GSEB Solutions for ધોરણ ૧૦ English

જવાબ : Hence, coordinates of point P are (16, 8).

જવાબ : Let Q divide AB in the ratio of p : 1

જવાબ : Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) be the vertices of a parallelogram ABCD.
Since, the diagonals of a parallelogram bisect each other. Hence, x = 6 and y = 3.

જવાબ : Let the point on y-axis be P(0, y) and AP : PB = k : 1

જવાબ :

જવાબ : Let A (5,-2), B (6, 4) and C (7, -2) be the vertices of a triangle

Here, AB = BC
∴ ∆ABC is an isosceles triangle.

જવાબ : The point P (0, 2) lies on y-axis

AP ≠ BP
∴ P(0, 2) does not lie on the perpendicular bisector of AB. So, given statement is false.

જવાબ : Points of trisection of line segment AB are given by

∴ Given statement is true.

જવાબ :

જવાબ : Let the required ratio be λ : 1

Given that this point lies on the x-axis

Thus, the required ratio is 2 : 1.

જવાબ :

જવાબ : Given, A(2, 3)= (x₁, y₁) B(4, k) = (x₂, y₂) C(6, -3) = (x₃, y₃) If the given points are collinear, the area of the triangle formed by them will be 0. ½ [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0 ½ [2(k + 3) + 4(-3 -3) + 6(3 – k)] = 0 ½ [2k + 6 – 24 + 18 – 6k] = 0 ½ (-4k) = 0 4k = 0 k = 0

જવાબ : Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5). Then, AP = BP AP² = BP² Using distance formula, (x – 7)² + (y – 1)² = (x – 3)² + (y – 5)² x² – 14x + 49 + y² – 2y + 1 = x² – 6x + 9 + y² – 10y + 25 x – y = 2 Hence, the relation between x and y is x – y = 2.

જવાબ : As we know, a₂ – a₁ = a₃ – a₂
2k – 1 – (k + 9) = 2k + 7 – (2k – 1)
2k – 1 – k – 9 = 2k + 7 – 2k + 1
k – 10 = 8 ∴ k = 8 + 10 = 18

જવાબ : When the points are collinear,
x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0
x(-4 – (-5)) + (-3)(-5 – 2) + 7(2 – (-4)) = 0
x(1) + 21 + 42 = 0
x + 63 = 0 ∴ x = -63

જવાબ : B(-3, 0), A (-3, 4)

જવાબ :

જવાબ : AB = 4 units
BC = 3 units
AC² = AB² + BC² …[Pythagoras’ theorem
= (4)² + (3)²
= 16 + 9 = 25
∴ AC = 5 cm

જવાબ : Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7).
∴ AP = BP …[Given
AP² = BP² …[Squaring both sides
(x – 2)² + (y – 5)² = (x + 3)² + (y – 7)²
⇒ x² – 4x + 4 + y² – 10y + 25
⇒ x² + 6x + 9 + y² – 14y + 49
⇒ -4x – 10y – 6x + 14y = 9 +49 – 4 – 25
⇒ -10x + 4y = 29
∴ 10x + 29 = 4y is the required relation.

જવાબ : When points are collinear,
∴ Area of ∆ABC = 0
= (x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)) = 0
= x (7 – 5) – 5 (5 – y) -4 (y – 7) = 0
= 2x – 25 + 5y – 4y + 28 = 0
∴ 2x + y + 3 = 0 is the required relation.

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જવાબ :

જવાબ :

 

જવાબ : Let A(-1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other.
∴ Coordinates of the mid-point of AC = Coordinates of the mid-point of BD
Hence, coordinates of the fourth vertex, D(-2, 1).

જવાબ :
We know that at y-axis coordinates of points are (0, y) and at x-axis (x, 0).
Let P(0, b) and Q(a,0)
Mid point of PQ = (2, -5) …[Given

જવાબ : Suppose the line 3x + y – 9 = 0 divides the line segment joining A(1, 3) and B(2, 7) in the ratio k : 1 at point C.

⇒ 6k + 3 + 7k + 3 – 9k – 9 = 0
⇒ 4k – 3 = 0
⇒ 4k = 3 ∴ k = 8
So, the required ratio is 3 : 4 internally.

જવાબ : We have A(4, 3), B(-1, y) and C(3, 4). In right angled triangle ABC,
(BC)² = (AB)² + (AC)…. [Pythagoras theorem
⇒ (-1 – 3)² + (y – 4)² = (4 + 1)² + (3 – y)² + (4 – 3)² + (3 – 4)² …(using distance formula
⇒ (-4)² + (y² – 8y + 16)
⇒ (5)² + (9 – 6y + y²) + (1)² + (-1)²
⇒ y² – 8y + 32 = y² – 6y + 36 = 0
⇒ -8y + 6y + 32 – 36
⇒ -2y – 4 = 0 ⇒ -2y = 4
∴ y = -2

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

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

જવાબ : PA = QA …[Given
PA² = QA² … [Squaring both sides
(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 sq. root of both sides
y – 5 = y + 3 y – 5 = -y – 3
0 = 8 … which is not possible ∴ y = 1

જવાબ : PA = PB …Given
PA² = PB² … [Squaring both sides
⇒ (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
⇒ 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 : PA = PB … [Given
PA² = PB² … [Squaring both sides
⇒ [(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 (Hence proved)

જવાબ : AB = AC … [Given
∴ AB² = AC² …[Squaring both sides
⇒ (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 : Let A (-2, 3), B(8,3), C(6, 7).
(AB)² = (8 + 2)² + (3 – 3)² = 10² + 0² = 100
(BC)² = (6 – 8)² + (7 – 3)² = (-2)² + 4² = 20
(AC)² = (6 + 2)² + (7 – 3)² = 8² + 4² = 80
Now, (BC)² + (AC)²= 20 + 80 = 100 = (AB)²
…[By converse of Pythagoras’ theorem
Therefore, Points A, B, C are the vertices of a right triangle.

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જવાબ :

જવાબ : The sum of the lengths of any two line segments is equal to the length of the third line segment then all three points are collinear. Consider, A = (1, 5) B = (2, 3) and C = (-2, -11) Find the distance between points; say AB, BC and CA Since AB + BC ≠ CA Therefore, the points (1, 5), (2, 3), and ( – 2, – 11) are not collinear.

જવાબ : Since two sides of any isosceles triangle are equal. To check whether given points are vertices of an isosceles triangle, we will find the distance between all the points. Let the points (5, – 2), (6, 4), and (7, – 2) are representing the vertices A, B, and C respectively. This implies, whether given points are vertices of an isosceles triangle.

જવાબ : Given: Distance between (2, – 3) and (10, y) is 10. Using distance formula, Simplify the above equation and find the value of y. Squaring both sides, 64 +(y+3)² = 100 (y+3)² = 36 y + 3 = ±6 y + 3 = +6 or y + 3 = −6 Therefore, y = 3 or -9.

જવાબ : Area of ∆OAB = 1/2 [0(0 – 1) – 3(0 – 0) + 5(0 – 0)] = 0
⇒ Given points are collinear

જવાબ :

જવાબ : Since (3, a) lies on the line 2x – 3y = 5
Then 2(3) – 3(a) = 5
– 3a = 5 – 6
– 3a = -1
⇒ a = 1/3

જવાબ :

જવાબ :

જવાબ : Here x₁ = -6, y₁ = 8
x₂ = 0, y₂ = 0
 

જવાબ : Points A, B and C are collinear
⇒ 1(0 – b) + 0 (b – 2) + a(2 – 0) = 0
⇒ -b + 2a = 0 or 2a = b

જવાબ : Using distance formula

જવાબ : In Fig. 6.6, let the point P(-1, 6) divides the line joining A(-3, 10) and B (6, -8) in the ratio k : 1

Hence, the point P divides AB in the ratio 2 : 7.

જવાબ :

જવાબ : Let the required point be (x, 0).
Since, (x, 0) is equidistant from the points (-3, 4) and (2, 5) .

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જવાબ : Given: 

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

જવાબ : Let A (7, 10), B(-2, 5), C(3, -4) be the vertices of a triangle.

… [By converse of Pythagoras theorem
∆ABC is an isosceles right angled triangle. …(ii) From (i) & (ii), Points A, B, C are the vertices of an isosceles right triangle.

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જવાબ : (i) Let A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) be the four given points.
Then, using distance formula, we have,

(ii) Let A (4, 5), B (7, 6), C (4, 3) and D (1, 2) be the given points. Then,

∴ ABCD is a parallelogram.

જવાબ :

જવાબ :

 

Since AB = BC = AC = 3 units
∴ Co-ordinates of B are (5, 0)
Let co-ordinates of C be (x, y)
AC² = BC²
[∵ ∆ABC is an equilateral triangle]
Using distance formula
⇒ (x – 2)² + (y – 0)² = (x – 5)² + (y – 0)²
⇒ x²+ 4 – 4x + y = x²+ 25 – 10x + y² 6x = 21
⇒ x = 216 = 72
Again (x – 2)² + (y – 0)² = 9

જવાબ : Let A (x₁, y₁) = (0, -1), B (x₂, y₂) = (2, 1), C (x₃, y₃) = (0, 3) be the vertices of ∆ABC.
Now, let P, Q, R be the mid-points of BC, CA and AB, respectively.
So, coordinates of P, Q, R are

જવાબ :

જવાબ :

જવાબ : Given: AD is the median on BC.
⇒ BD = DC
The coordinates of midpoint D are given by.

Hence, AD divides ∆ABC into two equal areas.

જવાબ : Given points are A(2, 4), P(3, 8) and Q(-10, y)
According to the question,

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જવાબ : ∵ O is the mid-point of the base BC.
∴ Coordinates of point B are (0, 3). So,
BC = 6 units Let the coordinates of point A be (x, 0).
Using distance formula,

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જવાબ : Let O(x, y) be the centre of circle. Given points are A(6, -6), B(3, -7) and C(3, 3).

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જવાબ : Let P(x₁, y₁) be a common point of both lines and divide the line segment joining A(2, -2) and B(3, 7) in ratio k : 1.

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