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Coordinate Geometry

Coordinate Geometry. Please choose a question to attempt from the following:. 1. 2. 3. 4. 5. STRAIGHT LINE : Question 1. Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). Reveal answer only.

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Coordinate Geometry

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  1. Coordinate Geometry Please choose a question to attempt from the following: 1 2 3 4 5

  2. STRAIGHT LINE : Question 1 Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). Reveal answer only Go to full solution Go to Marker’s Comments Go to Main Menu

  3. y = -5/3x - 6 STRAIGHT LINE : Question 1 Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). Reveal answer only Go to full solution Go to Marker’s Comments Go to Main Menu

  4. Question 1 y = -5/3x - 6 Back to Home 3x – 5y = 4 Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). 3x - 4 = 5y (5) 5y = 3x - 4 y = 3/5x - 4/5 Using y = mx + c , gradient of line is 3/5 So required gradient = -5/3 , ( m1m2 = -1) We now have (a,b) = (-6,4) & m =-5/3 . Using y – b = m(x – a) We get y – 4 = -5/3 (x – (-6)) y – 4 = -5/3 (x + 6) Begin Solution y – 4 = -5/3x - 10 Continue Solution Markers Comments

  5. Markers Comments -5 3 m1 = m2 = 3 5 y = -5/3x - 6 Back to Home • An attempt must be made to put the original equation into the form • y = mx + c to read off the gradient. 3x – 5y = 4 3x - 4 = 5y (5) 5y = 3x - 4 • State the gradient clearly. • State the condition for perpendicular lines m1 m2 = -1. y = 3/5x - 4/5 Using y = mx + c , gradient of line is 3/5 • When finding m2 simply invert and change the sign on m1 So required gradient = -5/3 , ( m1m2 = -1) We now have (a,b) = (-6,4) & m =-5/3 . Using y – b = m(x – a) • Use the y - b = m(x - a) form to obtain the equation of the line. We get y – 4 = -5/3 (x – (-6)) y – 4 = -5/3 (x + 6) Next Comment y – 4 = -5/3x - 10

  6. STRAIGHT LINE : Question 2 Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3). Reveal answer only Go to full solution Go to Marker’s Comments Go to Main Menu

  7. STRAIGHT LINE : Question 2 Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3). y = -2x + 7 Reveal answer only Go to full solution Go to Marker’s Comments Go to Main Menu

  8. Question 2 Back to Home 8x + 4y – 7 = 0 Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3). 4y = -8x + 7 (4) y = -2x + 7/4 Using y = mx + c , gradient of line is -2 So required gradient = -2 as parallel lines have equal gradients. We now have (a,b) = (5,-3) & m = -2. Using y – b = m(x – a) We get y – (-3) = -2(x – 5) Begin Solution y + 3 = -2x + 10 Continue Solution Markers Comments y = -2x + 7

  9. Markers Comments Back to Home • An attempt must be made to • put the original equation into • the form y = mx + c to • read off the gradient. 8x + 4y – 7 = 0 4y = -8x + 7 (4) y = -2x + 7/4 • State the gradient clearly. Using y = mx + c , gradient of line is -2 • State the condition for • parallel lines m1 = m2 So required gradient = -2 as parallel lines have equal gradients. • Use the y - b = m(x - a) form • to obtain the equation of • the line. We now have (a,b) = (5,-3) & m = -2. Using y – b = m(x – a) We get y – (-3) = -2(x – 5) y + 3 = -2x + 10 Next Comment y = -2x + 7

  10. Y C X A B STRAIGHT LINE : Question 3 In triangle ABC, A is (2,0), B is (8,0) and C is (7,3). (a) Find the gradients of AC and BC. (b) Hence find the size of ACB. Reveal answer only Go to full solution Go to Marker’s Comments Go to Main Menu

  11. Y C X A B STRAIGHT LINE : Question 3 In triangle ABC, A is (2,0), B is (8,0) and C is (7,3). (a) Find the gradients of AC and BC. (b) Hence find the size of ACB. mAC = 3/5 (a) Reveal answer only mBC = - 3 Go to full solution Go to Marker’s Comments (b) = 77.4° Go to Main Menu

  12. Question 3 Y C X B A Back to Home • Using the gradient formula: In triangle ABC, A is (2,0), B is (8,0) and C is (7,3). mAC = 3 – 0 7 - 2 = 3/5 • Find the gradients of AC • and BC. mBC = 3 – 0 7 - 8 = - 3 (b) Hence find the size of ACB. (b) Using tan = gradient If tan = 3/5 then CAB = 31.0° If tan = -3 then CBX = (180-71.6)° = 108.4 o so ABC = 71.6° Begin Solution Hence : ACB = 180°– 31.0° – 71.6° Continue Solution Markers Comments = 77.4°

  13. Markers Comments B Ø ° A Back to Home • If no diagram is given draw a • neat labelled diagram. • In calculating gradients state • the gradient formula. • Using the gradient formula: mAC = 3 – 0 7 - 2 = 3/5 • Must use the result that the • gradient of the line is equal • to the tangent of the angle • the line makes with the • positive direction of the x-axis. • Not given on the formula sheet. mBC = 3 – 0 7 - 8 = - 3 (b) Using tan = gradient If tan = 3/5 then CAB = 31.0° mAB = tanØ ° Ø ° = tan-1 mAB If tan = -3 then CBX = (180-71.6)° = 108.4 o so ABC = 71.6° Hence : ACB = 180°– 31.0° – 71.6° Next Comment = 77.4°

  14. STRAIGHT LINE : Question 4 Y In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Q(2,3) Find (a) the equation of the line e, the median from R of triangle PQR. X R(10,-1) P(4,-5) (b) the equation of the line f, the perpendicular bisector of QR. Reveal answer only (c) The coordinates of the point of intersection of lines e & f. Go to full solution Go to Marker’s Comments Go to Main Menu

  15. STRAIGHT LINE : Question 4 Y In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Q(2,3) Find (a) the equation of the line e, the median from R of triangle PQR. X R(10,-1) P(4,-5) (b) the equation of the line f, the perpendicular bisector of QR. Reveal answer only (c) The coordinates of the point of intersection of lines e & f. Go to full solution Go to Marker’s Comments (a) y = -1 (b) y = 2x – 11 Go to Main Menu (5,-1) (c)

  16. Question 4 (a) Y Q(2,3) X R(10,-1) P(4,-5) Back to Home • Midpoint of PQ is (3,-1): let’s call this S In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Using the gradient formula m = y2 – y1 x2 – x1 Find (a) the equation of the line e, the median from R of triangle PQR. mSR = -1 – (-1) 10 - 3 = 0 (ie line is horizontal) Since it passes through (3,-1) equation of e is y = -1 Solution to 4 (b) Begin Solution Continue Solution Markers Comments

  17. Question 4 (b) (b) Midpoint of QR is (6,1) Y Q(2,3) mQR = 3 – (-1) 2 - 10 = 4/-8 X R(10,-1) P(4,-5) so f is y = 2x – 11 Back to Home In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). = - 1/2 Find required gradient = 2 (m1m2 = -1) (b) the equation of the line f, the perpendicular bisector of QR. Using y – b = m(x – a) with (a,b) = (6,1) & m = 2 we get y – 1 = 2(x – 6) Solution to 4 (c) Begin Solution Continue Solution Markers Comments

  18. Question 4 (c) Y Q(2,3) X R(10,-1) P(4,-5) Back to Home (c) e & f meet when y = -1 & y = 2x -11 In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). so 2x – 11 = -1 Find ie 2x = 10 (c) The coordinates of the point of intersection of lines e & f. ie x = 5 Point of intersection is (5,-1) Begin Solution Continue Solution Markers Comments

  19. Markers Comments y Q x R P Back to Home • If no diagram is given draw a neat labelled diagram. • Midpoint of PQ is (3,-1): let’s call this S Using the gradient formula m = y2 – y1 x2 – x1 • Sketch the median and the • perpendicular bisector mSR = -1 – (-1) 10 - 3 (ie line is horizontal) median Perpendicular bisector Since it passes through (3,-1) equation of e is y = -1 Next Comment Comments for 4 (b)

  20. Markers Comments (b) Midpoint of QR is (6,1) mQR = 3 – (-1) 2 - 10 = 4/-8 required gradient = 2 (m1m2 = -1) , 2 + 10 3 + (-1) 2 2 Using y – b = m(x – a) with (a,b) = (6,1) & m = 2 y we get y – 1 = 2(x – 6) Q so f is y = 2x – 11 x R P Back to Home • To find midpoint of QR • Look for special cases: = - 1/2 Horizontal lines in the form y = k Vertical lines in the form x = k Next Comment Comments for 4 (c)

  21. Markers Comments Back to Home • To find the point of intersection of the two lines solve the two equations: (c) e & f meet when y = -1 & y = 2x -11 so 2x – 11 = -1 ie 2x = 10 y = -1 y = 2x - 11 ie x = 5 Point of intersection is (5,-1) Next Comment

  22. Y X E(6,-3) F(12,-5) G(2,-5) STRAIGHT LINE : Question 5 In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). Find (a) the equation of the altitude from vertex E. (b) the equation of the median from vertex F. (c) The point of intersection of the altitude and median. Reveal answer only Go to full solution Go to Marker’s Comments Go to Main Menu

  23. Y X E(6,-3) F(12,-5) G(2,-5) STRAIGHT LINE : Question 5 In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). Find (a) the equation of the altitude from vertex E. (b) the equation of the median from vertex F. (c) The point of intersection of the altitude and median. Reveal answer only Go to full solution (a) x = 6 Go to Marker’s Comments (b) x + 8y + 28 = 0 (c) Go to Main Menu (6,-4.25)

  24. Question 5(a) Back to Home • Using the gradient formula In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). mFG = -5 – (-5) 12 - 2 = 0 Find (a) the equation of the altitude from vertex E. (ie line is horizontal so altitude is vertical) Y Altitude is vertical line through (6,-3) ie x = 6 X E(6,-3) F(12,-5) Solution to 5 (b) G(2,-5) Begin Solution Continue Solution Markers Comments

  25. Question 5(b) Back to Home • Midpoint of EG is (4,-4)- let’s call this H In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). mFH = -5 – (-4) 12 - 4 = -1/8 Find Using y – b = m(x – a) with (a,b) = (4,-4) & m =-1/8 (b) the equation of the median from vertex F. Y we get y – (-4) = -1/8(x – 4) (X8) X E(6,-3) or 8y + 32 = -x + 4 F(12,-5) G(2,-5) Median is x + 8y + 28 = 0 Begin Solution Solution to 5 (c) Continue Solution Markers Comments

  26. Question 5(c) Back to Home (c) Lines meet when x = 6 & x + 8y + 28 = 0 In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). put x =6 in 2nd equation 8y + 34 = 0 Find ie 8y = -34 (c) The point of intersection of the altitude and median. ie y = -4.25 Y Point of intersection is (6,-4.25) X E(6,-3) F(12,-5) G(2,-5) Begin Solution Continue Solution Markers Comments

  27. Markers Comments y E x F G Back to Home • If no diagram is given draw a • neat labelled diagram. • Sketch the altitude and • the median. • Using the gradient formula mFG = -5 – (-5) 12 - 2 = 0 (ie line is horizontal so altitude is vertical) Altitude is vertical line through (6,-3) ie x = 6 median Comments for 5 (b) altitude Next Comment

  28. Markers Comments • To find midpoint of EG 2 + 6 -3 + (-5) 2 2 , H y E x F G Back to Home • Midpoint of EG is (4,-4)- call this H • Look for special cases: mFH = -5 – (-4) 12 - 4 = -1/8 Horizontal lines in the form y = k Vertical lines in the form x = k Using y – b = m(x – a) with (a,b) = (4,-4) & m =-1/8 we get y – (-4) = -1/8(x – 4) (X8) or 8y + 32 = -x + 4 Median is x + 8y + 28 = 0 Next Comment Comments for 5 (c)

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