Download 2y + 4x – 6 = 0 2y = -4x + 6 Y = -2x + 3 b

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