# GSEB Solutions for ધોરણ ૧૦ English

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

The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, -5) and R(-3, 6), find the coordinates of P.

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

Find the ratio in which the point (-3, k) divides the line-segment joining the points (-5, 4) and (-2, 3). Also find the value of k.

જવાબ : Let Q divide AB in the ratio of p : 1 If (1, 2), (4, y), (x, 6) and (3,5) are the vertices of a parallelogram taken in order, find x and y.

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.

Find the ratio in which y-axis divides the line segment joining the points A(5, -6) and B(-1, 4). Also, find the coordinates of the point of division.

જવાબ : Let the point on y-axis be P(0, y) and AP : PB = k : 1 Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.

જવાબ : Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

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.

Point P (0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points, A (-1, 1) and B (3, 3). State true or false and justify your answer.

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.

Point P (5, -3) is one of the two points of trisection of the line segment joining the points A (7, -2) and B (1, -5). State true or false and justify your answer.

જવાબ : Points of trisection of line segment AB are given by Given statement is true.

Show that ∆ABC, where A(-2, 0), B(2, 0), C(0, 2) and APQR where P(-4, 0), Q(4, 0), R(0,4) are similar triangles.
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 : Find the ratio in which the line segment joining the points P (3, -6) and Q (5,3) is divided by the x-axis.

જવાબ : Let the required ratio be λ : 1 Given that this point lies on the x-axis Thus, the required ratio is 2 : 1.

In Fig. 6.8, if A(-1, 3), B(1, -1) and C (5, 1) are the vertices of a triangle ABC, what is the length of the median through vertex A?

જવાબ : Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.

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

Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

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.

For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P.? (2016OD)

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

If the points A(x, 2), B(-3, 4) and C(7, -5) are collinear, then find the value of x. (2014D)

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

Find the distance of the point (-3, 4) from the x-axis. (2012OD)

જવાબ : B(-3, 0), A (-3, 4) In which quadrant the point P that divides the line segment joining the points A(2, -5) and B(5,2) in the ratio 2 : 3 lies? (2011D)

જવાબ : ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Calculate the length of one of its diagonals. (2014OD)

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

AC = 5 cm

Find a relation between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7). (2011D)

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.

Find the relation between x and y if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear. (2015OD)

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.

Find the ratio in which the point P(34,512) divides the line segment joining the points A(12,32) and B(2, -5). (2015D)

જવાબ : Find the ratio in which y-axis divides the line segment joining the points A(5, -6), and B(-1, -4). Also find the coordinates of the point of division. (2016D)

જવાબ : Let P and Q be the points of trisection of the line segment joining the points A (2, -2) and B (-7, 4) such that P is nearer to A. Find the coordinates of P and Question (2016D)

જવાબ : Three vertices of a parallelogram taken in order are (-1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of fourth vertex. (2011D)

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).

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 find the coordinates of P and Question (2017OD)

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 Determine the ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7). (2011D)

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.

If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y. (2015OD)

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

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
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

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
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

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
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

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
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

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
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)

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

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.

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 : Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) such that AP = AB. (2015OD)

જવાબ : Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.

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.

Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

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.

Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.

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)= 36 y + 3 = ±6 y + 3 = +6 or y + 3 = −6 Therefore, y = 3 or -9.

What is the area of the triangle formed by the points 0 (0, 0), A (-3, 0) and B (5, 0)?

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

If the centroid of triangle formed by points P (a, b), Q (b, c) and R (c, a) is at the origin, what is the value of a + b + c?

જવાબ : Find the value of a, so that the point (3, a) lie on the line 2x – 3y = 5.

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

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

જવાબ : Find distance between the points (0, 5) and (-5, 0).

જવાબ : Find the distance of the point (-6,8) from the origin.

જવાબ : Here x1 = -6, y1 = 8
x2 = 0, y2 = 0 If the points A (1, 2), B (0, 0) and C (a, b) are collinear, then what is the relation between a and b?

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

If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k?

જવાબ : Using distance formula Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).

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.

The coordinates of the points P and Q are respectively (4, -3) and (-1, 7). Find the abscissa of a point R on the line segment PQ such that PR/PQ = 3/5.

જવાબ : Write the coordinates of a point on x-axis which is equidistant from the points (-3, 4) and (2, 5).

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

​​​​​​ Find the values of x for which the distance between the points P (2, -3) and Q (x, 5) is 10.

જવાબ : What is the distance between the points (10 cos 30°, 0) and (0, 10 cos 60°)?

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

Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle. (2013D)

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.

If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right angled triangle with B = 90°, then find the value of t. (2015D)

જવાબ : ABC is a right angled triangle,
AC2 = BC2 + AB2 …(i)… (Pythagoras theorem
Using distance formula,
AB2 = (5 – 2)2 + (2 + 2)
= 25
BC2 = (2 + 2)2 + (t + 2)2
= 16 + (t + 2)2
AC2 = (5 + 2)2 + (2 – t)2
= 49 + (2 – t)2
Putting values of AB2, AC2 and BC2 in equation (i), we get
49 + (2 – t)2 = 16 + (t + 2)2 + 25

49 + (2 – t)2 = 41 + (t + 2)2
(t + 2)2 – (2 – t)2 = 8
(t2 + 4 + 4t – 4 – t2 + 4t) = 8
8t = 8
t = 1

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 : 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 : Find the distance between the following pairs of points:
(i) (-5, 7), (-1, 3)
(ii) (a, b), (-a, -b)

જવાબ : Name the type of quadrilateral formed, if any, by the following points, and give reasons for
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, Hence, four sides of quadrilateral are equal and diagonals AC and BD are also equal.
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.

Find the ratio in which the line segment joining the points A(3, -3) and B(-2, 7) is divided by x-axis. Also find the coordinates of the point of division.

જવાબ : In the given figure, ∆ABC is an equilateral triangle of side 3 units. Find the coordinates of the other two vertices.

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 + y6x = 21
x = 216 = 72
Again (x – 2)2 + (y – 0)2 = 9 Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

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

A median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A (4,-6), B (3, -2) and C (5, 2).

જવાબ : Find the ratio in which the point P (x, 2), divides the line segment joining the points A (12, 5) and B (4, -3). Also, find the value of x.

જવાબ : If A (4, 2), B (7, 6) and C (1, 4) are the vertices of a ∆ABC and AD is its median, prove that the median AD divides into two triangles of equal areas.

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.

If the point A (2, -4) is equidistant from P (3, 8) and Q (-10, y), find the values of y. Also find distance PQ.

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

​​​​​​ The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point Care (0, -3). The origin is the mid-point of the base. Find the coordinates of the points A and B. Also find the coordinates of another point D such that BACD is a rhombus.

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, Coordinates of point A = (x, 0) = (3√3, 0)
Since BACD is a rhombus.

AB = AC = CD = DB
Coordinates of point D = (-3√3, 0).

The line joining the points (2, 1) and (5, -8) is trisected by the points P and Q. If the point P lies on the line 2x – y + k = 0, find the value of k.

જવાબ : Prove that the diagonals of a rectangle bisect each other and are equal.

જવાબ : In what ratio does the y-axis divide the line segment joining the point P (4, 5) and Q (3, -7)? Also, find the coordinates of the point of intersection.

જવાબ : Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).

જવાબ : Let O(x, y) be the centre of circle. Given points are A(6, -6), B(3, -7) and C(3, 3). If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, – 3) and (3, 4). Find its centroid.

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

જવાબ : Points A(-l, 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.

જવાબ : If A(-3,5), B(-2, -7), C(l, -8) and D(6,3) are the vertices of a quadrilateral ABCD, find its area.

જવાબ : The three vertices of a parallelogram ABCD are A(3, -4), B(-l, -3) and C(-6, 2). Find the coordinates of vertex D and find the area of parallelogram ABCD.

જવાબ : If the points A(l, -2), B(2,3), C(-3,2) and D(-4, -3) are the vertices of parallelogram ABCD, then taking AB as the base, find the height of this parallelogram.

જવાબ : For the AABC formed by the points A(4, -6), B(3, -2) and C(5,2), verify that median divides the triangle into two triangles of equal area.

જવાબ : A point P divides the line segment joining the points A(3, -5) and B(-4, 8) such AP/PB=k/1. If P lies on the line x + y = 0, then find the value of K.

જવાબ : If two vertices of an equilateral triangle are (3,0) and (6,0), find the third vertex.

જવાબ : Find the area of the quadrilateral ABCD, whose vertices are A(-3, -1), B(-2, -4), C(4, -1) and D(3, 4).

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gseb maths textbook std 10

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