LOADING . . .

CBSE Solutions for Class 10 English

GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is _________

Hide | Show

જવાબ : 2


The mid-point of the line segment joining the points A (–2, 8) and B (– 6, – 4) is ______

Hide | Show

જવાબ : (– 4, 2)                                            


The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a ____________

Hide | Show

જવાબ : Rectangle  


The distance of the point P (2, 3) from the x-axis is______

Hide | Show

જવાબ : 3  


The distance between the points A (0, 6) and B (0, –2) is ______

Hide | Show

જવાબ : 8  


AOBC is a rectangle whose three vertices are vertices A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is _______

Hide | Show

જવાબ : √34                                                 


If P (a/3, 4) is the mid-point of the line segment joining the points Q (– 6, 5) and R (– 2, 3), then the value of a is ______

Hide | Show

જવાબ : – 12


The coordinates of the point which is equidistant from the three vertices of the Δ AOB as shown in the figure is ________

CBSE Class 10 Maths MCQs Chapter 7 Coordinate Geometry

Hide | Show

જવાબ : (x, y)


A circle drawn with origin as the centre passes through The point which does not lie in the interior of the circle is ________

Hide | Show

જવાબ : (–6, 5/2)


If the distance between the points (4, p) and (1, 0) is 5, then the value of p is ______

Hide | Show

જવાબ : ± 4  


The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is _____

 

Hide | Show

જવાબ : 8


The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2 internally lies in the ___________

Hide | Show

જવાબ : IV quadrant


One of the two points of trisection of the line segment joining the points A (7, – 2) and B (1, – 5) which divides the line in the ratio 1:2 are _________

Hide | Show

જવાબ : (5, –3) 


A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, –5) is the mid - point of PQ, then the coordinates of P and Q are ____________

Hide | Show

જવાબ : (0, – 10) and (4, 0)  


The ratio in which the point P (3/4, 5/12)divides the line segment joining the Points A (1/2, 3/2) and B (2, –5) is _________

Hide | Show

જવાબ : 1:5


State whether True or False - ΔABC with vertices (– 2, 0), (2, 0) and (0, 2) is similar to ΔDEF with vertices (-4, 0), (4, 0) and (0, 4).

Hide | Show

જવાબ : True


State whether True or False - The point (-4, 2) lies on the line segment joining the points (-4, 6) and (-4, -6).Solution. 

Hide | Show

જવાબ : True


State whether True or False - The points (0, 5), (0, -9) and (3, 6) are collinear.

Hide | Show

જવાબ : False


State whether True or False - Point (0, 2) is the point of intersection of Y-axis and perpendicular bisector of line segment joining the points (‒1, 1) and (3, 3).

Hide | Show

જવાબ : False


State whether True or False - The points A (3, 1), (12, -2) and (0, 2) cannot be vertices of a triangle.

Hide | Show

જવાબ : True


State whether True or False - The points A (4, 3), (6, 4), (5, - 6) and (-3, 5) are vertices of a parallelogram.

Hide | Show

જવાબ : False


Calculate the Distance between the point  (1,3) and ( 2,4)

 

Hide | Show

જવાબ : 2


Calculate the Mid-point of line segment  AB where A(2,5) and B( -5,5)

Hide | Show

જવાબ : (-3/2, 5)


Calculate the Area of  the triangle  formed by joining the line segments (0,0)  ,( 2,0) and (3,0)

 

Hide | Show

જવાબ : 0


Calculate the Distance of point (5,0) from Origin

 

Hide | Show

જવાબ : 5


Calculate the Distance of point (5,-5) from Origin

 

Hide | Show

જવાબ : 5


Calculate the Coordinate of the point M which divided the line segment A(2,3) and B( 5,6) in the ratio 2:3

Hide | Show

જવાબ : (125, 275)  


Calculate the Quadrant of the Mid-point of the line segment A(2,3) and B( 5,6)

Hide | Show

જવાબ : First quadrant


Calculate the coordinates of a point A, where AB is the diameter of circle whose center is (2,−3) and B is (1, 4)

Hide | Show

જવાબ : (3,-10)


State whether true or false :-  Point A( 0,0) B( 0,3) ,C( 0,7) and D( 2,0) formed a quadrilateral

Hide | Show

જવાબ : False


State whether true or  false :- The point P (–2, 4) lies on a circle of radius 6 and center C (3, 5)

Hide | Show

જવાબ : False


State whether true or  false :-  Triangle PQR with vertices P (–2, 0), Q (2, 0) and R (0, 2) is similar to Δ XYZ with
Vertices X (–4, 0) Y (4, 0) and Z (0, 4).

Hide | Show

જવાબ : True


State whether true or false :-   The triangle formed by joining the point A( -3,0) , B( 0,0) and C( 0,2) is a right angle triangle

Hide | Show

જવાબ : True


State whether true or false :-  Point X (2, 2) Y (0, 0) and Z (3, 0) are not collinear

Hide | Show

જવાબ : True


State whether true or false :-  A circle has its center at the origin and a point A (5, 0) lies on it. The point B (6, 8) lies inside the circle

Hide | Show

જવાબ : False


State whether true or false :-   The points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) in that order form a rectangle.

Hide | Show

જવાબ : True


the centroid of the triangle XYZ whose vertices are X (3, - 5) Y (- 3, 4) and Z (9, - 2) is ______

Hide | Show

જવાબ : (3, -1)


The area of the triangle ABC with coordinates as A (1, 2) B (2, 5) and C (- 2, - 5) is_______

Hide | Show

જવાબ : 1


Find the value of  a for which these point are collinear  (7,-2) , (5,1) ,(3,a)?

Hide | Show

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


Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7).

Hide | Show

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


In which axis or quadrant do the following coordinates lie:-  (0, 3)

Hide | Show

જવાબ : Y axis  


In which axis or quadrant do the following coordinates lie:- (0, -3)

 

Hide | Show

જવાબ : Y’ axis


In which axis or quadrant do the following coordinates lie :- (4, -3)

Hide | Show

જવાબ : IV quadrant


In which axis or quadrant do the following coordinates lie :- (-4, -3)

Hide | Show

જવાબ : III quadrant


In which axis or quadrant do the following coordinates lie :-  (4, 3)

Hide | Show

જવાબ : I quadrant  


In which axis or quadrant do the following coordinates lie :-  (4, 0)

Hide | Show

જવાબ : X axis  


In which axis or quadrant do the following coordinates lie :- (-4, -3)

Hide | Show

જવાબ : III quadrant


In which axis or quadrant do the following coordinates lie :- (-4, 3)

Hide | Show

જવાબ : II quadrant


In which axis or quadrant do the following coordinates lie :- (0, -5)

Hide | Show

જવાબ : Y’ axis


In which axis or quadrant do the following coordinates lie :- (-1, 5)

Hide | Show

જવાબ : II quadrant


Find that value(s) of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units. (2011D)

Hide | Show

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


Find the value of y for which the distance between the points A (3,-1) and B (11, y) is 10 units. (2011OD)

Hide | Show

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


The point A(3, y) is equidistant from the points P(6, 5) and Q(0, -3). Find the value of y. (2011D)

Hide | Show

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


Find the value of k, if the point P(2, 4) is equidistant from the points A(5, k) and B(k, 7). (2012OD)

Hide | Show

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


If the point P(k – 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k. (2014OD)

Hide | Show

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


Find a point P on the y-axis which is equidistant from the points A(4, 8) and B(-6, 6). Also find the distance AP. (2014OD)

Hide | Show

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


Prove that the points A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2) are the vertices of a rhombus ABCD. Is ABCD a square? (2013OD)

 

Hide | Show

જવાબ :


If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b), prove that bx = ay. (2016OD)

Hide | Show

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


If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also find the length of AB. (2014D)

Hide | Show

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


If the point P(2, 2) is equidistant from the points A(-2, k) and B(-2k, -3), find k. Also find the length of AP. (2014D)

Hide | Show

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

AP = 5 units
When k = -3
AP2 = (2+2)2 + (2+3)2 = 41

AP = √41


Prove that the points A(0, -1), B(-2, 3), C(6, 7) and D(8, 3) are the vertices of a rectangle ABCD. (2013D)

 

Hide | Show

જવાબ :

Opposite sides BC = AD = 4/√5
Diag. AC = BD = 10
∴ ABCD is a rectangle.
… [ Opp. sides are equal & diagonals are also equal


Points A(-1, y) and B(5, 7) lie on a circle with centre 0(2, -3y). Find the values of y. Hence find the radius of the circle. (2014D)

 

Hide | Show

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


Prove that the points A(2, 3), B(-2, 2), C(-1, -2) and D(3, -1) are the vertices of a square ABCD. (2013OD)

Hide | Show

જવાબ : Given in the question : A(2, 3), B(-2, 2), C(-1, -2), D(3, -1)
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 21
AB = BC = CD = DA …[All four sides are equal
AC = BD …[ diagonals are equal
=> ABCD is a Square.


Show that the points (-2, 3), (8, 3) and (6, 7) are the vertices of a right triangle. (2013D)

 

Hide | Show

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.


Prove that the diagonals of a rectangle ABCD, with vertices A(2, -1), B(5, -1), C(5, 6) & D(2,6), are equal and bisect each other. (2014OD)

Hide | Show

જવાબ :
=> Diagonals also bisect each other.


If the points A(-2, 1), B(a, b) and C(4, -1) are collinear and a – b = 1, find the values of a and (2014D)

Hide | Show

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


If the points A(-1, -4), B(b, c) and C(5, -1) are collinear and 2b + c = 4, find the values of b and c. (2014D)

Hide | Show

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


Find the area of the triangle ABC with A(1, 4) and mid-points of sides through A being (2, -1) and (0, -1). (2015D)

Hide | Show

જવાબ : Given


Find the area of the triangle PQR with Q(3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2). (2015D)


 

Hide | Show

જવાબ : P(x, y), R(z, t).


There are No Content Availble For this Chapter

Match the distance

1

 A(9, 3) and B(15, 11)

A

17

2

A(7, −4) and B(−5, 1)

B

32

3

A(−6, −4) and B(9, −12)

C

10

4

A(1, −3) and B(4, −6)

D

13

Hide | Show

જવાબ :

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

Distance from origin

1

A(5, −12)

A

2√13

2

B(−5, 5)

B

13

3

C(−4, −6)

C

4

4

D(0,4)

D

5√2

Hide | Show

જવાબ :

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

Distance between the points

1

A(2, −1) and B(5, 3)

A

10

2

A(2,−3) and B(10,-9)

B

10

3

A(0, 2) and B(3, 1)

C

2√13

4

A(6, 5) and B(0,9)

D

5

Hide | Show

જવાબ :

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

Quadrant and point

1

I

A

(-9,8)

2

II

B

(9,8)

3

III

C

(9,-8)

4

IV

D

(-9,-8)

Hide | Show

જવાબ :

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

Quadrant and point

1

I

A

(-5,-6)

2

II

B

(5,6)

3

III

C

(5,-6)

4

IV

D

(-5,6)

Hide | Show

જવાબ :

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

Point equidistant from AB

1

A(5, 1) and B(− 1, 5)

A

(3,0)

2

A(6, −1) and B(2, 3)

B

(2,3)

3

A(5, 3) and B(5, −5)

C

(− 2, 3)

4

A(3, − 1) and B(2, 8)

D

(3, −1)

Hide | Show

જવાબ :

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

Point collinear with the following

1

(5, 2) and (9, 5)

A

(6, 9)

2

(2, 3) and (8, 11)

B

(1, −1)

3

(0, 1) and (−6, −7)

C

(−2, 5)

4

(0, 1) and (2, −3)

D

(−1, −1)

Hide | Show

જવાબ :

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

Distance between the points

1

B(5, 2) and C(9, 5)

A

10

2

A(1, −1) and C(9, 5)

B

5

3

A(6, 9) and C(−6, −7)

C

15

4

A(−1, −1) and C(8, 11)

D

20

Hide | Show

જવાબ :

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

Distance between the points

1

A(1, −1) and B(5, 2)

A

25

2

B(0, 1) and C(−6, −7)

B

4√5

3

B(0, 1) and C(2, −3)

C

5

4

A(−2, 5) and C(2, −3)

D

10

Hide | Show

જવાબ :

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

Coordinates and Type of triangle

1

A(3, 0), B(6, 4) and C(− 1, 3)

A

 Equilateral triangle

2

 A(2, 4), B(2, 6) and C(2+3, 5)

B

Isosceles right triangle

3

A(13,-2), B(9-8) and C(5-2)

C

Scalene Triangle

4

A(−2, k), B(−2k, −3) and C (2, 1)

D

Isosceles triangle

Hide | Show

જવાબ :

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

Download PDF

Take a Test

Choose your Test :

Coordinate Geometry

Coordinate Geometry

.

આ પ્રકરણને લગતા વિવિધ એનિમેશન વિડીયો, હેતુલક્ષી પ્રશ્નો, ટૂંકા પ્રશ્નો, લાંબા પ્રશ્નો, પરિક્ષામાં પુછાઈ ગયેલા પ્રશ્નો તેમજ પરિક્ષામાં પુછાઈ શકે તેવા અનેક મુદ્દાસર પ્રશ્નો જોવા અમારી વેબસાઈટ પર રજીસ્ટર થાઓ અથવા અમારી App ફ્રી માં ડાઉનલોડ કરો.

Browse & Download CBSE Books For Class 10 All Subjects

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.