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.

OR

Show that ∆ABC with vertices A(-2, 0), B(0, 2) and C(2, 0) is similar to ∆DEF with vertices D(-4, 0), F(4,0) and E(0, 4).

[∆PQR is replaced by ∆DEF]

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)= (x_{1}, y_{1})
B(4, k) = (x_{2}, y_{2})
C(6, -3) = (x_{3}, y_{3})
If the given points are collinear, the area of the triangle formed by them will be 0.
½ [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 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
AP^{2} = BP^{2}
Using distance formula,
(x – 7)^{2} + (y – 1)^{2} = (x – 3)^{2} + (y – 5)^{2}
x^{2} – 14x + 49 + y^{2} – 2y + 1 = x^{2} – 6x + 9 + y^{2} – 10y + 25
x – y = 2
Hence, the relation between x and y is x – y = 2.

જવાબ : As we know, a_{2} – a_{1} = a_{3} – a_{2}

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,

x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}) = 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

AC^{2} = AB^{2} + BC^{2} …[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

AP^{2} = BP^{2} …[Squaring both sides

(x – 2)^{2} + (y – 5)^{2} = (x + 3)^{2} + (y – 7)^{2}

⇒ x^{2} – 4x + 4 + y^{2} – 10y + 25

⇒ x^{2} + 6x + 9 + y^{2} – 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

= (x_{1} (y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})) = 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.

જવાબ :

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} + (y^{2} – 8y + 16)

⇒ (5)^{2} + (9 – 6y + y^{2}) + (1)^{2} + (-1)^{2}

⇒ y^{2} – 8y + 32 = y^{2} – 6y + 36 = 0

⇒ -8y + 6y + 32 – 36

⇒ -2y – 4 = 0 ⇒ -2y = 4

∴ y = -2

જવાબ : PQ = 10 …Given

PQ^{2} = 10^{2} = 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

AB^{2} = 10^{2} = 100 … [Squaring both sides

(11 – 3)^{2} + (y + 1)^{2} = 100

8^{2} + (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

PA^{2} = QA^{2} … [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

PA^{2} = PB^{2} … [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}

k^{2} – 8k + 16 + 4 + k^{2} – 4k = 1 + 9

2k^{2} – 12k + 20 – 10 = 0

2k^{2} – 12k + 10 = 0

⇒ k^{2} – 6k + 5 = 0 …[Dividing by

⇒ k^{2} – 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

PA^{2} = PB^{2} … [Squaring both sides

⇒ [(a + b) – x]^{2} + [(b a) – y)^{2} = [(a – b) – x]^{2} + [(a + b) – y]^{2}

⇒ (a + b)^{2} + x^{2} – 2(a + b)x + (b – a)^{2} + y^{2} – 2(b – a)y = (a – b)^{2} + x^{2} – 2(a – b)x + (a + b)^{2} + y^{2} – 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

∴ AB^{2} = AC^{2} …[Squaring both sides

⇒ (3 – 0)^{2} + (p – 2)^{2}= (p – 0)^{2} + (5 – 2)^{2}

⇒ 9+ (p – 2)^{2} = p^{2} + 9

⇒ p^{2} – 4p + 4 – p^{2} = 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} = 10^{2} + 0^{2} = 100

(BC)^{2} = (6 – 8)^{2} + (7 – 3)^{2} = (-2)^{2} + 4^{2} = 20

(AC)^{2} = (6 + 2)^{2} + (7 – 3)^{2} = 8^{2} + 4^{2} = 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.

જવાબ :

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 x_{1} = -6, y_{1} = 8

x_{2} = 0, y_{2} = 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) .

જવાબ :

જવાબ :

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.

જવાબ :

ABC is a right angled triangle,∴ AC

Using distance formula,

AB

= 25

BC

= 16 + (t + 2)

AC

= 49 + (2 – t)

Putting values of AB

49 + (2 – t)

∴ 49 + (2 – t)

⇒ (t + 2)

⇒ (t

8t = 8 ⇒ t = 1

જવાબ :

જવાબ :

(i) (-5, 7), (-1, 3)

(ii) (a, b), (-a, -b)

જવાબ :

your answer: (i) (-1, -2), (1, 0), (- 1, 2), (-3, 0) (ii) (4, 5), (7, 6), (4, 3), (1, 2)

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,

∴ Quadrilateral ABCD is a square. (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 :

∴ Co-ordinates of B are (5, 0)

Let co-ordinates of C be (x, y)

AC

[∵ ∆ABC is an equilateral triangle]

Using distance formula

⇒ (x – 2)

⇒ x

⇒ x = 216 = 72

Again (x – 2)

જવાબ : Let A (x_{1}, y_{1}) = (0, -1), B (x_{2}, y_{2}) = (2, 1), C (x_{3}, y_{3}) = (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

Ratio of ar (∆PQR) to the ar (∆ABC) = 1 : 4.

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,

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,

Since BACD is a rhombus.

∴ AB = AC = CD = DB

∴ Coordinates of point D = (-3√3, 0).

જવાબ :

જવાબ :

જવાબ :

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

જવાબ :

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

D(4, 0) E (4, 0) and F (0, 4).

Solution:

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ :

gseb maths textbook std 10

- Math Book for GSEB ધોરણ ૧૦
- Chemistry Book for GSEB ધોરણ ૧૦
- Biology Book for GSEB ધોરણ ૧૦
- Physics Book for GSEB ધોરણ ૧૦
- History Book for GSEB ધોરણ ૧૦
- Geography Book for GSEB ધોરણ ૧૦
- Economics Book for GSEB ધોરણ ૧૦
- Political Science Book for GSEB ધોરણ ૧૦

The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.