π Coordinate Geometry (Chapter 1) CBSE Class 9
Ganita Manjari | Complete Solutions with Detailed Explanation
π Exercise Set 1.1 (Based on Fig. 1.3)
Q1 Fig. 1.3 shows Reiaan’s room with points OABC marking its corners.
The x- and y-axes are marked in the figure.
Point O is the origin.
Referring to Fig. 1.3, answer the following questions:
The x- and y-axes are marked in the figure.
Point O is the origin.
Referring to Fig. 1.3, answer the following questions:
Q1(i) If D1R1 represents the door to Reiaan’s room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?
✅ Detailed Explanation: The door lies on the x-axis (horizontal line).
Distance from x-axis = 0 (since on axis).
From figure, D₁ = (8,0).
Distance from y-axis = x-coordinate = 8 units.
Left wall is the y-axis.
Distance from x-axis = 0 (since on axis).
From figure, D₁ = (8,0).
Distance from y-axis = x-coordinate = 8 units.
Left wall is the y-axis.
Q1(ii) What are the coordinates of D1?
Answer: D₁ = (8, 0).
It is 8 units right of origin on x-axis.
It is 8 units right of origin on x-axis.
Q1(iii) If R1 is the point (11.5, 0), how wide is the door? Do you think this is
a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?
Explanation: Width = |11.5 – 8| = 3.5 units.
Assuming 1 unit = 1 ft → 3.5 ft = 42 inches.
Standard room door: 30–36 inches.
42" is wider than average → comfortable.
Wheelchair needs ≥32 inches (≈2.7 ft).
Since 3.5 > 2.7,
yes, wheelchair accessible easily.
Assuming 1 unit = 1 ft → 3.5 ft = 42 inches.
Standard room door: 30–36 inches.
42" is wider than average → comfortable.
Wheelchair needs ≥32 inches (≈2.7 ft).
Since 3.5 > 2.7,
yes, wheelchair accessible easily.
Q1(iv) If B1 (0, 1.5) and B2 (0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?
Solution: Bathroom width = 4 – 1.5 = 2.5 units.
Room door = 3.5 units.
2.5 is less than 3.5 → bathroom door is narrower
Room door = 3.5 units.
2.5 is less than 3.5 → bathroom door is narrower
π Think and Reflect (Real-world Observation)
π Question 1: What are the standard widths for a room door? Look around your home and in school.
✅ Detailed Answer:
| Location | Typical Width | Notes |
|---|---|---|
| Home (internal rooms) | 30–36 inches (2.5–3 ft) | Common in Indian homes |
| Home (main entrance) | 36–42 inches (3–3.5 ft) | Wider for furniture movement |
| School (classroom) | 36–48 inches (3–4 ft) | Must accommodate student flow |
| School (office/staff) | 30–36 inches (2.5–3 ft) | Standard internal door |
| International Standard (Wheelchair) | 32 inches (≈2.7 ft) minimum | ADA / Accessibility guidelines |
Conclusion: Standard room doors in Indian homes are typically 30–36 inches.
School doors are often wider at 36–48 inches.
International accessibility standards recommend a minimum of 32 inches clear width for wheelchair passage.
♿ Question 2: Are the doors in your school suitable for people in wheelchairs?
✅ Detailed Answer:
This depends on the specific school. Many modern schools in India now follow accessibility norms under the Rights of Persons with Disabilities Act (RPWD) 2016 and CBSE inclusive education guidelines.
✅ Suitable features (good schools):
- Door width ≥ 32 inches (2.7 ft) – allows wheelchair passage
- Ramps instead of steps at entrances
- Door handles at accessible height (not too high)
- Lightweight doors or automatic openers
- Thresholds that are flat or have small ramps
⚠️ Challenges in older schools:
- Doors may be only 24–28 inches wide – too narrow for wheelchairs
- Steps at entrance without ramps
- Heavy wooden doors difficult to open from a wheelchair
- Door handles placed too high
π Conclusion: While accessibility is improving across India, not all school doors are currently wheelchair suitable.
Students should observe their own school and suggest improvements if needed.
π Think and Reflect (Page 7 - Coordinate Concepts)
Q1 What is the x-coordinate of a point on the y-axis?
Answer: The x-coordinate is always 0.
Explanation: Any point on the y-axis has the form (0, y). Examples: (0, 5), (0, -3), (0, 0).
Explanation: Any point on the y-axis has the form (0, y). Examples: (0, 5), (0, -3), (0, 0).
Q2 Is there a similar generalisation for a point on the x-axis?
Answer: Yes. The y-coordinate of any point on the x-axis is always 0.
Explanation: Any point on the x-axis has the form (x, 0). Examples: (4, 0), (-7, 0).
Explanation: Any point on the x-axis has the form (x, 0). Examples: (4, 0), (-7, 0).
Q3 Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.
Answer: Yes, but only when x = y.
Justification: For two ordered pairs to be equal, their first coordinates must be equal AND their second coordinates must be equal. So x = y and y = x → both give x = y.
Example: If x = 3 and y = 3, then P(3,3) and Q(3,3) coincide. If x = 2 and y = 5, then P(2,5) and Q(5,2) are different points.
Justification: For two ordered pairs to be equal, their first coordinates must be equal AND their second coordinates must be equal. So x = y and y = x → both give x = y.
Example: If x = 3 and y = 3, then P(3,3) and Q(3,3) coincide. If x = 2 and y = 5, then P(2,5) and Q(5,2) are different points.
Q4 If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true?
Answer: Yes, the claim is true.
Proof:
• If x ≠ y, then the ordered pairs differ in at least one coordinate → (x,y) ≠ (y,x).
• If (x,y) = (y,x), then equating first coordinates gives x = y.
• If x = y, then both pairs become (x,x) → they are equal.
Key insight: Ordered pairs are order-sensitive. The point (y,x) is generally the reflection of (x,y) across the line y = x.
Proof:
• If x ≠ y, then the ordered pairs differ in at least one coordinate → (x,y) ≠ (y,x).
• If (x,y) = (y,x), then equating first coordinates gives x = y.
• If x = y, then both pairs become (x,x) → they are equal.
Key insight: Ordered pairs are order-sensitive. The point (y,x) is generally the reflection of (x,y) across the line y = x.
✍️ Exercise Set 1.2 (Study table, bathroom, dining room)
Q1 On a graph sheet, mark the x-axis and y-axis and the origin O.
Mark points from (– 7, 0) to (13, 0) on the x-axis and from (0, – 15) to (0, 12) on the y-axis.
(Use the scale 1 cm = 1 unit.)
Using Fig. 1.5, answer the given questions
Mark points from (– 7, 0) to (13, 0) on the x-axis and from (0, – 15) to (0, 12) on the y-axis.
(Use the scale 1 cm = 1 unit.)
Using Fig. 1.5, answer the given questions
1(i) Place Reiaan’s rectangular study table with three of its feet at the
points (8, 9), (11, 9) and (11, 7). (i)Where will the fourth foot of the table be?
i.The given three points form three corners of a rectangle:
A =(8,9)
B = (11, 9)
C=(11,7)
To complete the rectangle, the fourth point must have:
- same x-coordinate as A- 8
- same y-coordinate as C -> 7
Therefore, the fourth foot is at: (8,7) Reason: i.e Rectangle: x-coordinate same as A(8) and y same as C(7) → (8,7).
A =(8,9)
B = (11, 9)
C=(11,7)
To complete the rectangle, the fourth point must have:
- same x-coordinate as A- 8
- same y-coordinate as C -> 7
Therefore, the fourth foot is at: (8,7) Reason: i.e Rectangle: x-coordinate same as A(8) and y same as C(7) → (8,7).
1(ii) (ii) Is this a good spot for the table?
Answer: Yes, placed against the wall, does not block movement.because:
. The table is placed neatly inside the room.
. It does not block doors or pathways.
. It is positioned near the wall, which is practical for study.
. The table is placed neatly inside the room.
. It does not block doors or pathways.
. It is positioned near the wall, which is practical for study.
1(iii) What is the width of the table? The length? Can you make out
the height of the table?
ANSWER Width = 11-8 = 3 units;
Length = 9-7 = 2 units. Height cannot be known (2D top view only).
Length = 9-7 = 2 units. Height cannot be known (2D top view only).
2 If the bathroom door has a hinge at B1
and opens into the bedroom,
will it hit the wardrobe? Are there any changes you would suggest
if the door is made wider?
From Fig. 1.5:- BΔ± = (0, 1.5)
- B2 = (0, 4)
So, the bathroom door has width: =4-1.5 = 2.5 units
If the door is hinged at BI and opens into the bedroom, it will sweep an arc of radius 2.5 units from B1.
Now the wardrobe begins at:
W := (3,0)
W4 =(3,2)
The nearest point of the wardrobe from B:(0, 1.5) is around x = 3, which is farther than the door width 2.5 units.
Therefore, the bathroom door will not hit the wardrobe.
Door width = 2.5 units. Wardrobe starts at x=3 (>2.5) → will not hit. If made wider, shift wardrobe or make door open inward.
Suggestion if the door is made wider:
. If the door becomes much wider, it may come close to or hit the wardrobe.
. In that case, the door could be made to open inward into the bathroom, or
. the wardrobe could be shifted slightly to the right or
. the door width could be kept limited for comfortable movement.
- B2 = (0, 4)
So, the bathroom door has width: =4-1.5 = 2.5 units
If the door is hinged at BI and opens into the bedroom, it will sweep an arc of radius 2.5 units from B1.
Now the wardrobe begins at:
W := (3,0)
W4 =(3,2)
The nearest point of the wardrobe from B:(0, 1.5) is around x = 3, which is farther than the door width 2.5 units.
Therefore, the bathroom door will not hit the wardrobe.
Door width = 2.5 units. Wardrobe starts at x=3 (>2.5) → will not hit. If made wider, shift wardrobe or make door open inward.
Suggestion if the door is made wider:
. If the door becomes much wider, it may come close to or hit the wardrobe.
. In that case, the door could be made to open inward into the bathroom, or
. the wardrobe could be shifted slightly to the right or
. the door width could be kept limited for comfortable movement.
3(i) Look at Reiaan’s bathroom.
(i) What are the coordinates of the four corners O, F, R, and P of the bathroom?
From Fig 1.5: O(0,9), F(0,0), R(-6,9), P(-6,0).
3(ii) (ii) What is the shape of the showering area SHWR in Reiaan’s
bathroom? Write the coordinates of the four corners.
Since one pair of opposite sides is parallel, SHWR is a trapezium.
Shape of SHWR = Trapezium
Coordinates of the corners:
Shape = Trapezium. S(-6,5), H(-3,5), W(-2,9), R(-6,9).
Shape of SHWR = Trapezium
Coordinates of the corners:
Shape = Trapezium. S(-6,5), H(-3,5), W(-2,9), R(-6,9).
3(iii) (iii) Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft
space for the toilet. Write the coordinates of the corners of these spaces.
Washbasin space (3 ft × 2 ft):Take the rectangle at the bottom-left corner of the bathroom.
Coordinates of Corners:(-6, 0), (-3, 0), (-3, 2) and (-6, 2)
Toilet space (2 ft × 3 ft):Take a rectangle above the washbasin.
Coordinates of Corners:(-6, 2), (-4, 2), (-4, 5) and (-6, 5)
Coordinates of Washbasin corners:(-6, 0), (-3, 0), (-3, 2) and (-6, 2)
Coordinates of Toilet corners:(-6, 2), (-4, 2), (-4, 5) and (-6, 5)
..Washbasin: (-6,0), (-3,0), (-3,2), (-6,2). Toilet: (-6,2), (-4,2), (-4,5), (-6,5).
Coordinates of Corners:(-6, 0), (-3, 0), (-3, 2) and (-6, 2)
Toilet space (2 ft × 3 ft):Take a rectangle above the washbasin.
Coordinates of Corners:(-6, 2), (-4, 2), (-4, 5) and (-6, 5)
Coordinates of Washbasin corners:(-6, 0), (-3, 0), (-3, 2) and (-6, 2)
Coordinates of Toilet corners:(-6, 2), (-4, 2), (-4, 5) and (-6, 5)
..Washbasin: (-6,0), (-3,0), (-3,2), (-6,2). Toilet: (-6,2), (-4,2), (-4,5), (-6,5).
4(i) 4. Other rooms in the house:(i) Reiaan’s room door leads from the dining room which has
the length 18 ft and width 15 ft. The length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners.
From Fig. 1.5:Coordinates of P =(-6,0)Coordinates of A = (12, 0)
So, the length of PA = 12 - (-6) = 18 ft, which matches the given length.
If the dining room is 15 ft wide and lies below PA, then its upper side is PA and it extends downward 15 units.
Hence the coordinates of four corners are:
. P=(-6, 0)
· A=(12, 0)
. Q= (12,-15)
. S = (-6, -15)
The coordinates of the dining room corners are:(-6, 0), (12, 0), (12, -15), (-6, -15)
So, the length of PA = 12 - (-6) = 18 ft, which matches the given length.
If the dining room is 15 ft wide and lies below PA, then its upper side is PA and it extends downward 15 units.
Hence the coordinates of four corners are:
. P=(-6, 0)
· A=(12, 0)
. Q= (12,-15)
. S = (-6, -15)
The coordinates of the dining room corners are:(-6, 0), (12, 0), (12, -15), (-6, -15)
4(ii) (ii)
Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.
The dining room extends:
From x = -6 to x = 12
From y = 0 to y= -15
Centre of the dining room:
x-coordinate of centre = (-6 + 12)/2 = 3
y-coordinate of centre = (0 + (-15))/2 = -7.5
Now place a 5 ft x 3 ft table at the centre.
Taking length = 5 units along the x-axis and width = 3 units along the y-axis:
Half-length = 2.5
Half-width = 1.5
So the coordinates of corners (feet) of the table are:
· (3 - 2.5, -7.5 - 1.5) = (0.5, -9)
· (3 + 2.5, -7.5 - 1.5) = (5.5, -9)
· (3 + 2.5, -7.5 + 1.5) = (5.5, -6)
· (3 - 2.5, -7.5 + 1.5) = (0.5, -6)
The coordinates of the four feet of the dining table are:(0.5, -9), (5.5, -9), (5.5, -6), (0.5, -6)
P(-6,0), A(12,0), Q(12,-15), S(-6,-15)Centre (3, -7.5).
Feet: (0.5,-9), (5.5,-9), (5.5,-6), (0.5,-6).
From x = -6 to x = 12
From y = 0 to y= -15
Centre of the dining room:
x-coordinate of centre = (-6 + 12)/2 = 3
y-coordinate of centre = (0 + (-15))/2 = -7.5
Now place a 5 ft x 3 ft table at the centre.
Taking length = 5 units along the x-axis and width = 3 units along the y-axis:
Half-length = 2.5
Half-width = 1.5
So the coordinates of corners (feet) of the table are:
· (3 - 2.5, -7.5 - 1.5) = (0.5, -9)
· (3 + 2.5, -7.5 - 1.5) = (5.5, -9)
· (3 + 2.5, -7.5 + 1.5) = (5.5, -6)
· (3 - 2.5, -7.5 + 1.5) = (0.5, -6)
The coordinates of the four feet of the dining table are:(0.5, -9), (5.5, -9), (5.5, -6), (0.5, -6)
P(-6,0), A(12,0), Q(12,-15), S(-6,-15)Centre (3, -7.5).
Feet: (0.5,-9), (5.5,-9), (5.5,-6), (0.5,-6).
π Think and Reflect (Page 9 - Distance using Baudhayana-Pythagoras)
Q1 In moving from A (3, 4) to D (7, 1), what distance has been covered
along the x-axis? What about the distance along the y-axis?
Answer:
• Distance along x-axis: 7 − 3 = 4 units (moving right)
• Distance along y-axis: 4 − 1 = 3 units (moving down)
Note: We take absolute values, so direction doesn't matter for distance.
• Distance along x-axis: 7 − 3 = 4 units (moving right)
• Distance along y-axis: 4 − 1 = 3 units (moving down)
Note: We take absolute values, so direction doesn't matter for distance.
Q2 Can these distances help you find the distance AD?
Answer: Yes! Using the Baudhayana-Pythagoras Theorem:
AD = √[(x-distance)² + (y-distance)²] = √[(4)² + (3)²] = √[16 + 9] = √25 = 5 units.
The horizontal and vertical distances form the legs of a right triangle, and AD is the hypotenuse.
AD = √[(x-distance)² + (y-distance)²] = √[(4)² + (3)²] = √[16 + 9] = √25 = 5 units.
The horizontal and vertical distances form the legs of a right triangle, and AD is the hypotenuse.
π Think and Reflect (Page 11 - Reflection in Axes)
Q1 What has remained the same and what has changed with this reflection?
Answer:
✓ Remained the same: y-coordinates, side lengths (AD, DM, MA), shape, size, area, congruence.
✗ Changed: x-coordinates change sign (positive ↔ negative), triangle flips horizontally, position/quadrant changes.
✓ Remained the same: y-coordinates, side lengths (AD, DM, MA), shape, size, area, congruence.
✗ Changed: x-coordinates change sign (positive ↔ negative), triangle flips horizontally, position/quadrant changes.
Q2 Would these observations be the same if ΞADM is reflected in the x-axis (instead of the y-axis)?
Answer: Yes, the observations are similar, but the roles swap:
• Reflection in y-axis: x-coordinates change sign; y-coordinates unchanged.
• Reflection in x-axis: y-coordinates change sign; x-coordinates unchanged.
• In both cases: side lengths, shape, size, area, and congruence are preserved.
• Reflection in y-axis: x-coordinates change sign; y-coordinates unchanged.
• Reflection in x-axis: y-coordinates change sign; x-coordinates unchanged.
• In both cases: side lengths, shape, size, area, and congruence are preserved.
π End-of-Chapter Exercises (Full Solutions)
1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
Origin (0,0).
2. Point W has x-coordinate equal to – 5.
Can you predict the coordinates of point H which is on the line through W parallel to the y-axis?
Which quadrants can H lie in?
Can you predict the coordinates of point H which is on the line through W parallel to the y-axis?
Which quadrants can H lie in?
H = (-5, y).
QII if (y>0), QIII if (y<0 axis="" if="" on="" y="0.</div">
QII if (y>0), QIII if (y<0 axis="" if="" on="" y="0.</div">
3. Consider the points R (3, 0), A (0, – 2), M (– 5, – 2) and P (– 5, 2). If they are joined in the same order, predict:
(i) Two sides of RAMP that are perpendicular to each other.
(ii) One side of RAMP that is parallel to one of the axes.
(iii) Two points that are mirror images of each other in one axis.
Which axis will this be? Now plot the points and verify your predictions.
(ii) One side of RAMP that is parallel to one of the axes.
(iii) Two points that are mirror images of each other in one axis.
Which axis will this be? Now plot the points and verify your predictions.
Perpendicular: AM ⟂ MP.
.Parallel to axis: AM ∥ x-axis.
Mirror: M and P about x-axis.
.Parallel to axis: AM ∥ x-axis.
Mirror: M and P about x-axis.
4. Plot point Z (5, – 6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.
(Comment: Answers may differ from person to person.)
Point Z is (5, -6).
To form a right-angled triangle easily, take:
I = (5, 0) on the x-axis
N = (0, -6) on the y-axis
Then triangle IZN is right-angled at Z? Let's check:
IZ is vertical
ZN is horizontal
So, triangle IZN is right-angled at Z.
Coordinates of the points:
I = (5,0)
Z= (5, -6)
N = (0, -6)
Now find the lengths of the sides:
1. IZ: Distance between (5, 0) and (5, -6)= 0 - (-6) = 6 units
2. ZN:Distance between (5, -6) and (0, -6)= 5 - 0 = 5 units
3. IN:Using distance formula:IN = √[(5-0)² +(0-(-6))²]= √(5²+6²)= √(25+36)= √61 units
I = (5, 0)
Z= (5, -6)
N =(0,-6)
Lengths of the sides: IZ = 6 units
ZN = 5 units
IN = √61 units
(Activity) Example: I(5,0) → IZ=6, ZN depends on N. Students verify.
To form a right-angled triangle easily, take:
I = (5, 0) on the x-axis
N = (0, -6) on the y-axis
Then triangle IZN is right-angled at Z? Let's check:
IZ is vertical
ZN is horizontal
So, triangle IZN is right-angled at Z.
Coordinates of the points:
I = (5,0)
Z= (5, -6)
N = (0, -6)
Now find the lengths of the sides:
1. IZ: Distance between (5, 0) and (5, -6)= 0 - (-6) = 6 units
2. ZN:Distance between (5, -6) and (0, -6)= 5 - 0 = 5 units
3. IN:Using distance formula:IN = √[(5-0)² +(0-(-6))²]= √(5²+6²)= √(25+36)= √61 units
I = (5, 0)
Z= (5, -6)
N =(0,-6)
Lengths of the sides: IZ = 6 units
ZN = 5 units
IN = √61 units
(Activity) Example: I(5,0) → IZ=6, ZN depends on N. Students verify.
5.What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?
Only Quadrant I and positive axes. Cannot locate QII, QIII, QIV.
6. Are the points M (– 3, – 4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.
M (-3, -4), A (0, 0), and G (6, 8) lie on the same straight line
Distance formula:d=√[(X2-X1)²+(y2-y1)²]
MA= √[(0+3)² + (0+4)²]= √(3² + 4²)= √(9+16) = √25 = 5
AG= √[(6-0)²+(8-0)²]= √(36+64)= √100=10
MG =√[(6+3)² + (8+4)²]= √(9² + 12²)= √(81 +144)= √225 =15
MA + AG = 5 + 10 = 15 = MG
MA=5, AG=10, MG=15 → MA+AG=MG →The sum of two distances equals the third, the points M, A and G lie on the same straight Line is collinear.
Distance formula:d=√[(X2-X1)²+(y2-y1)²]
MA= √[(0+3)² + (0+4)²]= √(3² + 4²)= √(9+16) = √25 = 5
AG= √[(6-0)²+(8-0)²]= √(36+64)= √100=10
MG =√[(6+3)² + (8+4)²]= √(9² + 12²)= √(81 +144)= √225 =15
MA + AG = 5 + 10 = 15 = MG
MA=5, AG=10, MG=15 → MA+AG=MG →The sum of two distances equals the third, the points M, A and G lie on the same straight Line is collinear.
7. Use your method (from Problem 6) to check if the points
R (– 5, – 1), B (– 2, – 5) and C (4, – 12) are on the same straight line.
Now plot both sets of points and check your answers.
RB= √[(-2+5)² + (-5+1)²]=√(3²+(-4)²)= √(9 +16) = √25 = 5
BC = √[(4 + 2)² + (-12 + 5)²]= √(6²+(-7)²)= √(36 +49) = √85
RC= √[(4+5)² + (-12 + 1)²]= √(9²+(-11)²)=√(81+121)= √202
RB + BC= 5 + √85 ,
RB=5, BC=√85≈9.22,
RC=√202≈14.21 → 5+9.22≠14.21 → not collinear.
BC = √[(4 + 2)² + (-12 + 5)²]= √(6²+(-7)²)= √(36 +49) = √85
RC= √[(4+5)² + (-12 + 1)²]= √(9²+(-11)²)=√(81+121)= √202
RB + BC= 5 + √85 ,
RB=5, BC=√85≈9.22,
RC=√202≈14.21 → 5+9.22≠14.21 → not collinear.
8(i)Using the origin as one vertex, plot the vertices of:(i) A right-angled isosceles triangle.
O(0,0), A(4,0), B(0,4).
8 (ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV
O(0,0), P(-3,-4), Q(3,-4).
OP=OQ=5.
OP=OQ=5.
9. The following table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment
ST. Justify your answer When M is the mid-point of ST, can you find any connection
between the coordinates of M, S and T?
Question 9: Midpoint Verification Table
| S | M | T | Is M the midpoint of ST? (Yes/No) | Reason for your answer |
|---|---|---|---|---|
| (-3, 0) | (0, 0) | (3, 0) | Yes ✓ | SM = 3, MT = 3 → SM = MT |
| (2, 3) | (3, 4) | (4, 5) | Yes ✓ | SM = √2, MT = √2 → SM = MT |
| (0, 0) | (0, 5) | (0, -10) | No ✗ | SM = 5, MT = 15 → SM ≠ MT |
| (-8, 7) | (0, -2) | (6, -3) | No ✗ | SM = √145, MT = √37 → SM ≠ MT |
Follow-up .
Answer: Yes! When M is the midpoint of ST:
• M's x-coordinate = (x-coordinate of S + x-coordinate of T) / 2
• M's y-coordinate = (y-coordinate of S + y-coordinate of T) / 2
Midpoint Formula: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Example: For S(-3,0) and T(3,0): ((-3+3)/2, (0+0)/2) = (0,0) = M ✓
• M's x-coordinate = (x-coordinate of S + x-coordinate of T) / 2
• M's y-coordinate = (y-coordinate of S + y-coordinate of T) / 2
Midpoint Formula: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Example: For S(-3,0) and T(3,0): ((-3+3)/2, (0+0)/2) = (0,0) = M ✓
(-3,0)(0,0)(3,0)→Yes; (2,3)(3,4)(4,5)→Yes; (0,0)(0,5)(0,-10)→No; (-8,7)(0,-2)(6,-3)→No.
10. Use the connection you found to find the coordinates of B given that M (–7, 1) is the midpoint of A (3, – 4) and B (x, y).
x = -17, y = 6 → B(-17,6).
11. Let P, Q be points of trisection of AB, with P closer to A, and Q
closer to B. Using your knowledge of how to find the coordinates
of the midpoint of a segment, how would you find the coordinates
of P and Q? Do this for the case when the points are A (4, 7) and
B (16, –2).
P(8,4), Q(12,1).
12.(i) Given the points A (1, – 8), B (– 4, 7) and C (–7, – 4), show that
they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K?
(ii) Given the points D (– 5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle, or outside the circle K.
Radius = √65 ≈8.06.
D(-5,6): √61 LESS THAN √65 → inside.
E(0,9): 9 GREATER THAN √65 → outside.
D(-5,6): √61 LESS THAN √65 → inside.
E(0,9): 9 GREATER THAN √65 → outside.
13. The midpoints of the sides of triangle ABC are the points D, E, and
F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and
(0, 3), respectively, find the coordinates of A, B and C.
A(1,-1), B(11,3), C(-5,7).
14. A city has two main roads which cross each other at the centre
of the city. These two roads are along the North–South (N–S)
direction and East–West (E–W) direction. All the other streets of
the city run parallel to these roads and are 200 m apart. There are
10 streets in each direction.
(i) Using 1 cm = 200 m, draw a model of the city in your notebook.
Represent the roads/streets by single lines.
(ii) There are street intersections in the model. Each street
intersection is formed by two streets — one running in the
N–S direction and another in the E–W direction. Each street
intersection is referred to in the following manner: If the
second street running in the N–S direction and 5th street in
the E–W direction meet at some crossing, then we call this
street intersection (2, 5). Using this convention, find:
(a) how many street intersections can be referred to as (4, 3).
(b) how many street intersections can be referred to as (3, 4).
Both refer to exactly 1 unique crossing each.
15. A computer graphics program displays images on a rectangular
screen whose coordinate system has the origin at the
bottom-left corner. The screen is 800 pixels wide and 600 pixels
high. A circular icon of radius 80 pixels is drawn with its centre at
the point A (100, 150). Another circular icon of radius 100 pixels is
drawn with its centre at the point B (250, 230). Determine:
(i) whether any part of either circle lies outside the screen.
(ii) whether the two circles intersect each other.
Both circles fully inside (check edges).
Distance AB = 170;
sum radii =180
→ 170 less than 180
so yes intersect each other.
Distance AB = 170;
sum radii =180
→ 170 less than 180
so yes intersect each other.
16. Plot the points A (2, 1), B (–1, 2), C (–2, –1), and D (1, –2) in the
coordinate plane. Is ABCD a square? Can you explain why? What
is the area of this square?
Plot \(A(2, 1), B(-1, 2), C(-2, -1), D(1, -2)\). Is it a square? Area?
Plot A(2,1), B(-1,2), C(-2,-1), D(1,-2). Is it a square? Area?
Plot A(2,1), B(-1,2), C(-2,-1), D(1,-2). Is it a square? Area?
Plot A(2,1), B(-1,2), C(-2,-1), D(1,-2). Is it a square? Area?
Answer:
Sides: $AB = \sqrt{(-1-2)^2 + (2-1)^2} = \sqrt{9+1} = \sqrt{10}$
All sides equal $\sqrt{10}$.
Diagonals: $AC = \sqrt{(-2-2)^2 + (-1-1)^2} = \sqrt{16+4} = \sqrt{20}$
Both diagonals equal $\sqrt{20}$.
Conclusion: Yes, it is a square.
Area: $(\text{side})^2 = (\sqrt{10})^2 = 10$ sq. units
π Chapter Summary (as per Ganita Manjari)
- Two perpendicular lines: x-axis (horizontal) and y-axis (vertical).
- Origin = (0,0). Quadrants: I(+,+), II(-,+), III(-,-), IV(+,-).
- By the BaudhΔyana–Pythagoras Theorem, the distance between points (x1 , y1 ) and (x2 , y2 ) is Distance formula: √[(x₂−x₁)² + (y₂−y₁)²] (Baudhayana-Pythagoras).
- Midpoint formula: ((x₁+x₂)/2 , (y₁+y₂)/2).
- If x=y then (x,y)=(y,x); otherwise ordered pair matters.
