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Chapter 4 Exercises page 49 1a- 2y + 4x – 6 = 0 2y = -4x + 6 Y = -2x + 3 b- 3y – 2x + 5 = 2y + x + 7 3y – 2y = 3x + x + 7 – 5 Y = 4x + 12 c- 6y – 5(x – 2) = 2(3y – x) + 1 6y – 5x + 10 = 6y – 2x + 1 6y – 6y = 5x – 2x + 1 – 10 0 = 3x – 9 -3x = -9 x=3 d- 3(y – 2) = 2(x – 3) 3y – 6 = 2x – 6 3y = 2x – 6 + 6 Y = 2/3 x e- x = 5y + 10 5y = x – 10 y = 1/5 x – 2 f- 2(3x – 4y) + 5 = 3(2y + 3x + 2) 6x – 8y + 5 = 6y + 9x + 6 -8y – 6y = 9x – 6x + 6 – 5 -14y = 3x + 1 y = -3/14 x – 1/14 2- A belongs to (D) if its coordinates verify its equation a- A(2, 11) and (D): y = 5x + 1 5xA + 1 = 5(2) + 1 = 11 = yA So, A belongs to (D) b- A(-1, 5) and (D): y - 5x = 10 yA – 5xA = 5 – 5(-1) = 5 + 5 = 10 true So, A belongs to (D) c- A(-8, 2) and (D): y = 2 yA = 2 true So, A belongs to (D) d- A(√3, 2) and (D): y = √3x - 1 √3xA - 1 = √3(√3) - 1 = 3 –1 = 2 = yA So, A belongs to (D) e- A(1, 2) and (D): x = 2 xA = 1 ≠ 2 So, A doesn’t belong to (D) f- A(-1, 1) and (D): y = 2x – 10y – 2 = 0 2xA – 10yA – 2 = 2(-1) -10(1) – 2 = -14 ≠ 0 So, A doesn’t belong to (D) 3a- y – 2x = 0 → y = 2x a=2 b=0 line passes through the origin since it is of the form y = ax b- 3x – y – 4 = 0 → y = 3x – 4 a=3 b = -4 c- y = -4 a=0 b = -4 line is parallel to (x’x) since it is of the form y = b d- 0 = 5x – 4 → x = 4/5 This line has no slope and no y-intercept Line is parallel to (y’y) since it is of the form x = c e- 2y = -x + 3 → y = -1/2x + 3/2 a = -1/2 b = 3/2 f- 2y = 4x → y = 2x a=2 b=0 line passes through the origin since it is of the form y = ax g- 2y – 1 – 4x = 0 → y = 2x + 1/2 a=2 b=½ h- 2x + y – 6 = 0 → y = -2x + 6 a = -2 b=6 i- 3y – 6 = 0 → y = 2 a=0 b=2 line parallel to (x’x) j- 3(2x – y + 1) = 4x + y 6x – 3y + 3 = 4x + y 4y = 2x + 3 y = 1/2x + 3/4 a = 1/2 b = 3/4 4a- a = -2 b=3 equation of line is of the form y = ax + b → y = -2x + 3 b- a = 3 b=4 y = 3x + 4 c- a = 2 passing through point (1, 0) equation of form y = ax+ b → y = 2x + b Substitute coordinates of point → 0 = 2(1) + b → b = -2 So: y = 2x – 2 d- a = -0.5 passing through point (0, 1.5) equation of form y = ax+ b → y = -0.5x + b Substitute coordinates of point → 1.5 = -0.5(0) + b → b = 1.5 So: y = -0.5x + 1.5 5a- (D) passes through the origin → its equation is of the form y = ax A(-1, 0.4) belongs to (D), then its coordinates verify the equation of (D) → yA = axA 0.4 = a(-1) → a = -0.4 Then (D): y = -0.4x b- B(2, -5) B belongs to (D) if its coordinates verify the equation of (D) -0.4xB = -0.4(2) = -0.8 ≠ yB Then B doesn’t belong to (D) C(1, -0.4) C belongs to (D) if its coordinates verify the equation of (D) -0.4xC = -0.4(1) = -0.4 = yC Then C belongs to (D) D(-2.5, 0.5) D belongs to (D) if its coordinates verify the equation of (D) -0.4xD = -0.4(-2.5) = 1 ≠ yD Then D doesn’t belong to (D) E(2.5, -1) E belongs to (D) if its coordinates verify the equation of (D) -0.4xE = -0.4(2.5) = -1 = yE Then E belongs to (D) F(-1, 0.4) F belongs to (D) if its coordinates verify the equation of (D) -0.4xF = -0.4(-1) = 0.4 = yF Then F belongs to (D) G(5, -2) G belongs to (D) if its coordinates verify the equation of (D) -0.4xG = -0.4(5) = -2 = yG Then G belongs to (D) 6a- 3y + x – 3 = 0 x = 0 → 3y + 0 – 3 = 0 → 3y = 3 → y = 1 y = 0 → 3(0) + x – 3 = 0 → x = 3 x = -3 → 3y – 3 – 3 = 0 → y = 2 y = 3 → 3(3) + x – 3 = 0 → x = -6 x = 1/3 → 3y + 1/3 – 3 = 0 → 3y = 8/3 → y = 8/9 y = 2/3 → 3(2/3) + x – 3 = 0 → x = 1 X Y 0 1 3 0 -3 2 -6 3 b- 7- A belongs to (D), then its coordinates verify the equation of (D) a- A(-3, 0) yA = -2xA + m → 0 = -2(-3) + m →m = -6 b- A(3, -2) yA = -2xA + m → -2 = -2(3) + m → m = 4 c- A(-1, 3) yA = -2xA + m → 3 = -2(-1) + m → m = 1 d- A(1, -1) yA = -2xA + m → -1 = -2(1) + m → m = 1 1/3 8/9 1 2/3 8- A belongs to (D) so its coordinates verify the equation of (D) a- A(-1, 3) yA = axA – 1 → 3 = a(-1) – 1 → a = -4 b- A(0.5, 0) yA = axA – 1 → 0 = a(0.5) – 1 → a = 2 c- A(3, 2) yA = axA – 1 → 2 = a(3) – 1 → a = 1 d- A(1, -2) yA = axA – 1 → -2 = a(1) – 1 → a = -1 9a- (D) // (x’x) → its equation is of the form y = b A(1, 2) belongs to (D) → yA = b → 2 = b Therefore, (D): y = 2 b- (D) // (y’y) → its equation is of the form x = c A(1, 2) belongs to (D) → xA = c → 1 = c Therefore, (D): x = 1 c- (D) passes through the origin → its equation is of the form y = ax A(1, 2) belongs to (D) → yA = axA → 2 = a(1) → a = 2 Therefore, (D): y = 2x d- a = -1 → (D): y = ax + b → y = -x + b A(1, 2) belongs to (D) → yA = -xA + b → b = 3 Therefore, (D): y = -x + 3 10a- (D): y = 2x + 3 X Y b- (D): y = 3 0 3 -1 1 c- (D): y = -2x X Y 0 0 1 -2 d- (D): x = -1.5y + 2 1.5y = - x + 2 Y = -2/3x + 4/3 X Y 0 4/3 -1 2 e- (D): 2y + 4x – 5 = 0 2y = - 4x +5 Y = -2x + 5/2 X Y 0 5/2 1 1/2 f- (D): x/2 + y/2 = 3 → ( x 2) → x + y = 6 X Y 2 4 3 3 g- (D): 2(3y – x) + 6 = 3(2y + 2x) – 3 6y – 2x + 6 = 6y + 6x – 3 6y – 6y – 2x – 6x = -3 – 6 -8x = -9 → x = 9/8 h- (D): 2y + 4x + 6 = 3(y – 2x + 3) 2y + 4x + 6 = 3y – 6x + 9 2y – 3y = - 6x – 4x + 9 – 6 -y = -10x + 3 y = 10x- 3 X Y 0 -3 1/2 2 11a- A(0, 1) and m = 2 (D): y = mx + b → y = 2x + b A belongs to (D) → yA = 2xA + b 1 = 2(0) + b → b = 1 Therefore (D): y = 2x + 1 b- A(2, 1) and m = 0 (D): y = mx + b → y = 0x + b = b A belongs to (D) → yA = b → b = 1 Therefore (D): y = 1 c- A(-1, -2) and m = 0.5 (D): y = mx + b → y = 0.5x + b A belongs to (D) → yA = 0.5xA + b -2 = 0.5(-1) + b → b = -1.5 Therefore (D): y = 0.5x – 1.5 d- A(3, 5) and m = -2.5 (D): y = mx + b → y = -2.5x + b A belongs to (D) → yA = -2.5xA + b 5 = -2.5(3) + b → b = 12.5 Therefore (D): y = -2.5x + 12.5 e- A(1, -1) and m is not defined → (D) // (y’y) → (D): x = c A belongs to (D) → xA = c → c = 1 Therefore (D): x = 1 12a- (D): y = -1.5x + 1 X Y 0 1 2 -2 b- A(0, y) belongs to (D) → yA = -1.5xA + 1 = -1.5(0) + 1 = 1 → A(0, 1) B(x, -1) belongs to (D) → yB = -1.5xB + 1 -1 = -1.5xB + 1 → xB = 4/3 → B(4/3, -1) C(-2, y) belongs to (D) → yC = -1.5xC + 1 = -1.5(-2) + 1 = 4 → C(-2, 4) D(1, y) belongs to (D) → yD = -1.5xD + 1 = -1.5(1) + 1 = -0.5 → D(1, -0.5) E(-1, y) belongs to (D) → yE = -1.5xE + 1 = -1.5(-1) + 1 = 2.5 → E(-1, 2.5) F(10, y) belongs to (D) → yF = -1.5xF + 1 = -1.5(10) + 1 = -14 → F(10, -14) 13A(2, 1) and B(-3, 6) (AB): y = ax + b a(AB) = (yB – yA) / (xB – xA) = (6-1) / (-3 -2) = -1 A belongs to (D) → yA = -xA + b 1 = -2 + b → b = 3 Therefore (D): y = -x + 3 14a- a(u) = a(v) = 2 → (u) // (v) b- a(u) = 1 and a(v) = -1 a(u) x a(v) = (1)(-1) = -1 therefore (u) perpendicular to (v) c- (u): y = -3 is parallel to (x’x) (v): 0 = 2x + 2 → x = -1 is parallel to (y’y) Therefore (u) perpendicular to (v) d- (u): y = 3x – 1 → a(u) = 3 (v): -3y = 3x + 2 → y = - x – 2/3 → a(v) = -1 Then (u) and (v) are neither parallel nor equal e- (u): 0 = 4x + 3 → 4x = -3 → x = -3/4 (v): 0 = 8x – 2 → 8x = 2 → x = ¼ (u) and (v) are both parallel to (y’y) Therefore (u) and (v) are parallel 15- each unit on (t’t) = 1 Each unit on (q’q) = 20 16a- On (u’u): 1 unit → 1000 On (v’v): 1 unit → 1 b- I unit of u corresponds to v = 2 V = 2 → v = 2 x 1000 = 2000 Chapter 4 Problems page 52 2a- A(4, 0) B(0, 8) C(2, 4) b- R.T.P : C midpoint of [AB] (xA + xB) / 2 = (4 + 0) / 2 = 2 = xC (yA + yB) / 2 = (0 + 8) / 2 = 4 = yC Therefore, C is the midpoint of [AB] c- A belongs to (x’x) B belongs to (y’y) Then (AO) perpendicular to (BO) So, triangle AOB is right at O Then, [AB] is the diameter of the circle circumscribed about triangle AOB So, C(2, 4) is the center of the circle. Radius AC = √((xA – xC)2 + (yA – yC)2) = √(4 + 16) = √20 = 2√5 d- (OC) passes through the origin, then equation of form y = ax C belongs to (OC) then, yC = axC 4 = 2a then a = 2 So: (OC): y = 2x e- (d) perpendicular (OC) then a(d) x a(OC) = -1 So a(d) = - 1/2 Equation of (d) is of the form y = ax + b B belongs to (d) so: yB = axB + b 8 = -1/2(0) + b then b = 8 Therefore, (d): y = -1/2x + 8 3a- (u): y = 2x + 4 X Y 0 4 1 6 1 6 2 3 (v): y = -3x + 9 X Y b- ABCD rectangle AB = 2x + 4 BC = 9 – 3x AB = ?? if x = 0.5 AB = 2(0.5) + 4 = 5 X = ?? if BC = 3 3 = 9 – 3x so x = 6/3 = 2 c- ABCD becomes a square if AB = BC ( 2 consecutive sides equal) So: 2x + 4 = 9 – 3x 2x + 3x = 9 – 4 x=1 4a- m = (5x + 2y) / 7 = ( 5(9.5) + 2(12)) / 7 = 10.214 b- m = (5x +2y) / 7 → 7 x 10 = 5(9) + 2y 70 – 45 = 2y → y = 12.5 c- 12 = (5x + 2y) / 7 → 12 x 7 = 5x + 2y 5x + 2y = 84 → y = -5/2 x + 42 X Y 10 17 12 12 5a- (u): y = x since (u) is passing through the origin (0, -2) belongs to y = x – 2 → (0, -2) belongs to (v) → (v): y = x – 2 (0, 1) belongs to (s) → y = -0.5x + 1 1 = 1 true → (s): y = -0.5x + 1 (d): y = 2x + 1 b- (u): y = x → a(u) = 1 (v): y = x – 2 → a(v) = 1 → (u) // (v) (s): y = -0.5x + 1 (d): y = 2x + 1 a(s) x a(d) = 2(-0.5) = -1 → (s) is perpendicular to (d) 6a- (D): y = -3/4x + 6 X Y 0 6 4 3 b- M(2, yM) belongs to (D) so its coordinates verify its equation yM = -3/4xM + 6 = -3/4(2) + 6 = -6/4 + 6 = 9/2 then M(2, 9/2) c- B(8, yB) belongs to (D) So yB = -3/4xB + 6 = -3/4 (8) + 6 = 0 Then B(8, 0) d- C(xC, 6) belongs to (D) So yC = -3/4xC + 6 6 = -3/4xC + 6 then: xC = 0 Then C(0, 6) e- Since yB = 0, then B belongs to (x’x) Since xC = 0, then C belongs to (y’y) S0 (AB) perpendicular to (AC) Then triangle ABC is right at A AB = √((xA – xB)2 + (yA – yB)2) = √64 = 8 AC = √((xA – xc)2 + (yA – yc)2) = √36 = 6 BC = √((xB – xC)2 + (yB – yC)2) = √100 = 10 f- [AH] altitude in triangle ABC so: (AH) perpendicular to (BC) So: a(AH) x a(BC) = -1 but a(BC) = a(D) = -3/4 then a(AH) = 4/3 equation of (AH) is of the form y = ax since it passes through the origin so: (AH): y = 4/3 x 7a- b- A belongs to (D) if its coordinates verify the equation of (D) 2xA – 1 = 2(2) – 1 = 3 = yA true So, A belongs to (D) c- (D): y = 2x – 1 X Y 0 -1 1 1 d- I midpoint of [AB] xI = (xA + xB) / 2 = (2+0)/2 = 1 yI = (yA + yB) / 2 = (3+5)/2 = 4 then I(1, 4) e- H is the orthogonal projection of A on (y’y) xH = 0 and xA = 2 so AH = 2 yH= yA = 3 and yB = 5 BH = yB – yH = 5 – 3 = 2 ABH is a right triangle at H Using Pythagoras theorem, AB2 = AH2 + BH2 = 22 + 22 = 8 So, AB = √8 = 2√2 f- (u) perpendicular to (D) So a(u) x a(D) = -1 a(u) x 2 = -1 a(u) = - 1/2 g- (u) perpendicular to (D) B belongs to (u) Equation of (u) is of the form y = ax + b a(u) = - 1/2 (proved) b = y-intercept = yB = 5 therefore (u): y = -1/2x + 5 8a- r = 23x d = 2600 + 3x b- profit = income – expenses 2000 = 23x – (2600 + 3x) 2000 + 2600 = 23x – 3x → x = 230 kg c- (D): y = 23x X Y 0 0 20 460 (D’): y = 3x + 2600 X Y 0 2600 2600 2660 9a- b- (AB): y = ax + b a(AB) = (yB – yA) / (xB – xA) = (6 – 3) / (1 + 2) = 1 → y=x+b A belongs to (AB) → yA = xA + b → 3 = -2 + b → b = 5 Then (AB): y = x + 5 c- y = -x + 1 X Y 0 1 1 0 a(AB) x a(D) = 1 x (-1)= -1 → (AB) perpendicular to (D) d- L belongs to (D) if its coordinates verify its equation yL = 101 -xL + 1 = -100 + 1 = -99 -99 ≠ 101 So L doesn’t belong to (D) e- (l): y = x + 4 M belongs to (l) 10a- b- (AB): y = ax + b a(AB) = (yB – yA) / (xB – xA) = (3 – 6) / (-3 -2) = 3/5 A belongs to (AB) → yA = 3/5 xA + b 6 = 3/5(2) + b 6 – 6/5 = b → b = 24/5 → (AB): y = 3/5 x + 24/5 → (x 5) → 5y – 3x = 24 (BC): y = ax + b a(BC) = (yC – yB) / (xC – xB) = (0 – 3) / (2 + 3) = -3/5 → y = -3/5x + b B belongs to (BC) → yB = -3/5 xB + b 3 = -3/5(-3) + b → b= 6/5 → (BC): y = -3/5 x + 6/5 → (x 5) → 5y + 3x = 6 (CD): y = ax + b a(CD) = (yD – yC) / (xD – xC) = (3 – 0) / (7 -2) = 3/5 C belongs to (CD) → yC = 3/5 xC + b 0 = 3/5(2) + b 0 – 6/5 = b → b = - 6/5 → (CD): y = 3/5 x - 6/5 → (x 5) → 3x – 5y = 6 (AD): y = ax + b a(AD) = (yD – yA) / (xD – xA) = (3 – 6) / (7 -2) = - 3/5 A belongs to (AD) → yA = - 3/5 xA + b 6 = - 3/5(2) + b 6 + 6/5 = b → b = 36/5 → (AD): y = - 3/5 x + 36/5 → (x 5) → 5y + 3x = 36 c- a(AB) = a(CD) = 3/5 → (AB) // (CD) a(AD) = a(BC) = - 3/5 → (AD) // (BC) → ABCD is a parallelogram (opposite sides are parallel) d- using the graph: [BH] = [DK] = 3 [CH] = [CK] = 5 Angles CHB = CKD = 90 Then CHB and CKD are congruent [CD] = [CB] (triangles CHB and CKD are congruent) So ABCD is a parallelogram with 2 consecutive sides equal Therefore ABCD is a rhombus. e- M midpoint of [AC] (diagonals bisect each other) → xM= (xA + xC) / 2 = (2 + 2) / 2 = 2 yM = (yA + yC) / 2 = (6 + 0) / 2 = 3 then M(2, 3) f- Triangle MBC g- ABC is the symmetric of ADC with axis of symmetry AC 11A- a-expenses = 90 x 20 = 1800 F b-expenses = 520 + 90 x 20 = 2320 F B- a-T = 20x b-cost = 20x – 20% = 20x – 20/100(20x) = 20x – 4x = 16x c-B = cost + installation = 16x + 520 C- (d): y = 20x X Y 0 0 10 200 0 520 10 680 (d’): y = 16x + 520 X Y D- From the graph, (d) is below (d’) → moquette-all is better than nice carpets 12a- b- (AB): y = ax + b a(AB) = (yB – yA) / (xB – xA) = (-2 + 4) / (4 + 1) = 2/5 = 0.4 A belongs to (AB) → yA = 0.4 xA + b -4 = 0.4(-1)+ b → b = - 3.6 Therefore (AB): y = 0.4 x – 3.6 c- (u): y = ax + b = -5/2x + b B belongs to (u) → yB = -5/2xB + b -2 = -5/2(4) + b → b = 8 Therefore, (u): y = -5/2x + 8 d- (u): y = -5/2x + 8 a(u) x a(AB) = -5/2 x 0.4 = -1 → (u) perpendicular to (AB) e- (u) passes through C if its coordinates verify its equation -5/2 xC + 8 = -5/2(2) + 8 = 3 = yC true So, C belongs to (u) f- (v) // (AB) → a(v) = a(AB) = 0.4 C belongs to (v) → yC = 0.4xC + b 3 = 0.4(2) + b b = 2.2 therefore, (v): y = 0.4x + 2.2 g- Vectors BA = CD (by translation) → xA –xB = xD – xC -1 -4 = xD – 2 → xD = -3 → yA – yB = yD – yC -4 + 2 = yD – 3 → yD = 1 Therefore: D(-3, 1) D belongs to (v) since (CD) // (AB) and (v) // (AB) (given) Verification: 0.4xD + 2.2 = 0.4(-3) + 2.2 = 1 = yD Therefore, D belongs to (v) 13a- (d1): (d2): (d3): (d4): (d5): y = 2x + 3 y=6 x=6 y = -1/3x + 3 y = 2x – 3 b- (-1.5, 0) belongs to (d1) from the graph And the y-intercept of (d1) is b = 3 So, its equation is either y = 2x + 3 or y = 3x+ 3 Substitute (-1.5, 0) is the first 2(-1.5) + 3 = -3 + 3 = 0 true Therefore, (d1): y = 2x + 3 14a- b- (OA): y = ax (passes through the origin) A belongs to (OA) → yA = axA -2 = a(-4) → a = -2/-4 = 1/2 Therefore: (OA): y = 1/2x A’ belongs to (OA) ?? yA’ = 1 1/2xA’ = 1/2(2) = 1 → yA’ = 1/2xA’ Then A’ belongs to (OA) Therefore, A, O and A’ are collinear. c- (OB): y = ax (passes through the origin) B belongs to (OB) → yB = axB 6 = a(-3) → a = 6/-3 = -2 Therefore: (OB): y = -2x B’ belongs to (OB) ?? YB’ = -3 -2xB’ = -2(1.5) = -3 → yB’ = -2xB’ Then B’ belongs to (OB) Therefore, B, O and B’ are collinear. d- A’ midpoint [BC] ?? (xB + xC) / 2 = (-3 + 7) / 2 = 4/2 = 2 = xA’ (yB + yC) / 2 = (6 – 4) / 2 = 2/2 = 1 = yA’ Then A’ midpoint of [BC] B’ midpoint [AC] ?? (xA + xC) / 2 = (-4 + 7) / 2 = 3/2 = 1.5 = xB’ (yA + yC) / 2 = (-2 – 4) / 2 = -6/2 = -3 = yB’ Then B’ midpoint of [AC] e- A’ midpoint of [BC] (proved) → (AA’) median B’ midpoint of [AC] (proved) → (BB’) median (AA’) and (BB’) intersect at O which becomes centroid (OC) passes through O → (OC) is the third median in triangle ABC f- Triangle ABO is right OA = √((xA – xO)2 + (yA – yO)2 = √((-4 -0)2 + (-2 -0)2 = √20 = 2√5 OB = √((xB – xO)2 + (yB – yO)2 = √((-3 -0)2 + (6 -0)2 = √45 = 3√5 AB = √((xB – xA)2 + (yB – yA)2 = √((-3 + 4)2 + (6 +2)2 = √65 OA2 + OB2 = (2√5)2 + (3√5)2 = 65 AB2 = (√65)2 = 65 → OA2 + OB2 = AB2 Therefore OAB is right at O (inverse of Pythagoras theorem) g- Using Pythagoras theorem: In triangle AOH: OA2 = OH2 + HA2 = 22+ 42 = 20 → OA = √20 =2√5 In triangle OBK: OB2 = OK2 + KB2 = 62+ 32 = 45 → OB = √45 =3√5 OAB is a right triangle → AB2 = OA2 + OB2 = (2√5)2 + (3√5)2 = 20 + 45 = 65 → AB = √65 h- Area(OAB) = (b x h) / 2 = (OA x OB) / 2 = (2√5 x 3√5) / 2 = 30/2 = 15cm2