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Unformatted text preview: Chapter 2 Analytic Geometry Chapter 2 Get Ready Chapter 2 Get Ready a) Question 1 Page 54 3x + 2 = 14 3x + 2 − 2 = 14 − 2 3x = 12 3x 12 = 3 3 x=4 b) 7 y − 5 = 2 y + 10 7 y − 5 − 2 y + 5 = 2 y + 10 − 2 y + 5 5 y = 15 5 y 15 = 5 5 y=3 c) 1 1 z = z−2 4 3 1 1 12 × z = 12 × z − 12 × 2 4 3 3z = 4 z − 24 3z − 4 z = 4 z − 24 − 4 z − z = −24 − z −24 = −1 −1 z = 24 d) t = 0.5 ( t) 2 = ( 0.5 ) 2 t = 0.25 MHR • Principles of Mathematics 10 Solutions 1 Chapter 2 Get Ready a) Question 2 Page 54 x− y+2=0 x− y +2− x−2 = 0− x−2 − y = −x − 2 − y −x 2 = − −1 −1 −1 y = x+2 b) 3x + y − 5 = 0 3x + y − 5 − 3x + 5 = 0 − 3x + 5 y = −3 x + 5 c) 2x − 4 y + 7 = 0 2x − 4 y + 7 − 2x − 7 = 0 − 2x − 7 −4 y = −2 x − 7 −4 y −2 x 7 = − −4 −4 −4 1 7 y = x+ 2 4 d) 1 x − 3y + 5 = 0 2 1 1 1 x − 3y + 5 − x − 5 = 0 − x − 5 2 2 2 1 −3 y = − x − 5 2 1 − −3 y 5 = 2 x− −3 −3 −3 1 5 y= x+ 6 3 2 MHR • Principles of Mathematics 10 Solutions Chapter 2 Get Ready a) ma = Question 3 Page 54 1 2 2 4 1 = 2 b) mb = −2 3 2 =− 3 c) mc = −2 8 1 =− 4 d) ma = MHR • Principles of Mathematics 10 Solutions 3 Chapter 2 Get Ready a) m = Question 4 Page 54 y2 − y1 x2 − x1 10 − 6 12 − 4 4 = 8 1 = 2 = b) m = = y2 − y1 x2 − x1 2−6 12 − ( −4 ) −4 16 1 =− 4 = c) m = = y2 − y1 x2 − x1 −8 − ( −4 ) 3 − ( −5 ) −4 8 1 =− 2 = d) m = y2 − y1 x2 − x1 7.6 − 6.4 9.8 − 2.5 1.2 = 7.3 12 = 73 = 4 MHR • Principles of Mathematics 10 Solutions Chapter 2 Get Ready Question 5 Page 55 a) y = mx + b = −2 x + 4 The equation of the line is y = –2x + 4. b) y = mx + b = 2 x − 14 7 The equation of the line is y = c) 2 x – 14. 7 y = mx + b 3 = 4 (6) + b 3 = 24 + b −21 = b y = mx + b = 4 x + ( −21) = 4 x − 21 The equation of the line is y = 4x + 21. d) y = mx + b 1 ( −2 ) + b 2 4 = 1+ b 3=b 4=− y = mx + b 1 = − x+3 2 1 The equation of the line is y = − x + 3. 2 MHR • Principles of Mathematics 10 Solutions 5 Chapter 2 Get Ready a) m = Question 6 Page 55 y2 − y1 x2 − x1 9 −1 5 −1 8 = 4 =2 Substitute m = 2 and the coordinates of one point, say (1, 1), to find b. y = mx + b = 1 = 2 (1) + b −1 = b The equation of the line is y = 2x – 1. b) m= = y2 − y1 x2 − x1 −2 − 1 −3 − ( −1) −3 −2 3 = 2 = Substitute m = y = mx + b 1= 1+ 3 and the coordinates of one point, say (–1, 1), to find b. 2 3 ( −1) + b 2 3 =b 2 5 =b 2 The equation of the line is y = 3 5 x+ . 2 2 6 MHR • Principles of Mathematics 10 Solutions c) m= = y2 − y1 x2 − x1 4 −1 2 − ( −4 ) 3 6 1 = 2 = Substitute m = y = mx + b 1 1 = ( −4 ) + b 2 3=b 1 and the coordinates of one point, say (–4, 1), to find b. 2 The equation of the line is y = d) m= = 1 x + 3. 2 y2 − y1 x2 − x1 4 − ( −8 ) −1 − 5 12 = −6 = −2 Substitute m = –2 and the coordinates of one point, say (5, –8), to find b. y = mx + b −8 = −2 ( 5) + b 2=b The equation of the line is y = –2x + 2. MHR • Principles of Mathematics 10 Solutions 7 Chapter 2 Get Ready Question 7 Page 55 a) The slope of the given line is 3. The slope of a line parallel to the given line is also 3. 1 1 b) The slope of the given line is − . The slope of a line parallel to the given line is also − . 6 6 c) The slope of the given line is –4. The slope of a line perpendicular to the given line is the 1 negative reciprocal, . 4 d) The slope of the given line is 3 . The slope of a line perpendicular to the given line is the 4 4 negative reciprocal, − . 3 8 MHR • Principles of Mathematics 10 Solutions Chapter 2 Get Ready Question 8 Page 55 a) The slope of the required line is –3. y = mx + b 5 = −3 ( −3) + b −4 = b The equation of the line is y = –3x – 4. b) The slope of the required line is y = mx + b 2 . 3 2 (2) + b 3 4 3= +b 3 9 4 − =b 3 3 5 =b 3 3= The equation of the line is y = 2 5 x+ . 3 3 3 c) The slope of the required line is − . 4 y = mx + b 3 ( −5 ) + b 4 15 1= +b 4 4 15 − =b 4 4 11 − =b 4 1= − 3 11 The equation of the line is y = − x − . 4 4 MHR • Principles of Mathematics 10 Solutions 9 Chapter 2 Get Ready Question 9 Page 55 a) The measure of ∠D is the same as the measure of ∠C , or 60º. b) EF = AB EF = 5 ED AC 3 6 5×3 EF = 6 EF = 2.5 The length of side EF is 2.5 cm. Chapter 2 Get Ready Question 10 Page 55 If P is any point on the right bisector of line segment AB and Q is the point of intersection of AB and the right bisector, then AQ = QB and ∠ PQA = ∠ PQB = 90°. Side PQ is common to ΔPQA and ΔPQB. Therefore, ΔPQA is congruent to ΔPQB (side-angle-side). PA and PB are corresponding sides, so PA = PB. 10 MHR • Principles of Mathematics 10 Solutions Chapter 2 Section 1: Midpoint of a Line Segment Chapter 2 Section 1 Question 1 Page 66 a) The midpoint of AB is at (4, 6). b) The midpoint of CD is at (1, 3). c) The midpoint of EF is at (2, 2). MHR • Principles of Mathematics 10 Solutions 11 ⎛ 1 ⎞ d) The midpoint of GH is at ⎜ − , −2 ⎟ . 2 ⎝ ⎠ 12 MHR • Principles of Mathematics 10 Solutions Chapter 2 Section 1 Question 2 Page 66 ⎛ x + x y + y2 ⎞ a) ( x, y ) = ⎜ 1 2 , 1 2 ⎟⎠ ⎝ 2 ⎛5+3 7+9⎞ , =⎜ 2 ⎟⎠ ⎝ 2 ⎛ 8 16 ⎞ =⎜ , ⎟ ⎝2 2 ⎠ = ( 4,8 ) The coordinates of the midpoint of J(5, 7) and K(3, 9) are (4, 8). ⎛ x + x y + y2 ⎞ b) ( x, y ) = ⎜ 1 2 , 1 2 ⎟⎠ ⎝ 2 ⎛ −1 + 1 0 + ( −6 ) ⎞ , = ⎜⎜ ⎟⎟ 2 ⎝ 2 ⎠ ⎛ 0 −6 ⎞ =⎜ , ⎟ ⎝2 2 ⎠ = ( 0, −3) The coordinates of the midpoint of L(–1, 0) and M(1, –6) are (0, –3). ⎛ x + x y + y2 ⎞ c) ( x, y ) = ⎜ 1 2 , 1 2 ⎟⎠ ⎝ 2 ⎛ −2 + ( −2 ) −4 + 8 ⎞ , = ⎜⎜ ⎟ 2 2 ⎟⎠ ⎝ ⎛ −4 4 ⎞ =⎜ , ⎟ ⎝ 2 2⎠ = ( −2, 2 ) The coordinates of the midpoint of N(–2, –4) and P(–2, 8) are (–2, 2). ⎛ x + x y + y2 ⎞ d) ( x, y ) = ⎜ 1 2 , 1 2 ⎟⎠ ⎝ 2 ⎛ −3 + ( −1) −3 + ( −7 ) ⎞ , = ⎜⎜ ⎟⎟ 2 2 ⎝ ⎠ ⎛ −4 −10 ⎞ =⎜ , ⎟ ⎝ 2 2 ⎠ = ( −2, −5 ) The coordinates of the midpoint of Q(–3, –3) and R(–1, –7) are (–2, –5). MHR • Principles of Mathematics 10 Solutions 13 Chapter 2 Section 1 Question 3 Page 66 ⎛ x + x y + y2 ⎞ a) ( x, y ) = ⎜ 1 2 , 1 2 ⎟⎠ ⎝ 2 ⎛ 0.2 + 3.6 1.5 + 0.2 ⎞ =⎜ , ⎟ 2 2 ⎝ ⎠ ⎛ 3.8 1.7 ⎞ =⎜ , ⎟ ⎝ 2 2 ⎠ = (1.9,0.85 ) The coordinates of the midpoint of J(0.2, 1.5) and K(3.6, 0.2) are (1.9, 0.85). ⎛ x + x y + y2 ⎞ b) ( x, y ) = ⎜ 1 2 , 1 2 ⎟⎠ ⎝ 2 ⎛ −1.4 + 0.6 −3.2 + ( −5.3) ⎞ , = ⎜⎜ ⎟⎟ 2 2 ⎝ ⎠ ⎛ −0.8 −8.5 ⎞ , =⎜ ⎟ 2 ⎠ ⎝ 2 = ( −0.4, −4.25 ) The coordinates of the midpoint of N(–1.4, –3.2) and P(0.6, –5.3) are (–0.4, –4.25). c) x1 + x2 y1 + y2 ⎞ , 2 ⎟⎠ ⎝ 2 ⎛ 1 3 5 ⎛ 5⎞⎞ +⎜− ⎟ ⎟ ⎜ + 2 ⎝ 2⎠⎟ =⎜ 2 2 , 2 ⎜ 2 ⎟ ⎜ ⎟ ⎝ ⎠ ⎛2 0⎞ =⎜ , ⎟ ⎝2 2⎠ = (1,0 ) ( x, y ) = ⎛⎜ ⎛1 5⎞ ⎛ 3 5⎞ The coordinates of the midpoint of L ⎜ , ⎟ and M ⎜ , − ⎟ are (1, 0). 2 2 ⎝ ⎠ ⎝2 2⎠ 14 MHR • Principles of Mathematics 10 Solutions d) x1 + x2 y1 + y2 ⎞ , 2 ⎟⎠ ⎝ 2 ⎛ 3 1 ⎛ 7⎞⎞ ⎜ − + 2 +⎜− ⎟ ⎟ 8 ⎝ 8⎠⎟ , =⎜ 8 2 ⎜ 2 ⎟ ⎜ ⎟ ⎝ ⎠ ⎛ 13 6 ⎞ − ⎟ ⎜ =⎜ 8 , 8 ⎟ ⎜⎜ 2 2 ⎟⎟ ⎝ ⎠ ⎛ 13 3 ⎞ = ⎜ ,− ⎟ ⎝ 16 8 ⎠ ( x, y ) = ⎛⎜ 7⎞ ⎛ 13 3 ⎞ ⎛ 3 1⎞ ⎛ The coordinates of the midpoint of Q ⎜ − , ⎟ and R ⎜ 2, − ⎟ are ⎜ , − ⎟ . 8⎠ ⎝ 8 8⎠ ⎝ ⎝ 16 8 ⎠ MHR • Principles of Mathematics 10 Solutions 15 Chapter 2 Section 1 Question 4 Page 66 a) Find the coordinates of the midpoint of AB. x1 + x2 y1 + y2 ⎞ , 2 ⎟⎠ ⎝ 2 ⎛ 2+6 5+9⎞ =⎜ , 2 ⎟⎠ ⎝ 2 ⎛ 8 14 ⎞ =⎜ , ⎟ ⎝2 2 ⎠ = ( 4,7 ) ( x, y ) = ⎛⎜ Find the slope of the median. m= y2 − y1 x2 − x1 2−7 8−4 −5 = 4 5 =− 4 = 5 The slope of the median is − . 4 16 MHR • Principles of Mathematics 10 Solutions b) Find the coordinates of the midpoint of FG. x1 + x2 y1 + y2 ⎞ , 2 ⎟⎠ ⎝ 2 ⎛ 9 + 15 −4 + 14 ⎞ =⎜ , 2 ⎟⎠ ⎝ 2 ⎛ 24 10 ⎞ =⎜ , ⎟ ⎝ 2 2⎠ = (12, 5 ) ( x, y ) = ⎛⎜ Find the slope of the median. m= = = y2 − y1 x2 − x1 5 − ( −7 ) 12 − ( −5 ) 12 17 The slope of the median is Chapter 2 Section 1 12 . 17 Question 5 Page 67 x1 + x2 y1 + y2 ⎞ , 2 ⎟⎠ ⎝ 2 ⎛ 23.6 + 79.4 38.0 + 43.8 ⎞ =⎜ , ⎟ 2 2 ⎝ ⎠ ( x, y ) = ⎛⎜ ⎛ 103 81.8 ⎞ =⎜ , ⎟ 2 ⎠ ⎝ 2 = ( 51.5, 40.9 ) The checkpoint should be at (51.5, 40.9). MHR • Principles of Mathematics 10 Solutions 17 Chapter 2 Section 1 Question 6 Page 67 x1 + x2 y1 + y2 ⎞ , 2 ⎟⎠ ⎝ 2 ⎛ −7 + ( −1) −4 + 10 ⎞ = ⎜⎜ , ⎟ 2 2 ⎟⎠ ⎝ ⎛ −8 6 ⎞ =⎜ , ⎟ ⎝ 2 2⎠ = ( −4,3) ( x, y ) = ⎛⎜ The centre of the circle is at (–4, 3). Chapter 2 Section 1 Question 7 Page 67 Answers may vary. For example: Using The Geometer’s Sketchpad®: Plot the endpoints, and construct the line segment between them. Construct the midpoint of this line segment. Then, select the midpoint and choose Coordinates from the Measure menu. Using Cabri® Jr.: Choose Point from the F2 menu to plot the endpoints. Choose Coord. & Eq. from the F5 menu, and check the placement of the endpoints. Adjust the endpoints if necessary. Choose Segment from the F2 menu, and construct the line segment between the endpoints. Choose Midpoint from the F3 menu, and construct the midpoint. Then, choose Coord. & Eq. again to display the coordinates of the midpoint. 18 MHR • Principles of Mathematics 10 Solutions Chapter 2 Section 1 Question 8 Page 67 Find the coordinates of the midpoint, M, of BC. x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 ⎛ −6 + 2 2 + 0 ⎞ =⎜ , 2 ⎟⎠ ⎝ 2 ⎛ −4 2 ⎞ =⎜ , ⎟ ⎝ 2 2⎠ = ( −2,1) Find the slope of the median. y − y1 mAM = 2 x2 − x1 = 4 −1 4 − ( −2 ) 3 6 1 = 2 = Substitute m = y = mx + b 1 1 = ( −2 ) + b 2 1 = −1 + b 2=b 1 and the coordinates of one endpoint, say (–2, 1), to find b. 2 The equation of the median from vertex A is y = 1 x+2. 2 MHR • Principles of Mathematics 10 Solutions 19 Chapter 2 Section 1 Question 9 Page 67 Answers may vary. For example: Using The Geometer’s Sketchpad®: Plot the vertices of ΔABC, and construct the midpoint, M, of side BC. Construct a line through AM. Select the line, and choose Equation from the Measure menu. Using Cabri® Jr.: Choose Point from the F2 menu, and plot the vertices of ΔABC. Choose Coord. & Eq. from the F5 menu, and check the placement of the vertices. Adjust the vertices if necessary. Choose Segment from the F2 menu, and construct the line segment between vertices B and C. Select this line segment and choose Midpoint from the F3 menu. Choose Line from the F2 menu, and construct the line through the midpoint and vertex A. Then, choose Coord. & Eq. again to display the equation of the line. 20 MHR • Principles of Mathematics 10 Solutions Chapter 2 Section 1 Question 10 Page 67 a) Find the coordinates of the midpoint, M, of QR. x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 ⎛ 4 + 5 6 + ( −3) ⎞ = ⎜⎜ , ⎟⎟ 2 2 ⎝ ⎠ 9 3 ⎛ ⎞ =⎜ , ⎟ 2 2 ⎝ ⎠ Find the slope of the median. y − y1 mPM = 2 x2 − x1 = 3 −0 2 9 − ( −2 ) 2 3 = 2 13 2 3 = 13 Substitute m = y = mx + b 3 and the coordinates of one endpoint, say (–2, 0), to find b. 13 3 ( −2 ) + b 13 6 0= − +b 13 6 =b 13 0= The equation of the median from vertex P is y = 3 6 x+ . 13 13 MHR • Principles of Mathematics 10 Solutions 21 b) Find the coordinates of the midpoint, N, of PR. x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 ⎛ −2 + 5 0 + ( −3) ⎞ = ⎜⎜ , ⎟⎟ 2 ⎝ 2 ⎠ ⎛3 3⎞ = ⎜ ,− ⎟ ⎝2 2⎠ Find the slope of the median. y − y1 mQN = 2 x2 − x1 3 − −6 = 2 3 −4 2 15 − = 2 5 − 2 =3 Substitute m = 3 and the coordinates of one endpoint, say (4, 6), to find b. y = mx + b 6 = 3( 4) + b 6 = 12 + b −6 = b The equation of the median from vertex Q is y = 3 x − 6 . 22 MHR • Principles of Mathematics 10 Solutions Chapter 2 Section 1 Question 11 Page 67 Answers may vary. For example: Using The Geometer’s Sketchpad®: a) Plot the vertices of ΔPQR. Construct the midpoint, S, of side QR. Construct a line through points P and S. Select the line, and choose Equation from the Measure menu. b) Construct the midpoint, T, of side PR, and the line though points Q and T. Select the line, and choose Equation from the Measure menu. Using Cabri® Jr.: a) Choose Point from the F2 menu, and plot the vertices of ΔPQR. Choose Coord. & Eq. from the F5 menu, and check the placement of the vertices. Adjust the vertices if necessary. Choose Segment from the F2 menu, and construct the line segment between vertices Q and R. Select this line segment, and choose Midpoint from the F3 menu. Choose Line from the F2 menu, and construct the line through the midpoint and vertex P. Then, choose Coord. & Eq. again to display the equation of the line. b) Use the method in part a) to construct the midpoint T of side PR and the line through points Q and T. Then, choose Coord. & Eq. from the F5 menu to display the equation of the line. Chapter 2 Section 1 Question 12 Page 67 x1 + x2 y1 + y2 ⎞ , 2 ⎟⎠ ⎝ 2 ⎛ a + 3a b + 2b ⎞ =⎜ , ⎟ 2 ⎠ ⎝ 2 ( x, y ) = ⎛⎜ 3b ⎞ ⎛ = ⎜ 2a , ⎟ 2 ⎠ ⎝ These coordinates are the mean of the x-coordinates of the endpoints and the mean of the y-coordinates of the endpoints. MHR • Principles of Mathematics 10 Solutions 23 Chapter 2 Section 1 Question 13 Page 67 a) x1 + x2 y1 + y2 ⎞ , 2 ⎟⎠ ⎝ 2 ( x, y ) = ⎛⎜ x1 + 6 y1 + 5 ⎞ , 2 ⎟⎠ ⎝ 2 Set up equations to solve for the coordinates. x1 + 6 y1 + 5 =4 =2 2 2 x1 + 6 = 8 y1 + 5 = 4 ( 4, 2 ) = ⎛⎜ x1 = 2 y1 = −1 The coordinates of the other endpoint are D(2, –1). b) Answers may vary. For example: Refer to part a). Let the coordinates of the other endpoint be D(x1, y1). Solving the equations gives x1 = 2 and y1 = –1. Alternative method: Since the run from C to M is –2, subtract 2 from the x-coordinate of M to find the x-coordinate of D. Since the rise from C to M is –3, subtract 3 from the y-coordinate of M to find the y-coordinate of D. c) Answers may vary. Substitute the coordinates of points C and D into the midpoint formula to confirm that M is the midpoint of CD. Chapter 2 Section 1 Question 14 Page 67 The midpoint of the diameter is at (0, 0). Let the coordinates of the other endpoint be (x1, y1). x1 + x2 y1 + y2 ⎞ , 2 ⎟⎠ ⎝ 2 ( x, y ) = ⎛⎜ ⎛ x1 + ( −3) y1 + 4 ⎞ , ⎟ 2 2 ⎠ ⎝ ( 0, 0 ) = ⎜ Set up equations to solve for the coordinates. x1 − 3 y1 + 4 =0 =0 2 2 x1 − 3 = 0 y1 + 4 = 0 x1 = 3 y1 = −4 The coordinates of the other endpoint are (3, –4). 24 MHR • Principles of Mathematics 10 Solutions Chapter 2 Section 1 Question 15 Page 67 a) Case 1: The centre is D, and the endpoint is E. x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 ⎛ x1 + ( −1) y1 + 2 ⎞ , ⎟ 2 2 ⎠ ⎝ ( 2, 4 ) = ⎜ Set up equations to solve for the coordinates. x1 − 1 y1 + 2 =2 =4 2 2 x1 − 1 = 4 y1 + 2 = 8 x1 = 5 y1 = 6 Possible coordinates for the endpoint are (5, 6). Case 2: The centre is E, and the endpoint is D. x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 x +2 y +4 ( −1, 2 ) = ⎛⎜ 1 , 1 ⎞⎟ 2 ⎠ ⎝ 2 Set up equations to solve for the coordinates. x1 + 2 y1 + 4 = −1 =2 2 2 x1 + 2 = −2 y1 + 4 = 4 x1 = −4 y1 = 0 Possible coordinates for the endpoint are (–4, 0). b) There are two possible answers for part a) because either D or E could be the centre of the circle. MHR • Principles of Mathematics 10 Solutions 25 Chapter 2 Section 1 Question 16 Page 67 Find the coordinates of the midpoint of the line segment PQ. x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 ⎛ −5 + 3 −2 + 6 ⎞ =⎜ , 2 ⎟⎠ ⎝ 2 = ( −1, 2 ) Find the slope of the line through P and Q. y − y1 m= 2 x2 − x1 = 6 − ( −2 ) 3 − ( −5) 8 8 =1 = The slope of a line perpendicular to the line segment PQ is –1. Substitute m = 1 and the coordinates of the midpoint to find b. y = mx + b 2 = −1( −1) + b 2 = 1+ b 1= b The equation of the right bisector of PQ is y = − x + 1 . 26 MHR • Principles of Mathematics 10 Solutions Chapter 2 Section 1 Question 17 Page 67 a) Answers may vary. For example: Any point on the right bisector of a line segment is equidistant from the endpoints. Therefore, points on the right bisector of the line segment joining the two towns are possible locations for the relay tower. b) Find the midpoint of the line segment. x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 ⎛ 2 + 10 6 + 0 ⎞ =⎜ , 2 ⎟⎠ ⎝ 2 = ( 6,3) Find the slope of the line segment. y − y1 m= 2 x2 − x1 0−6 10 − 2 −6 = 8 3 =− 4 = The slope of a line perpendicular to the line segment is Substitute m = y = mx + b 4 3 = ( 6) + b 3 3= 8+b −5 = b 4 . 3 4 and the coordinates of the midpoint to find b. 3 The equation of the right bisector is y = 4 x −5. 3 MHR • Principles of Mathematics 10 Solutions 27 Chapter 2 Section 1 Question 18 Page 67 Answers may vary. For example: Using The Geometer’s Sketchpad®: Plot the points A(2, 6) and B(10, 0). Construct the line segment AB and the midpoint of AB. Then, construct a perpendicular line through the midpoint. Select the perpendicular line, and choose Equation from the Measure menu. Using Cabri® Jr.: Choose Segment from the F2 menu, and plot the endpoints at points (2, 6) and (10, 0). Use Coord. & Eq. from the F5 menu to check the placement of the endpoints, and adjust them if necessary. Select the line segment, and choose Midpoint from the F3 menu. Choose Perp. from the F3 menu, and construct the perpendicular line through the midpoint. Then, choose Coord. & Eq. again to display the equation of the line. Chapter 2 Section 1 Question 19 Page 67 a) 28 MHR • Principles of Mathematics 10 Solutions b) Find the coordinates of the midpoint, M, of BC. x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 ⎛ 4 + 8 8 + ( −2 ) ⎞ , = ⎜⎜ ⎟⎟ 2 ⎝ 2 ⎠ = ( 6,3) Find the slope of the median. y − y1 mAM = 2 x2 − x1 3− 0 6+2 3 = 8 = Substitute m = y = mx + b 3 and the coordinates of one endpoint, say (6, 3), to find b. 8 3 (6) + b 8 9 3= +b 4 3 =b 4 3= 3 3 The equation of the median from vertex A is y = x + . 8 4 MHR • Principles of Mathematics 10 Solutions 29 c) Find the slope of the line segment BC. y − y1 m= 2 x2 − x1 = 8 − ( −2 ) 8−4 10 = 4 5 = 2 2 The slope of a line perpendicular to BC is − . 5 Substitute m = − y = mx + b 2 and the coordinates of the midpoint to find b. 5 2 ( 6) + b 5 12 3= − +b 5 27 =b 5 3= − 2 27 The equation of the right bisector of BC is y = − x + . 5 5 d) Answers may vary. For example: Check that the slopes and y-intercepts on the drawing match those in the equations. 30 MHR • Principles of Mathematics 10 Solutions Chapter 2 Section 1 a), b), e) Question 20 c) Since U is the midpoint of PR, RU = UP = Page 68 1 PR. Since ST joins the midpoints of two sides of 2 1 PR. Therefore, ST = RU = UP. Similarly, UT = PS = SQ and RT = TQ = US. 2 Therefore, ΔRUT ≅ ΔUPS ≅ ΔSTU ≅ ΔTSQ (side-side-side). ΔPQR, ST = d) The area of ΔSTU is 1 the area of ΔPQR. 4 f) The area of one of the smallest triangles is 1 1 the area of ΔSTU and the area of ΔPQR. 4 16 MHR • Principles of Mathematics 10 Solutions 31 Chapter 2 Section 1 Question 21 Page 68 b) Answers may vary. For example: Join the midpoints of the sides of an equilateral triangle to form four equilateral triangles inside the original triangle. Shade the centre triangle. For each of the other three triangles, repeat the process of joining the midpoints to form smaller similar triangles, and shade the centre triangle. The procedure works with any triangle. The area relationships are the same as shown in question 20 since the line segment joining the midpoints of two sides of any triangle is half the length of the third side. c) d) Answers may vary. For example: Sierpinski’s triangle is a fractal since all of the smaller triangles in each step are similar to the original triangle. Chapter 2 Section 1 ED is Question 22 Page 68 1 of BC. If ED = 2, then BC is 8 × 2, or 16 units. 8 Chapter 2 Section 1 Question 23 Page 68 a) The run is 11 – 2 = 9. The rise is 19 – 1 = 18. The required points are (5, 7) and (8, 13). See part b) for the explanation. b) Answers may vary. For example: 1 1 of the run to the x-coordinate of the first endpoint and add For the first dividing point, add 3 3 2 of the rise to the y-coordinate of the first endpoint. For the second dividing point, add of the 3 2 run to the x-coordinate of the first endpoint and add of the rise to the y-coordinate of the first 3 endpoint. 32 MHR • Principles of Mathematics 10 Solutions Chapter 2 Section 1 Question 24 Page 68 a) Use grid paper or dynamic geometry software to plot the given midpoints. A line segment joining the midpoints of two sides of a triangle is parallel to the third side. Use this property to draw the sides of the triangle. b) Midpoint of AB: x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 ⎛ −1 + 1 −2 + 6 ⎞ =⎜ , 2 ⎟⎠ ⎝ 2 = ( 0, 2 ) Midpoint of BC: x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 ⎛1+ 3 6 + 2 ⎞ =⎜ , 2 ⎟⎠ ⎝ 2 = ( 2, 4 ) Midpoint of AC: x +x y +y ( x, y ) = ⎛⎜ 1 2 , 1 2 ⎞⎟ 2 ⎠ ⎝ 2 ⎛ −1 + 3 −2 + 2 ⎞ =⎜ , 2 ⎟⎠ ⎝ 2 = (1,0 ) The midpoint formula correctly predicts the coordinates of the midpoints. MHR • Principles of Mathematics 10 Solutions 33 Chapter 2...
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