Class 10

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.3 Geometry Class 10 Question 1. Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.
i. 45°
ii. 60°
iii. 90°
Solution:
i. Angle made with the positive direction of
X-axis (θ) = 45°
Slope of the line (m) = tan θ
∴ m = tan 45° = 1
∴ The slope of the line is 1.

ii. Angle made with the positive direction of X-axis (θ) = 60°
Slope of the line (m) = tan θ
∴ m = tan 60° = \(\sqrt { 3 }\)
∴ The slope of the line is \(\sqrt { 3 }\).

iii. Angle made with the positive direction of
X-axis (θ) = 90°
Slope of the line (m) = tan θ
∴ m = tan 90°
But, the value of tan 90° is not defined.
∴ The slope of the line cannot be determined.

Practice Set 5.3 Geometry Question 2. Find the slopes of the lines passing through the given points.
i. A (2, 3), B (4, 7)
ii. P(-3, 1), Q (5, -2)
iii. C (5, -2), D (7, 3)
iv. L (-2, -3), M (-6, -8)
v. E (-4, -2), F (6, 3)
vi. T (0, -3), s (0,4)
Solution:
i. A (x₁, y₁) = A (2, 3) and B (x₂, y₂) = B (4, 7)
Here, x₁ = 2, x₂ = 4, y₁ = 3, y₂ = 7

∴ The slope of line AB is 2.

ii. P (x₁, y₁) = P (-3, 1) and Q (x₂, y₂) = Q (5, -2)
Here, x₁ = -3, x₂ = 5, y₁ = 1, y₂ = -2

∴ The slope of line PQ is \(\frac { -3 }{ 8 } \)

iii. C (x₁, y₁) = C (5, -2) and D (x₂, y₂) = D (7, 3)
Here, x₁ = 5, x₂ = 7, y₁ = -2, y₂ = 3

∴ The slope of line CD is \(\frac { 5 }{ 2 } \)

iv. L (x₁, y₁) = L (-2, -3) and M (x₂,y₂) = M (-6, -8)
Here, x₁ = -2, x₂ = – 6, y₁ = – 3, y₂ = – 8

∴ The slope of line LM is \(\frac { 5 }{ 4 } \)

v. E (x₁, y₁) = E (-4, -2) and F (x₂, y₂) = F (6, 3)
Here,x₁ = -4, x₂ = 6, y₁ = -2, y₂ = 3

∴ The slope of line EF is \(\frac { 1 }{ 2 } \).

vi. T (x₁, y₁) = T (0, -3) and S (x₂, y₂) = S (0, 4)
Here, x₁ = 0, x₂ = 0, y₁ = -3, y₂ = 4

∴ The slope of line TS cannot be determined.

5.3.5 Practice Question 3. Determine whether the following points are collinear.
i. A (-1, -1), B (0, 1), C (1, 3)
ii. D (- 2, -3), E (1, 0), F (2, 1)
iii. L (2, 5), M (3, 3), N (5, 1)
iv. P (2, -5), Q (1, -3), R (-2, 3)
v. R (1, -4), S (-2, 2), T (-3,4)
vi. A(-4,4),K[-2,\(\frac { 5 }{ 2 } \)], N (4,-2)
Solution:


∴ slope of line AB = slope of line BC
∴ line AB || line BC
Also, point B is common to both the lines.
∴ Both lines are the same.
∴ Points A, B and C are collinear.

∴ slope of line DE = slope of line EF
∴ line DE || line EF
Also, point E is common to both the lines.
∴ Both lines are the same.
∴ Points D, E and F are collinear.

∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.

∴ slope of line PQ = slope of line QR
∴ line PQ || line QR
Also, point Q is common to both the lines.
∴ Both lines are the same.
∴ Points P, Q and R are collinear.

∴ slope of line RS = slope of line ST
∴ line RS || line ST
Also, point S is common to both the lines.
∴ Both lines are the same.
∴ Points R, S and T are collinear.

∴ slope of line AK = slope of line KN
∴ line AK || line KN
Also, point K is common to both the lines.
∴ Both lines are the same.
∴ Points A, K and N are collinear.

Practice Set 5.3 Geometry 9th Standard Question 4. If A (1, -1), B (0,4), C (-5,3) are vertices of a triangle, then find the slope of each side.
Solution:

∴ The slopes of the sides AB, BC and AC are -5, \(\frac { 1 }{ 5 } \) and \(\frac { -2 }{ 3 } \) respectively.

Geometry 5.3 Question 5. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.
Proof:


∴ Slope of side AB = Slope of side CD … [From (i) and (iii)]
∴ side AB || side CD
Slope of side BC = Slope of side AD … [From (ii) and (iv)]
∴ side BC || side AD
Both the pairs of opposite sides of ꠸ABCD are parallel.
꠸ABCD is a parallelogram.
Points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a parallelogram.

Question 6.
Find k, if R (1, -1), S (-2, k) and slope of line RS is -2.
Solution:
R(x₁, y₁) = R (1, -1), S (x₂, y₂) = S (-2, k)
Here, x₁ = 1, x₂ = -2, y₁ = -1, y₂ = k

But, slope of line RS is -2. … [Given]
∴ -2 = \(\frac { k+1 }{ -3 } \)
∴ k + 1 = 6
∴ k = 6 – 1
∴ k = 5

5.3 Class 10 Question 7. Find k, if B (k, -5), C (1, 2) and slope of the line is 7.
Solution:
B(x₁, y₁) = B (k, -5), C (x₂, y₂) = C (1, 2)
Here, x₁ = k, x₂ = 1, y₁ = -5, y₂ = 2

But, slope of line BC is 7. …[Given]
∴ 7 = \(\frac { 7 }{ 1-k } \)
∴ 7(1 – k) = 7
∴ 1 – k = \(\frac { 7 }{ 7 } \)
∴ 1 – k = 1
∴ k = 0

Question 8.
Find k, if PQ || RS and P (2, 4), Q (3, 6), R (3,1), S (5, k).
Solution:

But, line PQ || line RS … [Given]
∴ Slope of line PQ = Slope of line RS
∴ 2 = \(\frac { k-1 }{ 2 } \)
∴ 4 = k – 1
∴ k = 4 + 1
∴ k = 5

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
Find the co-ordinates of point P if P divides the line segment joining the points A (-1, 7) and B (4, -3) in the ratio 2:3.
Solution:
Let the co-ordinates of point P be (x, y) and A (x₁, y₁) B (x₂, y₂) be the given points.
Here, x₁ = -1, y₁ = 7, x₂ = 4, y₂ = -3, m = 2, n = 3
∴ By section formula,

∴ The co-ordinates of point P are (1,3).

Question 2.
In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.
i. P (-3, 7), Q (1, -4), a : b = 2 : 1
ii. P (-2, -5), Q (4, 3), a : b = 3 : 4
iii. P (2, 6), Q (-4, 1), a : b = 1 : 2
Solution:
Let the co-ordinates of point A be (x, y).
i. Let P (x₁, y₁), Q (x₂, y₂) be the given points.
Here, x₁ = -3, y₁ = 7, x₂ = 1, y₂ = -4, a = 2, b = 1
∴ By section formula,

∴ The co-ordinates of point A are (\(\frac { -1 }{ 3 } \),\(\frac { -1 }{ 3 } \)).

ii. Let P (x₁,y₁), Q (x₂, y₂) be the given points.
Here, x₁ = -2, y₁ = -5, x₂ = 4, y₂ = 3, a = 3, b = 4
By section formula,

∴ The co-ordinates of point A are (\(\frac { 4 }{ 7 } \),\(\frac { -11 }{ 7 } \))

iii. Let P (x₁, y₁), Q (x₂, y₂) be the given points.
Here,x₁ = 2,y₁ = 6, x₂ = -4, y₂ = 1, a = 1,b = 2
∴ By section formula,

∴ The co-ordinates of point A are (0,\(\frac { 13 }{ 3 } \))

Question 3.
Find the ratio in which point T (-1, 6) divides the line segment joining the points P (-3,10) and Q (6, -8).
Solution:
Let P (x₁, y₁), Q (x₂, y₂) and T (x, y) be the given points.
Here, x₁ = -3, y₁ = 10, x₂ = 6, y₂ = -8, x = -1, y = 6
∴ By section formula,

∴ Point T divides seg PQ in the ratio 2 : 7.

Question 4.
Point P is the centre of the circle and AB is a diameter. Find the co-ordinates of point B if co-ordinates of point A and P are (2, -3) and (-2,0) respectively.
Solution:
Let A (x₁, y₁), B (x₂, y₂) and P (x, y) be the given points.
Here, x₁ = 2, y₁ =-3,
x = -2, y = 0

Point P is the midpoint of seg AB.
∴ By midpoint formula,

∴ The co-ordinates of point B are (-6,3).

Question 5.
Find the ratio in which point P (k, 7) divides the segment joining A (8, 9) and B (1,2). Also find k.
Solution:
Let A (x₁, y₁), B (x₂, y₂) and P (x, y) be the given points.
Here, x₁ = 8, y₁ = 9, x₂ = 1, y₂ = 2, x = k, y = 7
∴ By section formula,

∴ Point P divides seg AB in the ratio 2 : 5, and the value of k is 6.

Question 6.
Find the co-ordinates of midpoint of the segment joining the points (22, 20) and (0,16).
Solution:
Let A (x₁, y₁) = A (22, 20),
B (x₂,y₂) = B (0, 16)
Let the co-ordinates of the midpoint be P (x,y).
∴ By midpoint formula,

The co-ordinates of the midpoint of the segment joining (22, 20) and (0, 16) are (11,18).

Question 7.
Find the centroids of the triangles whose vertices are given below.
i. (-7, 6), (2,-2), (8, 5)
ii. (3, -5), (4, 3), (11,-4)
iii. (4, 7), (8, 4), (7, 11)
Solution:
i. Let A (x₁, y₁) = A (-7, 6),
B (x₂, y₂) = B (2, -2),
C (x₃, y₃) = C(8, 5)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (1,3).

ii. Let A (x₁ y₁) = A (3, -5),
B (x₂, y₂) = B (4, 3),
C(x₃, y₃) = C(11,-4)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (6, -2).

iii. Let A (x₁, y₁) = A (4, 7),
B (x₂, y₂) = B (8,4),
C (x₃, y₃) = C(7,11)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (\(\frac { 19 }{ 3 } \),\(\frac { 22 }{ 3 } \))

Question 8.
In ∆ABC, G (-4, -7) is the centroid. If A (-14, -19) and B (3, 5), then find the co-ordinates of C.
Solution:
G (x, y) = G (-4, -7),
A (x₁, y₁) = A (-14, -19),
B(x₂, y₂) = B(3,5)
Let the co-ordinates of point C be (x₃, y₃).
G is the centroid.
By centroid formula,

∴ The co-ordinates of point C are (-1, – 7).

Question 9.
A (h, -6), B (2, 3) and C (-6, k) are the co-ordinates of vertices of a triangle whose centroid is G (1,5). Find h and k.
Solution:
A(x₁,y₁) = A(h, -6),
B (x₂, y₂) = B(2, 3),
C (x₃, y₃) = C (-6, k)
∴ centroid G (x, y) = G (1, 5)
G is the centroid.
By centroid formula,

∴ 3 = h – 4
∴ h = 3 + 4
∴ h = 7

∴ 15 = -3 + k
∴ k = 15 + 3
∴ k = 18
∴ h = 7 and k = 18

Question 10.
Find the co-ordinates of the points of trisection of the line segment AB with A (2,7) and B (-4, -8).
Solution:
A (2, 7), B H,-8)
Suppose the points P and Q trisect seg AB.
∴ AP = PQ = QB

∴ Point P divides seg AB in the ratio 1:2.
∴ By section formula,

Co-ordinates of P are (0, 2).
Point Q is the midpoint of PB.
By midpoint formula,

Co-ordinates of Q are (-2, -3).
∴ The co-ordinates of the points of trisection seg AB are (0,2) and (-2, -3).

Question 11.
If A (-14, -10), B (6, -2) are given, find the co-ordinates of the points which divide segment AB into four equal parts.
Solution:
Let the points C, D and E divide seg AB in four equal parts.

Point D is the midpoint of seg AB.
∴ By midpoint formula,

∴ Co-ordinates of D are (-4, -6).
Point C is the midpoint of seg AD.
∴ By midpoint formula,

∴ Co-ordinates of C are (-9, -8).
Point E is the midpoint of seg DB.
∴ By midpoint formula,


∴ Co-ordinates of E are (1,-4).
∴ The co-ordinates of the points dividing seg AB in four equal parts are C(-9, -8), D(-4, -6) and E(1, – 4).

Question 12.
If A (20, 10), B (0, 20) are given, find the co-ordinates of the points which divide segment AB into five congruent parts.
Solution:
Suppose the points C, D, E and F divide seg AB in five congruent parts.
∴ AC = CD = DE = EF = FB

∴ co-ordinates of C are (16, 12).
E is the midpoint of seg CB.
By midpoint formula,

∴ co-ordinates of E are (8, 16).
D is the midpoint of seg CE.

∴ co-ordinates of F are (4, 18).
∴ The co-ordinates of the points dividing seg AB in five congruent parts are C (16, 12), D (12, 14), E (8, 16) and F (4, 18).

Maharashtra Board Class 10 Maths Chapter 5 Co-ordinate Geometry Intext Questions and Activities

Question 1.
A (15, 5), B (9, 20) and A-P-B. Find the ratio in which point P (11, 15) divides segment AB. Find the ratio using x and y co-ordinates. Write the conclusion. (Textbook pg. no. 113)
Solution:
Suppose point P (11,15) divides segment AB in the ratio m : n.
By section formula,

∴ Point P divides seg AB in the ratio 2 : 1.
The ratio obtained by using x and y co-ordinates is the same.

Question 2.
External division: (Textbook pg. no. 115)
Suppose point R divides seg PQ externally in the ratio 3:1.

Let the common multiple be k.
Let PR = 3k and QR = k
Now, PR = PQ + QR … [P – Q – R]
∴ 3k = PQ + k
∴ \(\frac { PQ }{ QR } \) = \(\frac { 2k }{ k } \) = \(\frac { 2 }{ 1 } \)
∴ Point Q divides seg PR in the ratio 2 : 1 internally.
Thus, we can find the co-ordinates of point R, when co-ordinates of points P and Q are given.

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.1 Geometry Class 10 Question 1. Find the distance between each of the following pairs of points.
i. A (2, 3), B (4,1)
ii. P (-5, 7), Q (-1, 3)
iii. R (0, -3), S (0,\(\frac { 5 }{ 2 } \))
iv. L (5, -8), M (-7, -3)
v. T (-3, 6), R (9, -10)
vi. W(\(\frac { -7 }{ 2 } \),4), X(11, 4)
Solution:
i. Let A (x₁, y₁) and B (x₂, y₂) be the given points.
∴ x₁ = 2, y₁ = 3, x₂ = 4, y₂ = 1
By distance formula,

∴ d(A, B) = 2\(\sqrt { 2 }\) units
∴ The distance between the points A and B is 2\(\sqrt { 2 }\) units.

ii. Let P (x₁, y₁ ) and Q (x₂, y₂) be the given points.
∴ x₁ = -5, y₁ = 7, x₂ = -1, y₂ = 3
By distance formula,

∴ d(P, Q) = 4\(\sqrt { 2 }\) units
∴ The distance between the points P and Q is 4\(\sqrt { 2 }\) units.

iii. Let R (x₁, y₁) and S (x₂, y₂) be the given points.
∴ x₁ = 0, y₁ = -3, x₂ = 0, y₂ = \(\frac { 5 }{ 2 } \)
By distance formula,

∴ d(R, S) = \(\frac { 11 }{ 2 } \) units
∴ The distance between the points R and S is \(\frac { 11 }{ 2 } \) units.

iv. Let L (x₁, y₁) and M (x₂, y₂) be the given points.
∴ x₁ = 5, y₁ = -8, x₂ = -7, y₂ = -3
By distance formula,

∴ d(L, M) = 13 units
∴ The distance between the points L and M is 13 units.

v. Let T (x₁,y₁) and R (x₂, y₂) be the given points.
∴ x₁ = -3, y₁ = 6,x₂ = 9,y₂ = -10
By distance formula,

∴ d(T, R) = 20 units
∴ The distance between the points T and R 20 units.

vi. Let W (x₁, y₁) and X (x₂, y₂) be the given points.

∴ d(W, X) = \(\frac { 29 }{ 2 } \) units
∴ The distance between the points W and X is \(\frac { 29 }{ 2 } \) units.

Practice Set 5.1 Geometry 10th Question 2. Determine whether the points are collinear.
i. A (1, -3), B (2, -5), C (-4, 7)
ii. L (-2, 3), M (1, -3), N (5, 4)
iii. R (0, 3), D (2, 1), S (3, -1)
iv. P (-2, 3), Q (1, 2), R (4, 1)
Solution:
i. By distance formula,


∴ d(A, B) = \(\sqrt { 5 }\) …(i)
On adding (i) and (iii),
d(A, B) + d(A, C)= \(\sqrt { 5 }\) + 5\(\sqrt { 5 }\) = 6\(\sqrt { 5 }\)
∴ d(A, B) + d(A, C) = d(B, C) … [From (ii)]
∴ Points A, B and C are collinear.

ii. By distance formula,

On adding (i) and (iii),
d(L, M) + d(L, N) = 3\(\sqrt { 5 }\) + 5\(\sqrt { 2 }\) ≠ \(\sqrt { 65 }\)
∴ d(L, M) + d(L, N) ≠ d(M, N) … [From (ii)]
∴ Points L, M and N are not collinear.

iii. By distance formula,

On adding (i) and (ii),
∴ d(R, D) + d(D, S) = \(\sqrt { 8 }\) + \(\sqrt { 5 }\) ≠ 5
∴ d(R, D) + d(D, S) ≠ d(R, S) … [From (iii)]
∴ Points R, D and S are not collinear.

iv. By distance formula,

On adding (i) and (ii),
d(P, Q) + d(Q, R) = \(\sqrt { 10 }\) + \(\sqrt { 10 }\) = 2\(\sqrt { 10 }\)
∴ d(P, Q) + d(Q, R) = d(P, R) … [From (iii)]
∴ Points P, Q and R are collinear.

Coordinate Geometry Class 10 Practice Set 5.1 Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).
Solution:
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
∴ its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
∴ AC = BC

∴ (x + 3)² + (-4)² = (x- 1)² + 4²
∴ x² + 6x + 9 + 16 = x² – 2x + 1 + 16
∴ 8x = – 8
∴ x = – \(\frac { 8 }{ 8 } \) = -1
∴ The point on X-axis which is equidistant from points A and B is (-1,0).

10th Geometry Practice Set 5.1 Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Solution:
Distance between two points

Consider, PQ² + QR² = 42 + 52 = 16 + 25 = 41 … [From (i) and (ii)]
∴ PR² = PQ² + QR² … [From (iii)]
∴ ∆PQR is a right angled triangle. … [Converse of Pythagoras theorem]
∴ Points P, Q and R are the vertices of a right angled triangle.

Question 5.
Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Proof:
Distance between two points

PQ = RS … [From (i) and (iii)]
QR = PS … [From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
∴ □ PQRS is a parallelogram.
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 6.
Show that points A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are vertices of rhombus ABCD.
Proof:
Distance between two points


∴ AB = BC = CD = AD …[From (i), (ii), (iii) and (iv)]
In a quadrilateral, if all the sides are equal, then it is a rhombus.
∴ □ ABCD is a rhombus.
∴ Points A, B, C and D are the vertices of rhombus ABCD.

Practice Set 5.1 Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.
Solution:
X₁ = x, y₁ = 7, x₂ = 1, y₂ = 15
By distance formula,

∴ 1 – x = ± 6
∴ 1 – x = 6 or l – x = -6
∴ x = – 5 or x = 7
∴ The value of x is – 5 or 7.

Geometry 5.1 Question 8. Show that the points A (1, 2), B (1, 6), C (1 + 2\(\sqrt { 3 }\), 4) are vertices of an equilateral triangle.
Proof:
Distance between two points

∴ AB = BC = AC … [From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.

Maharashtra Board Class 10 Maths Chapter 5 Coordinate Geometry Intext Questions and Activities

Question 1.
In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)

Solution:
In ∆ABC, ∠B = 900
∴ (AB)² + (BC)² = [(Ac)² …(i) … [Pythagoras theorem]
seg CB || X-axis
∴ y co-ordinate of B = 2
seg BA || Y-axis
∴ x co-ordinate of B = 2
∴ co-ordinate of B is (2, 2) = (x₁,y₁)
co-ordinate of A is (2, 3) = (x₂, Y₂)
Since, AB || to Y-axis,
d(A, B) = Y₂ – Y₁
d(A,B) = 3 – 2 = 1
co-ordinate of C is (-2,2) = (x₁,y₁)
co-ordinate of B is (2, 2) = (x₂, y₂)
Since, BC || to X-axis,
d(B, C) = x₂ – x₁
d(B,C) = 2 – -2 = 4
∴ AC² = 12 + 42 …[From (i)]
= 1 + 16 = 17
∴ AC = \(\sqrt { 17 }\) units …[Taking square root of both sides]

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.2 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.2 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.

Solution:
In ∆ PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 7 }{ 3 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 3.5 }{ 1.5 } \) = \(\frac { 35 }{ 15 } \) = \(\frac { 7 }{ 3 } \) (ii)
∴ \(\frac { PQ }{ PR } \) = \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR. [Converse of angle bisector theorem]

ii. In ∆PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 10 }{ 7 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 8 }{ 6 } \) = \(\frac { 4 }{ 3 } \) (ii)
∴ \(\frac { PQ }{ PR } \) ≠ \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is not the bisector of ∠QPR

iii. In ∆PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 9 }{ 10 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 3.6 }{ 4 } \) = \(\frac { 36 }{ 40 } \) = \(\frac { 9 }{ 10 } \) (ii)
∴ \(\frac { PQ }{ PR } \) = \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR [Converse of angle bisector theorem]

Question 2.
In ∆PQR PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.

Solution:
PN + NR = PR [P – N – R]
∴ PN + 8 = 20
∴ PN = 20 – 8 = 12
Also, PM + MQ = PQ [P – M – Q]
∴ 15 + MQ = 25

∴ line NM || side RQ [Converse of basic proportionality theorem]

Question 3.
In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7, MQ = 2.5, then find QP.

Solution:
In ∆MNP, NQ is the bisector of ∠N. [Given]
∴\(\frac { PN }{ MN } \) = \(\frac { QP }{ MQ } \) [Property of angle bisector of a triangle]
∴\(\frac { 7 }{ 5 } \) = \(\frac { QP }{ 2.5 } \)
∴ QP = \(\frac { 7\times 2.5 }{ 5 } \)
∴ QP = 3.5 units

Question 4.
Measures of some angles in the figure are given. Prove that \(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \)

Solution:
Proof
∠APQ = ∠ABC = 60° [Given]
∴ ∠APQ ≅ ∠ABC
∴ side PQ || side BC (i) [Corresponding angles test]
In ∆ABC,
sidePQ || sideBC [From (i)]
∴\(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \) [Basic proportionality theorem]

Question 5.
In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.

Solution:
side AB || side PQ || side DC [Given]
∴\(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \) [Property of three parallel lines and their transversals]
∴\(\frac { 15 }{ 12 } \) = \(\frac { BQ }{ 14 } \)
∴ BQ = \(\frac { 15\times 14 }{ 12 } \)
∴ BQ = 17.5 units

Question 6.
Find QP using given information in the figure.

Solution:
In ∆MNP, seg NQ bisects ∠N. [Given]
∴\(\frac { PN }{ MN } \) = \(\frac { QP }{ MQ } \) [Property of angle bisector of a triangle]
∴\(\frac { 40 }{ 25 } \) = \(\frac { QP }{ 14 } \)
∴ QP = \(\frac { 40\times 14 }{ 25 } \)
∴ QP = 22.4 units

Question 7.
In the adjoining figure, if AB || CD || FE, then find x and AE.

Solution:
line AB || line CD || line FE [Given]
∴\(\frac { BD }{ DF } \) = \(\frac { AC }{ CE } \) [Property of three parallel lines and their transversals]
∴\(\frac { 8 }{ 4 } \) = \(\frac { 12 }{ X } \)
∴ X = \(\frac { 12\times 4 }{ 8 } \)
∴ X = 6 units
Now, AE AC + CE [A – C – E]
= 12 + x
= 12 + 6
= 18 units
∴ x = 6 units and AE = 18 units

Question 8.
In ∆LMN, ray MT bisects ∠LMN. If LM = 6, MN = 10, TN = 8, then find LT.

Solution:
In ∆LMN, ray MT bisects ∠LMN. [Given]
∴\(\frac { LM }{ MN } \) = \(\frac { LT }{ TN } \) [Property of angle bisector of a triangle]
∴\(\frac { 6 }{ 10 } \) = \(\frac { LT }{ 8 } \)
∴ LT = \(\frac { 6\times 8 }{ 10 } \)
∴ LT = 4.8 units

Question 9.
In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.

Solution:
In ∆ABC, seg BD bisects ∠ABC. [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AD }{ CD } \) [Property of angle bisector of a triangle]
∴\(\frac { x }{ x+5 } \) = \(\frac { x-2 }{ x+2 } \)
∴ x(x + 2) = (x – 2)(x + 5)
∴ x2 + 2x = x2 + 5x – 2x – 10
∴ 2x = 3x – 10
∴ 10 = 3x – 2x
∴ x = 10

Question 10.
In the adjoining figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.

Solution:

Question 11.
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB = seg AC, then prove that ED || BC.

Solution:
In ∆ABC, ray BD bisects ∠ABC. [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (i) [Property of angle bisector of a triangle]
Also, in ∆ABC, ray CE bisects ∠ACB. [Given]
∴\(\frac { AC }{ BC } \) = \(\frac { AE }{ EB } \) (ii) [Property of angle bisector of a triangle]
But, seg AB = seg AC (iii) [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (iv) [From (ii) and (iii)]
∴\(\frac { AD }{ DC } \) = \(\frac { AE }{ EB } \) [From (i) and (iv)]
∴ ED || BC [Converse of basic proportionality theorem]

Question 1.
i. Draw a ∆ABC.
ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D.
iii. Measure the sides.

iv. Find ratios \(\frac { AB }{ BC } \) and \(\frac { AD }{ DC } \)
v. You will find that both the ratios are almost equal.
vi. Bisect remaining angles of the triangle and find the ratios as above. Verify that the ratios are equal. (Textbook pg. no. 8)
Solution:

Note: Students should bisect the remaining angles and verify that the ratios are equal.

Question 2.
Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof.
i. The areas of two triangles of equal height are proportional to their bases.
ii. Every point on the bisector of an angle is equidistant from the sides of the angle. (Textbook pg. no. 9)

Given: In ∆CAB, ray AD bisects ∠A.
To prove: \(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \)
Construction: Draw seg DM ⊥ seg AB A – M – B and seg DN ⊥ seg AC, A – N – C.
Solution:
Proof:
In ∆ABC,
Point D is on angle bisector of ∠A. [Given]
∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle]
\(\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B \times D M}{A C \times D N}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B}{A C}\) (ii) [From (i)]
Also, ∆ABD and ∆ACD have equal height.
∴ \(\frac{\mathrm{A}(\Delta \mathrm{ABD})}{\mathrm{A}(\Delta \mathrm{ACD})}=\frac{\mathrm{BD}}{\mathrm{CD}}\) (iii) [Triangles having equal height]
∴\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}\) [From (ii) and (iii)]

Question 3.
i. Draw three parallel lines.
ii. Label them as l, m, n.
iii. Draw transversals t₁ and t₂.
iv. AB and BC are intercepts on transversal t₁.
v. PQ and QR are intercepts on transversal t₂.
vi. Find ratios \(\frac { AB }{ BC } \) and \(\frac { PQ }{ QR } \). You will find that they are almost equal. Verify that they are equal.(Textbook pg, no. 10)
Solution:

(Students should draw figures similar to the ones given and verify the properties.)

Question 4.
In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF.(Textbook pg, no. 12)

Solution:

Question 5.
In ∆ABC, ray BD bisects ∠ABC. A – D – C, side DE || side BC, A – E – B, then prove that \(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (Textbook pg, no. 13)

Solution:

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.1 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.1 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Solution:
Let the base, height and area of the first triangle be b₁, h₁, and A₁ respectively.
Let the base, height and area of the second triangle be b₂, h₂ and A₂ respectively.

[Since Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ The ratio of areas of the triangles is 3:4.

Question 2.
In the adjoining figure, BC ± AB, AD _L AB, BC = 4, AD = 8, then find \(\frac{A(\Delta A B C)}{A(\Delta A D B)}\)

Solution:
∆ABC and ∆ADB have same base AB.

[Since Triangles having equal base]

Question 3.
In the adjoining figure, seg PS ± seg RQ, seg QT ± seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.

Solution:
In ∆PQR, PR is the base and QT is the corresponding height.
Also, RQ is the base and PS is the corresponding height.
\(\frac{A(\Delta P Q R)}{A(\Delta P Q R)}=\frac{P R \times Q T}{R Q \times P S}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{1}{1}=\frac{P R \times Q T}{R Q \times P S}\)
∴ PR × QT = RQ × PS
∴ 12 × QT = 6 × 6
∴ QT = \(\frac { 36 }{ 12 } \)
∴ QT = 3 units

Question 4.
In the adjoining figure, AP ⊥ BC, AD || BC, then find A(∆ABC) : A(∆BCD).

Solution:
Draw DQ ⊥ BC, B-C-Q.

AD || BC [Given]
∴ AP = DQ   (i)  [Perpendicular distance between two parallel lines is the same]
∆ABC and ∆BCD have same base BC.

Question 5.
In the adjoining figure, PQ ⊥ BC, AD ⊥ BC, then find following ratios.

Solution:
i. ∆PQB and tPBC have same height PQ.

ii. ∆PBC and ∆ABC have same base BC.

iii. ∆ABC and ∆ADC have same height AD.

Question 1.
Find \(\frac{A(\Delta A B C)}{A(\Delta A P Q)}\)

Solution:
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 5 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Problem Set 5 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Problem Set 5 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
Fill in the blanks using correct alternatives.

i. Seg AB is parallel to Y-axis and co-ordinates of point A are (1, 3), then co-ordinates of point B can be _______.
(A) (3,1)
(B) (5,3)
(C) (3,0)
(D) (1,-3)
Answer: (D)
Since, seg AB || Y-axis.
∴ x co-ordinate of all points on seg AB
will be the same,
x co-ordinate of A (1, 3) = 1
x co-ordinate of B (1, – 3) = 1
∴ Option (D) is correct.

ii. Out of the following, point lies to the right of the origin on X-axis.
(A) (-2,0)
(B) (0,2)
(C) (2,3)
(D) (2,0)
Answer: (D)

iii. Distance of point (-3, 4) from the origin is _________.
(A) 7
(B) 1
(C) 5
(D) -5
Answer: (C)
Distance of (-3, 4) from origin
\(\begin{array}{l}{=\sqrt{(-3)^{2}+(4)^{2}}} \\ {=\sqrt{9+16}} \\ {=\sqrt{25}=5}\end{array}\)

iv. A line makes an angle of 30° with the positive direction of X-axis. So the slope of the line is ________.
(A) \(\frac { 1 }{ 2 } \)
(B) \(\frac{\sqrt{3}}{2}\)
(C) \(\frac{1}{\sqrt{3}}\)
(D) \(\sqrt { 3 }\)
Answer: (C)

Question 2.
Determine whether the given points are collinear.
i. A (0, 2), B (1, -0.5), C (2, -3)
ii. P(1,2), Q(2,\(\frac { 8 }{ 5 } \)),R(3,\(\frac { 6 }{ 5 } \))
iii L (1, 2), M (5, 3), N (8, 6)
Solution:

∴ slope of line AB = slope of line BC
∴ line AB || line BC
Also, point B is common to both the lines.
∴ Both lines are the same.
∴ Points A, B and C are collinear.

∴ slope of line PQ = slope of line QR
∴ line PQ || line QR
Also, point Q is common to both the lines.
∴ Both lines are the same.
∴ Points P, Q and R are collinear.

∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.
[Note: Students can solve the above problems by using distance formula.]

Question 3.
Find the co-ordinates of the midpoint of the line segment joining P (0,6) and Q (12,20).
Solution:
P(x₁,y₁) = P (0, 6), Q(x₂, y₂) = Q (12, 20)
Here, x₁ = 0, y₁ = 6, x₂ = 12, y₂ = 20
∴ Co-ordinates of the midpoint of seg PQ

∴ The co-ordinates of the midpoint of seg PQ are (6,13).

Question 4.
Find the ratio in which the line segment joining the points A (3, 8) and B (-9, 3) is divided by the Y-axis.
Solution:
Let C be a point on Y-axis which divides seg AB in the ratio m : n.
Point C lies on the Y-axis
∴ its x co-ordinate is 0.
Let C = (0, y)
Here A (x₁,y₁) = A(3, 8)
B (x₂, y₂) = B (-9, 3)
∴ By section formula,


∴ Y-axis divides the seg AB in the ratio 1 : 3.

Question 5.
Find the point on X-axis which is equidistant from P (2, -5) and Q (-2,9).
Solution:
Let point R be on the X-axis which is equidistant from points P and Q.
Point R lies on X-axis.
∴ its y co-ordinate is 0.
Let R = (x, 0)
R is equidistant from points P and Q.
∴ PR = QR

∴ (x – 2)² + [0 – (-5)]² = [x – (- 2)]² + (0 – 9)² …[Squaring both sides]
∴ (x – 2)² + (5)² = (x + 2)² + (-9)²
∴ 4 – 4x + x² + 25 = 4 + 4x + x² + 81
∴ – 8x = 56
∴ x = -7
∴ The point on X-axis which is equidistant from points P and Q is (-7,0).

Question 6.
Find the distances between the following points.
i. A (a, 0), B (0, a)
ii. P (-6, -3), Q (-1, 9)
iii. R (-3a, a), S (a, -2a)
Solution:
i. Let A (x₁, y₁) and B (x₂, y₂) be the given points.
∴ x₁ = a, y₁ = 0, x₂ = 0, y₂ = a
By distance formula,

∴ d(A, B) = a\(\sqrt { 2 }\) units

ii. Let P (x₁, y₁) and Q (x₂, y₂) be the given points.
∴ x₁ = -6, y₁ = -3, x₂ = -1, y₂ = 9
By distance formula,

∴ d(P, Q) = 13 units

iii. Let R (x₁, y₁) and S (x₂, y₂) be the given points.
∴ x₁ = -3a, y₁ = a, x₂ = a, y₂ = -2a
By distance formula,

∴ d(R, S) = 5a units

Question 7.
Find the co-ordinates of the circumcentre of a triangle whose vertices are (-3,1), (0, -2) and (1,3).
Solution:
Let A (-3, 1), B (0, -2) and C (1, 3) be the vertices of the triangle.
Suppose O (h, k) is the circumcentre of ∆ABC.

∴ (h + 3)² + (k – 1)² = h² + (k + 2)²
∴ h² + 6h + 9 + k² – 2k + 1 = h² + k² + 4k + 4
∴ 6h – 2k + 10 = 4k + 4
∴ 6h – 2k – 4k = 4 – 10
∴ 6h – 6k = – 6
∴ h – k = -1 ,..(i)[Dividing both sides by 6]
OB = OC …[Radii of the same circle]

∴ h² + (k + 2)² = (h – 1)² + (k – 3)²
∴ h² + k² + 4k + 4 = h² – 2h + 1 + k² – 6k + 9
∴ 4k + 4 = -2h + 1 – 6k + 9
∴ 2h+ 10k = 6
∴ h + 5k = 3 …(ii)
Subtracting equation (ii) from (i), we get

∴ The co-ordinates of the circumcentre of the triangle are (\(\frac { -1 }{ 3 } \),\(\frac { 2 }{ 3 } \))

Question 8.
In the following examples, can the segment joining the given points form a triangle? If triangle is formed, state the type of the triangle considering sides of the triangle.
i. L (6, 4), M (-5, -3), N (-6, 8)
ii. P (-2, -6), Q (-4, -2), R (-5, 0)
iii. A(\(\sqrt { 2 }\),\(\sqrt { 2 }\)),B(-\(\sqrt { 2 }\),-\(\sqrt { 2 }\)),C(\(\sqrt { 6 }\),\(\sqrt { 6 }\))
Solution:
i. By distance formula,

∴ d(M, N) + d (L, N) > d (L, M)
∴ Points L, M, N are non collinear points.
We can construct a triangle through 3 non collinear points.
∴ The segment joining the given points form a triangle.
Since MN ≠ LN ≠ LM
∴ ∆LMN is a scalene triangle.
∴ The segments joining the points L, M and N will form a scalene triangle.

ii. By distance formula,


∴ d(P, Q) + d(Q, R) = d (P, R) …[From (iii)]
∴ Points P, Q, R are collinear points.
We cannot construct a triangle through 3 collinear points.
∴ The segments joining the points P, Q and R will not form a triangle.

iii. By distance formula,

∴ d(A, B) + d(B, C) + d(A, C) … [From (iii)]
∴ Points A, B, C are non collinear points.
We can construct a triangle through 3 non collinear points.
∴ The segment joining the given points form a triangle.
Since, AB = BC = AC
∴ ∆ABC is an equilateral triangle.
∴ The segments joining the points A, B and C will form an equilateral triangle.

Question 9.
Find k, if the line passing through points P (-12, -3) and Q (4, k) has slope \(\frac { 1 }{ 2 } \).
Solution:
P(x₁,y₁) = P(-12,-3),
Q(X₂,T₂) = Q(4, k)
Here, x₁ = -12, x₂ = 4, y₁ = -3, y₂ = k

But, slope of line PQ (m) is \(\frac { 1 }{ 2 } \) ….[Given]
∴ \(\frac { 1 }{ 2 } \) = \(\frac { k+3 }{ 16 } \)
∴ \(\frac { 16 }{ 2 } \) = k + 3
∴ 8 = k + 3
∴ k = 5
The value of k is 5.

Question 10.
Show that the line joining the points A (4,8) and B (5, 5) is parallel to the line joining the points C (2, 4) and D (1 ,7).
Proof:

∴ Slope of line AB = Slope of line CD
Parallel lines have equal slope.
∴ line AB || line CD

Question 11.
Show that points P (1, -2), Q (5, 2), R (3, -1), S (-1, -5) are the vertices of a parallelogram.
Proof:
By distance formula,

In ꠸PQRS,
PQ = RS … [From (i) and (iii)]
QR = PS … [From (ii) and (iv)]
∴ ꠸ PQRS is a parallelogram.
[A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent]
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 12.
Show that the ꠸PQRS formed by P (2, 1), Q (-1, 3), R (-5, -3) and S (-2, -5) is a rectangle.
Proof:
By distance formula,


In ꠸PQRS,
PQ = RS …[From (i) and (iii)]
QR = PS …[From (ii) and (iv)]
꠸PQRS is a parallelogram.
[A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent]

In parallelogram PQRS,
PR = QS … [From (v) and (vi)]
∴ ꠸PQRS is a rectangle.
[A parallelogram is a rectangle if its diagonals are equal]

Question 13.
Find the lengths of the medians of a triangle whose vertices are A (-1, 1), B (5, -3) and C (3,5).
Solution:

Suppose AD, BE and CF are the medians.
∴ Points D, E and F are the midpoints of sides BC, AC and AB respectively.
∴ By midpoint formula,


∴ The lengths of the medians of the triangle 5 units, 2\(\sqrt { 13 }\) units and \(\sqrt { 37 }\) units.

Question 14.
Find the co-ordinates of centroid of the triangle if points D (-7, 6), E (8, 5) and F (2, -2) are the mid points of the sides of that triangle.
Solution:

Suppose A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of the triangle.
D (-7, 6), E (8, 5) and F (2, -2) are the midpoints of sides BC, AC and AB respectively.
Let G be the centroid of ∆ABC.
D is the midpoint of seg BC.
By midpoint formula,

E is the midpoint of seg AC.
By midpoint formula,

Adding (i), (iii) and (v),
x₂ + x₃ + x₁ + x₃ + x₁ + x₂ = -14 + 16 + 4
∴ 2x₁ + 2x₂ + 2x₃ = 6
∴ x₁ + x₂ + x₃ = 3 …(vii)
Adding (ii), (iv) and (vi),
y₂ + y₃ + y₁ + y₃ + y₁ +y₂ = 12 + 10 – 4
∴ 2y₁ + 2y₂ + 2y₃ = 18
∴ y₁ + y₂ + y₃ = 9 …(viii)
G is the centroid of ∆ABC.
By centroid formula,

∴ The co-ordinates of the centroid of the triangle are (1,3).

Question 15.
Show that A (4, -1), B (6, 0), C (7, -2) and D (5, -3) are vertices of a square.
Proof:
By distance formula,


∴ □ABCD is a square.
[A rhombus is a square if its diagonals are equal]

Question 16.
Find the co-ordinates of circumcentre and radius of circumcircle of AABC if A (7, 1), B (3,5) and C (2,0) are given.
Solution:
Suppose, O (h, k) is the circumcentre of ∆ABC


∴ h² – 6h + 9 + k² – 10k + 25 = h² – 4h + 4 + k²
∴ 2h + 10k = 30
∴ h + 5k = 15 … (ii)[Dividing both sides by 2]
Multiplying equation (i) by 5, we get
25h + 5k = 115 …(iii)
Subtracting equation (ii) from (iii), we get

Substituting the value of h in equation (i), we get

∴ The co-ordinates of the circumcentre of the triangle are (\(\frac { 25 }{ 6 } \),\(\frac { 13 }{ 6 } \)) and radius of circumcircle is \(\frac{13 \sqrt{2}}{6}\) units.

Question 17.
Given A (4, -3), B (8, 5). Find the co-ordinates of the point that divides segment AB in the ratio 3:1.
Solution:
Suppose point C divides seg AB in the ratio 3:1.
Here; A(x₁, y₁) = A (4, -3)
B (x₂, y₂) = B (8, 5)
By section formula,

∴ The co-ordinates of point dividing seg AB in ratio 3 : 1 are (7, 3).

Question 18.
Find the type of the quadrilateral if points A (-4, -2), B (-3, -7), C (3, -2) and D (2, 3) are joined serially.
Solution:

Slope of AB = slope of CD
∴ line AB || line CD
slope of BC = slope of AD
∴ line BC || line AD
Both the pairs of opposite sides of ∆ABCD are parallel.
∴ ꠸ ABCD is a parallelogram.
∴ The quadrilateral formed by joining the points A, B, C and D is a parallelogram.

Question 19.
The line segment AB is divided into five congruent parts at P, Q, R and S such that A-P-Q-R-S-B. If point Q (12, 14) and S (4, 18) are given, find the co-ordinates of A, P, R, B.
Solution:

Points P, Q, R and S divide seg AB in five congruent parts.
Let A (x₁, y₁), B (x₂, y₂), P (x₃, y₃) and
R (x₄, y₄) be the given points.
Point R is the midpoint of seg QS.
By midpoint formula,
x co-ordinate of R = \(\frac { 12+4 }{ 2 } \) = \(\frac { 16 }{ 2 } \) = 8
y co-ordinate of R = \(\frac { 14+18 }{ 2 } \) = \(\frac { 32 }{ 2 } \) = 16
∴ co-ordinates of R are (8, 16).
Point Q is the midpoint of seg PR.
By midpoint formula,

∴ 28 = y₃ + 16
∴ y₃ = 12
∴ P(x₃,y₃) = (16, 12)
∴ co-ordinates of P are (16, 12).
Point P is the midpoint of seg AQ.
By midpoint formula,

∴ co-ordinates of A are (20, 10).
Point S is the midpoint of seg RB.
By midpoint formula,

∴ 36 = y₂ + 16
∴ y₂ = 20
∴ B(x₂, y₂) = (0, 20)
∴ co-ordinates of B are (0, 20).
∴ The co-ordinates of points A, P, R and B are (20, 10), (16, 12), (8, 16) and (0, 20) respectively.

Question 20.
Find the co-ordinates of the centre of the circle passing through the points P (6, -6), Q (3, -7) and R (3,3).
Solution:
Suppose O (h, k) is the centre of the circle passing through the points P, Q and R.

∴ (h – 6)² + (k + 6)² = (h – 3)² + (k + 7)²
∴ h² – 12h + 36 + k² + 12k + 36
= h² – 6h + 9 + k² + 14k + 49
∴ 6h + 2k = 14
∴ 3h + k = 7 …(i)[Dividing both sides by 2]
OP = OR …[Radii of the same circle]

∴ (h – 6)² + (k + 6)² = (h – 3)² + (k – 3)²
∴ h² – 12h + 36 + k² + 12k + 36
= h² – 6h + 9 + k² – 6k + 9
∴ 6h – 18k = 54
∴ 3h – 9k = 27 …(ii)[Dividing both sides by 2]
Subtracting equation (ii) from (i), we get

Substituting the value of k in equation (i), we get
3h – 2 = 7
∴ 3h = 9
∴ h = \(\frac { 9 }{ 3 } \) = 3
∴ The co-ordinates of the centre of the circle are (3, -2).

Question 21.
Find the possible pairs of co-ordinates of the fourth vertex D of the parallelogram, if three of its vertices are A (5, 6), B (1, -2) and C (3, -2).
Solution:

Let the points A (5, 6), B (1, -2) and C (3, -2) be the three vertices of a parallelogram.
The fourth vertex can be point D or point Di or point D₂ as shown in the figure.
Let D(x₁,y₁), D, (x₂, y₂) and D₂ (x₃,y₃).
Consider the parallelogram ACBD.
The diagonals of a parallelogram bisect each other.
∴ midpoint of DC = midpoint of AB


Co-ordinates of point D(x₁, y₁) are (3, 6).
Consider the parallelogram ABD₁C.
The diagonals of a parallelogram bisect each other.
∴ midpoint of AD₁ = midpoint of BC

∴ Co-ordinates of D₁(x₂,y₂) are (-1,-10).
Consider the parallelogram ABCD₂.
The diagonals of a parallelogram bisect each other.
∴ midpoint of BD₂ = midpoint of AC

∴ co-ordinates of point D₂ (x₃, y₃) are (7, 6).
∴ The possible pairs of co-ordinates of the fourth vertex D of the parallelogram are (3, 6), (-1,-10) and (7,6).

Question 22.
Find the slope of the diagonals of a quadrilateral with vertices A (1, 7), B (6,3), C (0, -3) and D (-3,3).
Solution:
Suppose ABCD is the given quadrilateral.

∴ The slopes of the diagonals of the quadrilateral are 10 and 0.

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers