the following form The line 4x – 3y – 40 = 0 touches the circle (x – 2)2 + (y – 6)2 = 100 at P = (10,0). Show that the radius at P is perpendicular to this line. 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒= 4 3 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑟𝑎𝑑𝑖𝑢𝑠=− 3 4 Multiply the gradients together… 4 3 ×− 3 4 =− 12 12 (x – 2)2 + (y – 6)2 = 100 4x – 3y – 40 = 0 =−1 (2,6) If two gradients multiply to make -1, they are perpendicular (this is another way of defining it!) (10,0) 4D

Teachings for Exercise 4E

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane 𝑥− 𝑦−4 2 =65 You need to be able to solve lots of problems linking circles and other aspects of geometry on a set of axes Find where the circle with the following equation meets the x-axis… Sub in y = 0 𝑥− −4 2 =65 Square the numerical bracket 𝑥− =65 Subtract 16 𝑥−5 2 =49 Square root the whole of both sides 𝑥− 𝑦−4 2 =65 𝑥−5=±7 Add 5 𝑥=5±7 𝑥=12 𝑜𝑟 𝑥=−2 So the coordinates are: 12,0 𝑎𝑛𝑑 (−2,0) In the places where the equation meets the x-axis, the y-coordinate is 0  Sub this into the equation… 4E

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You need to be able to solve lots of problems linking circles and other aspects of geometry on a set of axes Find the coordinates where the line y = x + 5 meets the circle x2 + (y – 2)2 = 29. This is effectively solving simultaneous equations, where one is a quadratic (although actually it is a circle) You can solve by substitution. Replace the y in the circle equation with x + 5 since we are told these are equivalent… We now know the x-coordinates where the lines meets are -5 and 2 Sub these into the linear equation to find the y-coordinates… 𝑥 2 + 𝑦−2 2 =29 Replace y with x + 5 𝑥 2 + 𝑥+5−2 2 =29 Simplify the bracket 𝑥 2 + 𝑥+3 2 =29 Expand the squared bracket 𝑥 2 + 𝑥 2 +6𝑥+9=29 Group terms on the left side 2𝑥 2 +6𝑥−20=0 Divide by 2 𝑥 2 +3𝑥−10=0 Factorise 𝑥+5 𝑥−2 =0 Find solutions 𝑥=−5 𝑜𝑟 𝑥=2 4E

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You need to be able to solve lots of problems linking circles and other aspects of geometry on a set of axes Find the coordinates where the line y = x + 5 meets the circle x2 + (y – 2)2 = 29. This is effectively solving simultaneous equations, where one is a quadratic (although actually it is a circle) You can solve by substitution. Replace the y in the circle equation with x + 5 since we are told these are equivalent… We now know the x-coordinates where the lines meets are -5 and 2 Sub these into the linear equation to find the y-coordinates… 𝑦=𝑥+5 x = -5 x = 2 𝑦=−5+5 𝑦=2+5 𝑦=0 𝑦=7 (−5,0) (2,7) (2,7) (-5,0) x2 + (y – 2)2 = 29 y = x + 5 4E

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You need to be able to solve lots of problems linking circles and other aspects of geometry on a set of axes Show that the line y = x – 7 does not touch the circle (x + 2)2 + y2 = 33.  Start in the same way as the last question, by replacing y with x – 7 in the circle equation… (𝑥+2) 𝑦 2 =33 Replace y with x - 7 (𝑥+2) (𝑥−7) 2 =33 Square both brackets 𝑥 2 +4𝑥+4+ 𝑥 2 −14𝑥+49=33 Group terms 2𝑥 2 −10𝑥+53=33 Subtract 33 2𝑥 2 −10𝑥+20=0 Divide by 2 𝑥 2 −5𝑥+10=0 We will use the Quadratic formula… 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 𝑎=1 𝑏=−5 𝑐=10 Replace a, b and c 𝑥= 5± (−5) 2 −(4×1×10) 2(1) Calculate parts As we cannot square root a negative number, this equation is unsolvable  The geometrical implication is that the lines do not meet! 𝑥= 5± −15 2 4E

𝑥−𝑎 𝑦−𝑏 2 = 𝑟 2 𝐶𝑒𝑛𝑡𝑟𝑒=(𝑎,𝑏) 𝑅𝑎𝑑𝑖𝑢𝑠=𝑟 Co-ordinate Geometry in the (x, y) Plane You need to be able to solve lots of problems linking circles and other aspects of geometry on a set of axes Show that the line y = x – 7 does not touch the circle (x + 2)2 + y2 = 33.  This is what the curves would look like if you drew them… y = x - 7 (x + 2)2 + y2 = 33 4E

Summary We have been reminded of how to calculate midpoints and distances between two coordinates We have looked at the equation for a circle We have seen lots of questions based on this, involving a lot of the geometry from C1 We have also seen how to involve the various methods for solving quadratics in these problems…

Co-ordinate Geometry in the (x, y) Plane.

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Co-ordinate Geometry in the (x, y) Plane.

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