# 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 _________

જવાબ : 2

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

જવાબ : (– 4, 2)

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

જવાબ : Rectangle

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

જવાબ : 3

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

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 _______

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 ______

જવાબ : – 12

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

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 ________

જવાબ : (–6, 5/2)

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

જવાબ : ± 4

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

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 ___________

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 _________

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 ____________

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 _________

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).

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.

જવાબ : True

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

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).

જવાબ : False

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

જવાબ : True

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

જવાબ : False

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

જવાબ : 2

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

જવાબ : (-3/2, 5)

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

જવાબ : 0

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

જવાબ : 5

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

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

જવાબ : (125, 275)

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

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)

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

જવાબ : False

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

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).

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

જવાબ : True

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

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

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.

જવાબ : True

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

જવાબ : (3, -1)

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

જવાબ : 1

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

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).

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)

જવાબ : Y axis

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

જવાબ : Y’ axis

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

જવાબ : IV quadrant

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

જવાબ : III quadrant

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

જવાબ : I quadrant

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

જવાબ : X axis

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

જવાબ : III quadrant

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

જવાબ : II quadrant

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

જવાબ : Y’ axis

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

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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.

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

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)

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)

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)

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)

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)

જવાબ : P(x, y), R(z, t).

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Match the distance

 1 A(9, 3) and B(15, 11) A 17 2 A(7, −4) and B(−5, 1) B 3√2 3 A(−6, −4) and B(9, −12) C 10 4 A(1, −3) and B(4, −6) D 13

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

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

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)

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)

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)

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)

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

જવાબ :

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

Distance between the points

 1 A(1, −1) and B(5, 2) A 2√5 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

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

જવાબ :

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

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