CBSE Solutions for Class 10 English

જવાબ : 2

જવાબ : (– 4, 2)                                            

જવાબ : Rectangle  

જવાબ : 3  

જવાબ : 8  

જવાબ : √34                                                 

જવાબ : – 12

જવાબ : (x, y)

જવાબ : (–6, 5/2)

જવાબ : ± 4  

જવાબ : 8

જવાબ : IV quadrant

જવાબ : (5, –3) 

જવાબ : (0, – 10) and (4, 0)  

જવાબ : 1:5

જવાબ : True

જવાબ : True

જવાબ : False

જવાબ : False

જવાબ : True

જવાબ : False

જવાબ : 2

જવાબ : (-3/2, 5)

જવાબ : 0

જવાબ : 5

જવાબ : 5

જવાબ : (125, 275)  

જવાબ : First quadrant

જવાબ : (3,-10)

જવાબ : False

જવાબ : False

જવાબ : True

જવાબ : True

જવાબ : True

જવાબ : False

જવાબ : True

જવાબ : (3, -1)

જવાબ : 1

જવાબ : For these points to be collinear
A=0
Or
12[7(1−a)+5(a+2)+3(−2−1)]=012[7(1−a)+5(a+2)+3(−2−1)]=0
7-7a+5a+10-9=0
a=2

જવાબ : Let the ratio be m: n
Now
Coordinate of the intersection
x=3m+3nm+nx=3m+3nm+n
y=7m−2nm+ny=7m−2nm+n
Now these points should lie of the line, So
2(3m+2nm+n)+(7m−nm+n)−4=02(3m+2nm+n)+(7m−nm+n)−4=0
m:n=2:9

જવાબ : Y axis  

જવાબ : Y’ axis

જવાબ : IV quadrant

જવાબ : III quadrant

જવાબ : I quadrant  

જવાબ : X axis  

જવાબ : III quadrant

જવાબ : II quadrant

જવાબ : Y’ axis

જવાબ : II quadrant

જવાબ : PQ = 10 …Given in question
PQ² = 10² = 100 … [Squaring
(9 – x)² + (10 – 4)² = 100… (using the distance formula
(9 – x)² + 36 = 100
(9 – x)² = 100 – 36 = 64
(9 – x) = ± 8 …[Taking square-root
9 – x = 8 or 9 – x = -8
9 – 8 = x or 9+ 8 = x
x = 1 or x = 17

જવાબ : AB = 10 units … [Given in the question
AB² = 10² = 100 … [Squaring
(11 – 3)² + (y + 1)² = 100
8² + (y + 1)² = 100
(y + 1)² = 100 – 64 = 36
y + 1 = ±6 … [Taking square-root
y = -1 ± 6 ∴ y = -7 or 5

જવાબ : PA = QA …[Given in the question
PA² = QA² … [Squaring  
(3 – 6)² + (y – 5)² = (3 – 0)² + (y + 3)²
9 + (y – 5)² = 9 + (y + 3)²
(y – 5)² = (y + 3)²
y – 5 = ±(y + 3) … [Taking square root  
y – 5 = y + 3 y – 5 = -y – 3
0 = 8 … which is not possible ∴ y = 1

જવાબ : Let P(2, 4), A(5, k) and B(k, 7).

PA = PB …[Given in the question
PA² = PB² … [Squaring
(5 – 2)² + (k – 4)² = (k – 2)² + (7 – 4)²
9 + (k – 4)² – (k – 2)² = 9
(k – 4 + k – 2) (k – 4 – k + 2) = 0
(2k – 6)(-2) = 0
2k – 6 = 0
2k = 6 ∴ k = 3

જવાબ : PA = PB …Given in the question
PA² = PB² … [Squaring
⇒ (k – 1 – 3)² + (2 – k)² = (k – 1 – k)² + (2 – 5)²
⇒ (k – 4)² + (2 – k)² = (-1)² + (-3)²
k² – 8k + 16 + 4 + k² – 4k = 1 + 9
2k² – 12k + 20 – 10 = 0
2k² – 12k + 10 = 0
⇒ k² – 6k + 5 = 0 …[Dividing by 2
⇒ k² – 5k – k + 5 = 0
⇒ k(k – 5) – 1(k – 5) = 0
⇒ (k – 5) (k – 1) = 0
⇒ k – 5 = 0 or k – 1 = 0
∴ k = 5 or k = 1

જવાબ :
Let P(0, y) be any point on y-axis.
PA = PB … [Given in the question
PA² = PB² … [Squaring
⇒ (4 – 0)² + (8 – y)² = (-6 – 0)² + (6 – y)²
16 + 64 – 16y + y² = 36 + 36 + y² – 12y
80 – 72 = -12y + 16y
8 = 4y ⇒ y = 2
∴ Point P (0, 2).
Now, AP = (4-0)² +(8-2)²
∴ Distance, AP = 52

જવાબ :

જવાબ : PA = PB … [Given in the question
PA² = PB² … [Squaring
⇒ [(a + b) – x]² + [(b a) – y)² = [(a – b) – x]² + [(a + b) – y]²
⇒ (a + b)² + x² – 2(a + b)x + (b – a)² + y² – 2(b – a)y = (a – b)² + x² – 2(a – b)x + (a + b)² + y² – 2(a + b)y …[∵ (a – b) ² = (b – a)²
⇒ -2(a + b)x + 2(a – b)x = -2(a + b)y + 2(b – a)y
⇒ 2x(-a – b + a – b) = 2y(-a – b + b – a)
⇒ -2bx = – 2ay
⇒ bx = ay

જવાબ : AB = AC … [Given in the question
∴ AB² = AC² …[Squaring
⇒ (3 – 0)² + (p – 2)²= (p – 0)² + (5 – 2)²
⇒ 9+ (p – 2)² = p² + 9
⇒ p² – 4p + 4 – p² = 0
⇒ -4p + 4 = 0
⇒ -4p = -4 ⇒ p = 1

જવાબ : PA = PB … [Given in the question
PA² = PB² … [Squaring
⇒ (2 + 2)² + (2 – k)² = (2 + 2k)² + (2 + 3)²
16 + 4 + k² – 4k = 4 + 4k² + 8k + 25
4k² + 8k + 25 – k² + 4k – 16 = 0
3k² + 12k + 9 = 0
⇒ k² + 4k + 3 = 0 …[Dividing by 3
k² + 3k + k + 3 = 0
k(k + 3) + 10k + 3) = 0
(k + 1) (k + 3) = 0
k + 1 = 0 or k + 3 = 0
 k = -1 or k = -3
AP² = (2+2)² + (2-k)²
When k = -1
AP² = (2+2)² + (2+1)² = 25

જવાબ :

જવાબ : Join OA and OB. …[radii of a circle
∴ OA = OB OA² = OB² …[Squaring both sides
⇒ (2 + 1)² + (-3y – y)² = (5 – 2)² + (7 + 3y)²
⇒ 9+ (-4y)² = 9 + (7 + 3y)²
⇒ 16y² = 49 + 42y + 9y²
⇒ 16y² – 9y² – 42y – 49 = 0
⇒ 7y² – 42y – 49 = 0
⇒ y² – 6y – 7 = 0 …[Dividing by 7
⇒ y² – 7y + y – 7 = 0
⇒ y(y – 7) + 1(y – 7) = 0
⇒ (y – 7) (y + 1) = 0
y – 7 = 0 or y + 1 = 0
y = 7 or y = -1
(i) Taking y = 7

જવાબ : Given in the question : A(2, 3), B(-2, 2), C(-1, -2), D(3, -1)

AB = BC = CD = DA …[All four sides are equal
AC = BD …[ diagonals are equal
=> ABCD is a Square.

જવાબ : Let p (-2, 3), q(8,3), r(6, 7).
(pq)² = (8 + 2)² + (3 – 3)² = 10² + 0² = 100
(qr)² = (6 – 8)² + (7 – 3)² = (-2)² + 4² = 20
(pr)² = (6 + 2)² + (7 – 3)² = 8² + 4² = 80
Now, (qr)² + (pr)²= 20 + 80 = 100 = (pq)²
…[By converse of Pythagoras’ theorem
Therefore, Points p, q, r are the vertices of a right triangle.

જવાબ :
=> Diagonals also bisect each other.

જવાબ : A(-2, 1), B(a, b) and C(4, -1) are collinear…..{given in the question
∴ x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0
⇒ -2[b – (-1)] + a(-1 – 1) + 4(1 – b) = 0
⇒ -2b – 2 – 2a + 4 – 4b = 0
⇒ -2a – 6b = -2
⇒ a + 3b = 1 … [Dividing by (-2)
⇒ a = 1 – 3 …..(A)
We have, a – b = 1 …[Given in the question
(1 – 3b) – b = 1 …[From (A)
1 – 3b – b = 1
-4b = 1 – 1 = 0 ∴ b = 0/-4 = 0
From (A), a = 1 – 3(0) = 1 – 0 = 1
∴ a = 1, b = 0

જવાબ : A(-1, -4), B(b, c), C(5, -1) are collinear     ……Given in the question
∴ x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0
⇒ -1(c + 1) + b(-1 + 4) + 5(-4 – c) = 0
⇒ -c – 1 – b + 4b – 20 – 5c = 0
⇒ 3b – 6c = 21
⇒ – b – 2c = 7 …[Dividing by 3
⇒ b = 7 + 2c   ……(A)
We have, 2b + c = 4 … [Given in the question
2(7 + 2c) + c = 4 … [From (A)
⇒ 14 + 4c + c = 4
⇒ 5c = 4 – 14 = -10 ⇒ c = -2
⇒ b = 7 + 2(-2) = 3 … [From (A)
∴ b = 3, c = -2

જવાબ : Given

જવાબ : P(x, y), R(z, t).

જવાબ :

1-C, 2-D, 3-A, 4-B

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-D, 2-A, 3-B, 4-C

જવાબ :

1-B, 2-A, 3-D, 4-C

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-B, 2-A, 3-D, 4-C

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-B, 2-A, 3-D, 4-C

જવાબ :

1-C, 2-D, 3-A, 4-B

જવાબ :

1-B, 2-A, 3-D,4-D