Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 1 Basic Concepts in Geometry Set 1.1 Solutions

Question 1. Find the distances with the help of the number line given below.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक संख्या रेखा दी गई है जिस पर Q, P, K, J, H, O, A, B, C, D, E बिंदु -5 से 6 तक पूर्णांकों पर अंकित हैं। प्रत्येक बिंदु की अपनी संगत संख्या रेखा पर एक विशिष्ट निर्देशांक है।
(i) d(B, E)
(ii) d(J, A)
(iii) d(P, C)
(iv) d(J, H)
(v) d(K, O)
(vi) d(O, E)
(vii) d(P, J)
(viii) d(Q, B)
Answer: i. Co-ordinate of the point B is 2. Co-ordinate of the point E is 5. Since, 5 > 2
\( \therefore \) d(B, E) = 5 - 2
\( \therefore \) d(B, E) = 3
ii. Co-ordinate of the point J is -2. Co-ordinate of the point A is 1. Since, 1 > -2
\( \therefore \) d(J, A) = 1 - (-2) = 1 + 2
\( \therefore \) d(J, A) = 3
iii. Co-ordinate of the point P is -4. Co-ordinate of the point C is 3. Since, 3 > -4
\( \therefore \) d(P,C) = 3 - (-4) = 3 + 4
\( \therefore \) d(P,C) = 7
iv. Co-ordinate of the point J is -2. Co-ordinate of the point H is -1. Since, -1 > -2
\( \therefore \) d(J,H) = - 1 - (-2) = -1 + 2
\( \therefore \) d(J,H) = 1
v. Co-ordinate of the point K is -3. Co-ordinate of the point O is 0. Since,0 > -3
\( \therefore \) d(K, O) = 0 - (-3) = 0 + 3
\( \therefore \) d(K, O) = 3
vi. Co-ordinate of the point O is 0. Co-ordinate of the point E is 5. Since, 5 > 0
\( \therefore \) d(O, E) = 5 - 0
\( \therefore \) d(O, E) = 5
vii. Co-ordinate of the point P is -4. Co-ordinate of the point J is -2. Since -2 > -4
\( \therefore \) d(P, J) = -2 - (-4) = -2 + 4
\( \therefore \) d(P, J) = 2
viii. Co-ordinate of the point Q is -5. Co-ordinate of the point B is 2. Since,2 > -5
\( \therefore \) d(Q,B) = 2 - (-5) = 2 + 5
\( \therefore \) d(Q, B) = 7
In simple words: The distance between two points on a number line is found by subtracting the smaller coordinate from the larger coordinate. This always results in a positive value for distance.

🎯 Exam Tip: Always remember that distance is a non-negative value. When finding the distance between two points, subtract the smaller coordinate from the larger one to avoid negative results.

Question 2. If the co-ordinate of A is x and that of B is . y, find d(A, B).
(i) x = 1, y = 7
(ii) x = 6, y = -2
(iii) x = -3, y = 7
(iv) x = -4, y = -5
(v) x = -3, y = -6
(vi) x = 4, y = -8
Answer: i. Co-ordinate of point A is x = 1. Co-ordinate of point B is y = 7 Since, 7 > 1
\( \therefore \) d(A, B) = 7 - 1
\( \therefore \) d(A, B) = 6
ii. Co-ordinate of point A is x = 6. Co-ordinate of point B is y = -2. Since, 6 > -2
\( \therefore \) d(A, B) = 6 - (-2) = 6 + 2
\( \therefore \) d(A, B) = 8
iii. Co-ordinate of point A is x = -3. Co-ordinate of point B is y = 7. Since, 7 > -3
\( \therefore \) d(A, B) = 7 - (-3) = 7 + 3
\( \therefore \) d(A, B) = 10
iv. Co-ordinate of point A is x = -4. Co-ordinate of point B is y = -5. Since, -4 > -5
\( \therefore \) d(A, B) = -4 - (-5) = -4 + 5
\( \therefore \) d(A, B) = 1
v. Co-ordinate of point A is x = -3. Co-ordinate of point B is y = -6. Since, -3 > -6
\( \therefore \) d(A, B) = -3 - (-6) = -3 + 6
\( \therefore \) d(A, B) = 3
vi. Co-ordinate of point A is x = 4. Co-ordinate of point B is y = -8. Since, 4 > -8
\( \therefore \) d(A, B) = 4 - (-8) = 4 + 8
\( \therefore \)d(A, B) = 12
In simple words: The distance between two points A(x) and B(y) on a number line is given by \( |x - y| \) or \( |y - x| \), which is simply the absolute difference between their coordinates. We always subtract the smaller coordinate from the larger one to get a positive distance.

🎯 Exam Tip: When given coordinates, always identify the larger coordinate first. The distance calculation is straightforward: (larger coordinate) - (smaller coordinate).

Question 3. From the information given below, find which of the point is between the other two. If the points are not collinear, state so.
(i) d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
(ii) d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
(iii) d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
(iv) d(L, M) = 11, d(M, N) = 12, d(N, L) = 8
(v) d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
(vi) d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
Answer: i. Given, d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3 d(P, Q) = 10 ...(i) d(P, R) + d(Q, R) = 7 + 3 = 10 ... (ii)
\( \therefore \) d(P, Q) = d(P, R) + d(Q, R) ...[From (i) and (ii)]
\( \therefore \) Point R is between the points P and Q i. e., P - R - Q or Q - R - P.
\( \therefore \) Points P, R, Q are collinear.
ii. Given, d(R, S) = 8, d(S, T) = 6, d(R, T) = 4 d(R, S) = 8 ...(i) d(S, T) + d(R, T) = 6 + 4 = 10 ...(h)
\( \therefore \) d(R, S) \( \neq \) d(S, T) + d(R, T) ... [From (i) and (ii)]
\( \therefore \) The given points are not collinear.
iii. Given, d(A, B) = 16, d(C, A) = 9, d(B, C) = 7 d(A, B) = 16 ...(i) d(C, A) + d(B, C) = 9 + 7 = 16 ...(ii)
\( \therefore \) d(A, B) = d(C, A) + d(B, C) ...[From(i) and (ii)]
\( \therefore \) Point C is between the points A and B. i. e., A - C - B or B - C - A.
\( \therefore \) Points A, C, B are collinear
iv. Given, d(L, M) = 11, d(M, N) = 12, d(N, L) = 8 d(M, N) = 12 ...(i) d(L, M) + d(N, L) = 11 + 8 = 19 ...(ii)
\( \therefore \)d(M, N) \( \neq \) d(L, M) + d(N, L) ... [From (i) and (ii)]
\( \therefore \) The given points are not collinear.
v. Given, d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8 d(X, Y) = 15 ...(i) d(X,Z) + d(Y, Z) = 8 + 7 = 15 ...(ii)
\( \therefore \) d(X, Y) = d(X, Z) + d(Y, Z) ...[From (i) and (ii)]
\( \therefore \) Point Z is between the points X and Y i. e.,X - Z - Y or Y - Z - X.
\( \therefore \) Points X, Z, Y are collinear.
vi. Given, d(D, E) = 5, d(E, F) = 8, d(D, F) = 6 d(E, F) = 8 ...(i) d(D, E) + d(D, F) = 5 + 6 = 11 ...(ii)
\( \therefore \) d(E, F) \( \neq \) d(D, E) + d(D, F) ... [From (i) and (ii)]
\( \therefore \) The given points are not collinear.
In simple words: To check collinearity and determine which point lies between the other two, we use the segment addition postulate. If the sum of the distances between two pairs of points equals the distance of the third pair (i.e., d(A,C) + d(C,B) = d(A,B)), then the points are collinear, and the common point (C) lies between the other two (A and B). If the sum is not equal, the points are not collinear.

🎯 Exam Tip: When evaluating collinearity, identify the largest distance first. Then, check if the sum of the other two distances equals this largest distance. If it does, the point common to the two smaller distances lies in between.

Question 4. On a number line, points A, B and C are such that d(A, C) = 10, d(C, B) = 8. Find d(A, B) considering all possibilities.
Answer: Solution: Given, d(A, C) = 10, d(C, B) = 8. Case I: Points A, B, C are such that, A - B - C.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस स्थिति में, बिंदु B, A और C के बीच में स्थित है। d(B, C) को 8 के रूप में दर्शाया गया है और d(A, C) को 10 के रूप में दर्शाया गया है, जहां A, B, C एक ही रेखा पर हैं।
\( \therefore \) d(A, C) = d(A, B) + d(B, C)
\( \therefore \) 10 = d(A, B) + 8
\( \therefore \) d(A, B) = 10 - 8
\( \therefore \) d(A, B) = 2
Case II: Points A, B, C are such that, A - C - B.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस स्थिति में, बिंदु C, A और B के बीच में स्थित है। d(A, C) को 10 के रूप में दर्शाया गया है और d(C, B) को 8 के रूप में दर्शाया गया है, जहां A, C, B एक ही रेखा पर हैं।
\( \therefore \) d(A, B) = d(A, C) + d(C, B) = 10 + 8
\( \therefore \) d(A, B) = 18
Case III: Points A, B, C are such that, B - A - C.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस स्थिति में, बिंदु A, B और C के बीच में स्थित है। d(A, C) को 10 के रूप में दर्शाया गया है और d(B, C) को 8 के रूप में दर्शाया गया है, जहां B, A, C एक ही रेखा पर हैं। From the diagram, d (A, C) > d(B, C) Which is not possible
\( \therefore \) Point A is not between B and C.
\( \therefore \) d(A, B) = 2 or d(A, B) = 18.
In simple words: When three points are collinear, there are three possible arrangements. We use the segment addition postulate (d(P,R) = d(P,Q) + d(Q,R)) to find unknown distances by considering which point lies between the other two for each possibility. Sometimes, one arrangement might be impossible if it contradicts given distance values.

🎯 Exam Tip: Always consider all possible arrangements of three collinear points (A-B-C, A-C-B, B-A-C). Sketching diagrams for each case helps visualize the relationships and apply the segment addition postulate correctly.

Question 5. Points X, Y, Z are collinear such that d(X, Y) = 17, d(Y, Z) = 8, find d(X, Z).
Answer: Solution: Given,d(X, Y) = 17, d(Y, Z) = 8 Case I: Points X, Y, Z are such that, X - Y - Z.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस स्थिति में, बिंदु Y, X और Z के बीच में स्थित है। d(X, Y) को 17 के रूप में दर्शाया गया है और d(Y, Z) को 8 के रूप में दर्शाया गया है, जहां X, Y, Z एक ही रेखा पर हैं।
\( \therefore \) d(X, Z) = d(X, Y) + d(Y, Z) = 17 + 8
\( \therefore \) d(X, Z) = 25
Case II: Points X, Y, Z are such that, X - Z - Y.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस स्थिति में, बिंदु Z, X और Y के बीच में स्थित है। d(Z, Y) को 8 के रूप में दर्शाया गया है और d(X, Y) को 17 के रूप में दर्शाया गया है, जहां X, Z, Y एक ही रेखा पर हैं।
\( \therefore \) d(X,Y) = d(X,Z) + d(Z,Y)
\( \therefore \) 17 = d(X, Z) + 8
\( \therefore \) d(X, Z) = 17 - 8
\( \therefore \) d(X, Z) = 9
Case III: Points X, Y, Z are such that, Z - X - Y.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस स्थिति में, बिंदु X, Z और Y के बीच में स्थित है। d(X, Y) को 17 के रूप में दर्शाया गया है और d(Z, Y) को 8 के रूप में दर्शाया गया है, जहां Z, X, Y एक ही रेखा पर हैं। From the diagram, d(X, Y) > d (Y, Z) Which is not possible
\( \therefore \) Point X is not between Z and Y.
\( \therefore \) d(X, Z) = 25 or d(X, Z) = 9.
In simple words: For three collinear points, there are three possible orderings. We apply the segment addition postulate for each valid ordering to find the unknown distance. If an ordering leads to a contradiction (like a segment being longer than the total distance it's part of), that ordering is invalid.

🎯 Exam Tip: Always analyze all three possible arrangements of collinear points (X-Y-Z, X-Z-Y, Z-X-Y). Carefully check if any arrangement contradicts the given distances, ruling out impossible scenarios.

Question 6. Sketch proper figure and write the answers of the following questions.
(i) If A - B - C and l(AC) = 11, l(BC) = 6.5, then l(AB) = ?
(ii) If R - S - T and l(ST) = 3.7, l(RS) = 2.5, then l(RT) = ?
(iii) If X - Y - Z and l(XZ) = \( 3\sqrt{7} \), l(XY) = \( \sqrt{7} \), then l(YZ) = ?
Answer: i. Given, l(AC) =11, l(BC) = 6.5
ℹ️ चित्र व्याख्या (Diagram Explanation): एक रेखाखंड A-B-C है जहाँ बिंदु B, A और C के बीच में स्थित है। A से C तक की कुल लंबाई 11 है, और B से C तक की लंबाई 6.5 है। l(AC) = l(AB) + l(BC) ... [A - B - C]
\( \therefore \) 11 = l(AB) + 6.5
\( \therefore \) l(AB) = 11 - 6.5
\( \therefore \) l(AB) = 4.5
ii. Given, l(ST) = 3.7, l(RS) = 2.5
ℹ️ चित्र व्याख्या (Diagram Explanation): एक रेखाखंड R-S-T है जहाँ बिंदु S, R और T के बीच में स्थित है। R से S तक की लंबाई 2.5 है, और S से T तक की लंबाई 3.7 है। l(RT) = l(RS) + l(ST) ... [R - S - T] = 2.5 + 3.7
\( \therefore \) (RT) = 6.2
iii. l(XZ) = \( 3\sqrt{7} \), l(XY) = \( \sqrt{7} \),
ℹ️ चित्र व्याख्या (Diagram Explanation): एक रेखाखंड X-Y-Z है जहाँ बिंदु Y, X और Z के बीच में स्थित है। X से Z तक की कुल लंबाई \( 3\sqrt{7} \) है, और X से Y तक की लंबाई \( \sqrt{7} \) है। l(XZ) = l(XY) + l(YZ) ... [X - Y - Z]
\( \therefore \) \( 3\sqrt{7} \) \( \implies \) \( \sqrt{7} \) + l(YZ)
\( \therefore \) l(YZ) = \( 3\sqrt{7} \) - \( \sqrt{7} \)
\( \therefore \) l(YZ) = \( 2\sqrt{7} \)
In simple words: When points are collinear and their order is given (e.g., A-B-C), we can use the segment addition postulate. This means the length of the total segment (AC) is equal to the sum of the lengths of the smaller segments (AB + BC). We can then use this equation to find any unknown segment lengths.

🎯 Exam Tip: Always draw a simple line segment diagram for each question part to visualize the point order. This helps in correctly applying the segment addition postulate (e.g., \( l(AC) = l(AB) + l(BC) \) for A-B-C).

Question 7. Which figure is formed by three non-collinear points?
Answer: Solution: Three non-collinear points form a triangle.
In simple words: If you connect three points that do not lie on the same straight line, the shape you create is a triangle.

🎯 Exam Tip: Understand the fundamental definition: "non-collinear" means the points do not lie on a single straight line. This basic concept is key to understanding various geometric figures.