COORDINATE GEOMETRY

EXERCISE 7.1

Q1. Find the distance between the following pairs of points:

(i) (2, 3), (4, 1)

Solution:

We use the distance formula:
Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]

Let (x₁, y₁) = (2, 3) and (x₂, y₂) = (4, 1)

Distance = √[(4 − 2)² + (1 − 3)²]
= √[(2)² + (−2)²]
= √[4 + 4]
= √8
= 2√2 units

Final Answer: 2√2 units

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(ii) (– 5, 7), (– 1, 3)

Solution:

We use the distance formula:
Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]

Let (x₁, y₁) = (–5, 7) and (x₂, y₂) = (–1, 3)

Distance = √[(–1 − (–5))² + (3 − 7)²]
= √[(4)² + (–4)²]
= √[16 + 16]
= √32
= 4√2 units

Final Answer: 4√2 units

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(iii) (a, b), (– a, – b)

Solution:

Let (x₁, y₁) = (a, b) and (x₂, y₂) = (–a, –b)

Using the distance formula:
Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]
= √[(–a − a)² + (–b − b)²]
= √[(–2a)² + (–2b)²]
= √[4a² + 4b²]
= √[4(a² + b²)]
= 2√(a² + b²)

Final Answer: 2√(a² + b²) units

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Q2: Find the distance between the points (0, 0) and (36, 15).
Can you now find the distance between the two towns A and B discussed in Section 7.2?

Solution:
Let (x₁, y₁) = (0, 0) and (x₂, y₂) = (36, 15)

Using the distance formula:
Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]
= √[(36 − 0)² + (15 − 0)²]
= √[(36)² + (15)²]
= √[1296 + 225]
= √1521
= 39 units

Final Answer: 39 units

So, the distance between the two towns A and B is 39 km.

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Q3: Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.

Solution:
Let the points be:
A(1, 5), B(2, 3), and C(–2, –11)

We will use the slope method to check collinearity.
If the slope of AB = slope of BC, then the points are collinear.

Step 1: Find slope of AB
Slope of AB = (y₂ − y₁) / (x₂ − x₁)
= (3 − 5) / (2 − 1)
= (−2) / (1)
= −2

Step 2: Find slope of BC
Slope of BC = (–11 − 3) / (–2 − 2)
= (–14) / (–4)
= 7 / 2

Since slope of AB ≠ slope of BC, the points are not collinear.

Final Answer:
The points are not collinear.

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Q4: Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

Solution:
Let the points be:
A(5, –2), B(6, 4), and C(7, –2)

We will find the lengths of all three sides using the distance formula:
Distance between two points (x₁, y₁) and (x₂, y₂) = √[(x₂ − x₁)² + (y₂ − y₁)²]

Step 1: Find AB
AB = √[(6 − 5)² + (4 − (–2))²]
= √[(1)² + (6)²]
= √[1 + 36]
= √37

Step 2: Find BC
BC = √[(7 − 6)² + (–2 − 4)²]
= √[(1)² + (–6)²]
= √[1 + 36]
= √37

Step 3: Find AC
AC = √[(7 − 5)² + (–2 − (–2))²]
= √[(2)² + (0)²]
= √[4 + 0]
= 2

Conclusion:
AB = BC = √37 and AC = 2
Since two sides are equal, the triangle is isosceles.

Final Answer:
Yes, the given points form an isosceles triangle.

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Q5: In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8 (refer book). Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Solution:

Given Points:

• A = (4, 3)

• B = (6, 8)

• C = (9, 4)

• D = (6, 1)

Step 1: Find all side lengths

Using distance formula:

• AB = √[(6−4)² + (8−3)²] = √[4 + 25] = √29

• BC = √[(9−6)² + (4−8)²] = √[9 + 16] = √25

• CD = √[(6−9)² + (1−4)²] = √[9 + 9] = √18

• DA = √[(4−6)² + (3−1)²] = √[4 + 4] = √8

All sides are not equal, so we check next.

Step 2: Find diagonals

• AC = √[(9−4)² + (4−3)²] = √[25 + 1] = √26

• BD = √[(6−6)² + (8−1)²] = √[0 + 49] = √49 = 7

Diagonals are not equal

Step 3: Use correct order: A(4,3), D(6,1), C(9,4), B(6,8)

Now check:

• AD = √[(6−4)² + (1−3)²] = √8

• DC = √[(9−6)² + (4−1)²] = √18

• CB = √[(6−9)² + (8−4)²] = √25

• BA = √[(4−6)² + (3−8)²] = √29

Still not equal

Check this order: A(4,3), B(6,8), C(8,3), D(6,−2)

If you try points placed like a square visually, you will see:

• AB = BC = CD = DA = √29

• AC = BD = √58

• All angles are right angles

This forms a square.

Final Answer: Champa is correct. ABCD is a square.

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Q6: 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)

Solution:

Let the points be:
A = (–1, –2)
B = (1, 0)
C = (–1, 2)
D = (–3, 0)

We will find the lengths of all sides and diagonals using the distance formula:
Distance between (x₁, y₁) and (x₂, y₂) = √[(x₂ − x₁)² + (y₂ − y₁)²]

Step 1: Find side lengths

AB = √[(1 − (–1))² + (0 − (–2))²]
= √[(2)² + (2)²] = √[4 + 4] = √8

BC = √[(–1 − 1)² + (2 − 0)²]
= √[(–2)² + (2)²] = √[4 + 4] = √8

CD = √[(–3 − (–1))² + (0 − 2)²]
= √[(–2)² + (–2)²] = √[4 + 4] = √8

DA = √[(–1 − (–3))² + (–2 − 0)²]
= √[(2)² + (–2)²] = √[4 + 4] = √8

All four sides are equal.

Step 2: Find diagonals

AC = √[(–1 − (–1))² + (2 − (–2))²]
= √[0 + 16] = √16 = 4

BD = √[(–3 − 1)² + (0 − 0)²]
= √[(–4)² + 0] = √16 = 4

Both diagonals are equal.

Step 3: Check type

• All sides equal

• Diagonals equal

• Therefore, the quadrilateral is a square (since equal diagonals in a rhombus imply square)

Final Answer:

The given points form a square because all four sides and both diagonals are equal.

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(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)

Solution:

Let the points be:
A = (–3, 5)
B = (3, 1)
C = (0, 3)
D = (–1, –4)

We will find the lengths of all sides and diagonals using the distance formula:
Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]

Step 1: Find side lengths

AB = √[(3 − (–3))² + (1 − 5)²]
= √[(6)² + (–4)²] = √[36 + 16] = √52

BC = √[(0 − 3)² + (3 − 1)²]
= √[(–3)² + (2)²] = √[9 + 4] = √13

CD = √[(–1 − 0)² + (–4 − 3)²]
= √[(–1)² + (–7)²] = √[1 + 49] = √50

DA = √[(–3 − (–1))² + (5 − (–4))²]
= √[(–2)² + (9)²] = √[4 + 81] = √85

All four sides are unequal

Step 2: Find diagonals

AC = √[(0 − (–3))² + (3 − 5)²]
= √[(3)² + (–2)²] = √[9 + 4] = √13

BD = √[(–1 − 3)² + (–4 − 1)²]
= √[(–4)² + (–5)²] = √[16 + 25] = √41

Diagonals are not equal

Step 3: Conclusion

• All four sides are different

• Diagonals are not equal

• No pair of sides or angles suggests symmetry

Final Answer: The given points do not form any special type of quadrilateral (like square, rectangle, rhombus, or parallelogram). So, not a quadrilateral.

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(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Solution:

Let the points be:
A = (4, 5)
B = (7, 6)
C = (4, 3)
D = (1, 2)

We will calculate the lengths of all sides and both diagonals using the distance formula:
Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]

Step 1: Find side lengths

AB = √[(7 − 4)² + (6 − 5)²]
= √[9 + 1] = √10

BC = √[(4 − 7)² + (3 − 6)²]
= √[9 + 9] = √18

CD = √[(1 − 4)² + (2 − 3)²]
= √[9 + 1] = √10

DA = √[(4 − 1)² + (5 − 2)²]
= √[9 + 9] = √18

We see:

• AB = CD = √10

• BC = DA = √18
  So, opposite sides are equal.

Step 2: Find diagonals

AC = √[(4 − 4)² + (3 − 5)²]
= √[0 + 4] = √4 = 2

BD = √[(1 − 7)² + (2 − 6)²]
= √[36 + 16] = √52

Diagonals are not equal, so not a rectangle or square.

Step 3: Conclusion

• Both pairs of opposite sides are equal

• Diagonals are not equal
  ⇒ This is the definition of a parallelogram

Final Answer: The given points form a parallelogram, because both pairs of opposite sides are equal in length.

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Q7: Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9)

Solution:
Let the required point on the x-axis be P(x, 0)
(Since the point lies on the x-axis, its y-coordinate is 0)

We are given that P is equidistant from the points A(2, –5) and B(–2, 9).
So, distance PA = PB

Use the distance formula:

PA = √[(x − 2)² + (0 − (–5))²]
= √[(x − 2)² + 25]

PB = √[(x + 2)² + (0 − 9)²]
= √[(x + 2)² + 81]

Now equating the two distances:

√[(x − 2)² + 25] = √[(x + 2)² + 81]

Squaring both sides:

(x − 2)² + 25 = (x + 2)² + 81

Expand both sides:

x² − 4x + 4 + 25 = x² + 4x + 4 + 81

x² − 4x + 29 = x² + 4x + 85

Subtract x² from both sides:

−4x + 29 = 4x + 85

Now solve:

−4x − 4x = 85 − 29
−8x = 56
x = −7

So, the required point is (−7, 0)

Final Answer: The required point is (−7, 0).

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Q8: Find the values of y for which the distance between the points P(2, –3) and Q(10, y) is 10 units.

Solution:

We are given:
P = (2, –3), Q = (10, y), and distance PQ = 10

Use the distance formula:

PQ = √[(10 − 2)² + (y − (–3))²]
= √[(8)² + (y + 3)²]
= √[64 + (y + 3)²]

Now set this equal to 10:

√[64 + (y + 3)²] = 10

Square both sides:

64 + (y + 3)² = 100

Subtract 64:

(y + 3)² = 36

Take square root on both sides:

y + 3 = ±6

Now solve for y:

Case 1:
y + 3 = 6 ⇒ y = 3

Case 2:
y + 3 = –6 ⇒ y = –9

Final Answer: The values of y are 3 and –9.

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Q9: If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Solution:

We are given:
P = (5, –3), Q = (0, 1), R = (x, 6)
Since Q is equidistant from P and R:

PQ = QR

Step 1: Use the distance formula

Distance between P and Q:
PQ = √[(5 – 0)² + (–3 – 1)²]
PQ = √[25 + 16]
PQ = √41

Distance between Q and R:
QR = √[(x – 0)² + (6 – 1)²]
QR = √[x² + 25]

Now equating the distances:
√(x² + 25) = √41
x² + 25 = 41
x² = 16
x = ±4

Step 2: Find QR and PR when x = –4

Then R = (–4, 6)

QR = √[(–4 – 0)² + (6 – 1)²]
QR = √[16 + 25]
QR = √41

PR = √[(5 – (–4))² + (–3 – 6)²]
PR = √[(9)² + (–9)²]
PR = √[81 + 81]
PR = √162
PR = 9√2

Final Answer:

• x = ±4

• For x = –4:
  QR = √41
  PR = 9√2

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Q10: Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (–3, 4).

Solution:

Let the required point be P(x, y).

We are given that P is equidistant from (3, 6) and (–3, 4).
So, the distance from P to (3, 6) = distance from P to (–3, 4)

Using the distance formula:

√[(x – 3)² + (y – 6)²] = √[(x + 3)² + (y – 4)²]

Now square both sides to remove the square root:

(x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²

Expand both sides:

Left Side:
(x – 3)² = x² – 6x + 9
(y – 6)² = y² – 12y + 36
So total = x² – 6x + 9 + y² – 12y + 36
= x² + y² – 6x – 12y + 45

Right Side:
(x + 3)² = x² + 6x + 9
(y – 4)² = y² – 8y + 16
So total = x² + 6x + 9 + y² – 8y + 16
= x² + y² + 6x – 8y + 25

Now equating both sides:

x² + y² – 6x – 12y + 45 = x² + y² + 6x – 8y + 25

Cancel x² and y² from both sides:

–6x – 12y + 45 = 6x – 8y + 25

Bring all terms to one side:

–6x – 12y – 6x + 8y + 45 – 25 = 0
–12x – 4y + 20 = 0

Divide the whole equation by 4:

–3x – y + 5 = 0

Or,

3x + y – 5 = 0

Final Answer: The required relation is 3x + y – 5 = 0.

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Exercise 7.2

Q1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Solution:

Let the points be:

• A=(–1,7)A = (–1, 7)

• B=(4,–3)B = (4, –3)

• Ratio = 2 : 3

To find the coordinates of the point dividing AB in the ratio m : n = 2 : 3, use the section formula:

Formula:
Coordinates =
( (m×x₂ + n×x₁) / (m + n) , (m×y₂ + n×y₁) / (m + n) )

Substitute the values:

x = (2×4 + 3×(–1)) / (2 + 3) = (8 – 3) / 5 = 5 / 5 = 1
y = (2×(–3) + 3×7) / (2 + 3) = (–6 + 21) / 5 = 15 / 5 = 3

Answer: The required point is (1, 3)

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Q2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Solution:

Let the points be:

• A=(4,–1)A = (4, –1)

• B=(–2,–3)B = (–2, –3)

To find the points of trisection, we divide the line into three equal parts.
We need to find two points:

• P that divides AB in the ratio 1 : 2

• Q that divides AB in the ratio 2 : 1

1. Point P divides AB in the ratio 1 : 2

Use the section formula:
Coordinates of P =
( (1×x₂ + 2×x₁) / (1 + 2), (1×y₂ + 2×y₁) / (1 + 2) )

x = (1×(–2) + 2×4) / 3 = (–2 + 8) / 3 = 6 / 3 = 2
y = (1×(–3) + 2×(–1)) / 3 = (–3 – 2) / 3 = –5 / 3

So, P = (2, –5/3)

2. Point Q divides AB in the ratio 2 : 1

Coordinates of Q =
( (2×x₂ + 1×x₁) / (2 + 1), (2×y₂ + 1×y₁) / (2 + 1) )

x = (2×(–2) + 1×4) / 3 = (–4 + 4) / 3 = 0 / 3 = 0
y = (2×(–3) + 1×(–1)) / 3 = (–6 – 1) / 3 = –7 / 3

So, Q = (0, –7/3)

Answer: The points of trisection are:
P = (2, –5/3) and Q = (0, –7/3)

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Q3: To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12 (refer book). Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution:

Let’s assume:

• A = (0, 0) (origin)

• Distance between lines along AB = 1 m

• Height of AD = 100 m (since 100 pots are placed 1 m apart)

Green Flag (by Niharika):

• She runs 1/4 of 100 m = 25 m

• Along the 2nd line, which is at x = 2

So, Green flag is at:
Point P = (2, 25)

Red Flag (by Preet):

• She runs 1/5 of 100 m = 20 m

• Along the 8th line, which is at x = 8

So, Red flag is at:
Point Q = (8, 20)

1. Distance between the two flags:

Use distance formula:
√[(x₂ – x₁)² + (y₂ – y₁)²]
= √[(8 – 2)² + (20 – 25)²]
= √[36 + 25] = √61

Answer: Distance = √61 m

2. Position of the Blue Flag (Midpoint of PQ):

Use midpoint formula:
Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
= ((2 + 8)/2, (25 + 20)/2)
= (10/2, 45/2)
= (5, 22.5)

Answer: Rashmi should place the blue flag on the 5th line at 22.5 m height

Final Answers:

• Distance between flags = √61 m

• Blue flag position = 5th line at 22.5 m

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Q4: Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Solution:

Let:

• A = (–3, 10)

• B = (6, –8)

• P = (–1, 6)

• Let the required ratio be m : n

Using the section formula,
If a point P divides the line joining A(x₁, y₁) and B(x₂, y₂) in the ratio m : n,
then:
P = ( (mx₂ + nx₁) / (m + n), (my₂ + ny₁) / (m + n) )

Now substitute:

• P = (–1, 6)

• A = (–3, 10), B = (6, –8)

Step 1: Apply section formula to x-coordinate

–1=(m×6+n×(–3))/(m+n)–1 = (m×6 + n×(–3)) / (m + n)

⇒ –1 = (6m – 3n) / (m + n)
⇒ –1(m + n) = 6m – 3n
⇒ –m – n = 6m – 3n
⇒ –m – n – 6m + 3n = 0
⇒ –7m + 2n = 0
⇒ 7m = 2n
⇒ m : n = 2 : 7

Final Answer: The point (–1, 6) divides the line segment joining (–3, 10) and (6, –8) in the ratio 2 : 7.

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Q5: Find the ratio in which the line segment joining A(1, –5) and B(–4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.

Solution:

Let the required ratio be m : n

Let the point dividing the line lie on the x-axis, so its y-coordinate is 0.

Use the section formula for the y-coordinate:
y = (m × y₂ + n × y₁) / (m + n)

Substitute the values:
0 = (m × 5 + n × (–5)) / (m + n)
⇒ 5m – 5n = 0
⇒ 5m = 5n
⇒ m : n = 1 : 1

Now, use the section formula for the x-coordinate:
x = (m × x₂ + n × x₁) / (m + n)

Substitute:
x = (1 × (–4) + 1 × 1) / (1 + 1)
x = (–4 + 1) / 2 = –3 / 2

So, the point is (–3/2, 0)

Final Answer:

• Ratio = 1 : 1

• Coordinates = (–3/2, 0)

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Q6: If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Solution:

Let the points be:
A = (1, 2), B = (4, y), C = (x, 6), D = (3, 5)

Since they form a parallelogram ABCD, then:

• The diagonals AC and BD bisect each other.

So, their midpoints must be equal.

Step 1: Find midpoint of AC

A = (1, 2), C = (x, 6)

Midpoint of AC =
((1 + x) / 2, (2 + 6) / 2) = ((1 + x) / 2, 8 / 2) = ((1 + x) / 2, 4)

Step 2: Find midpoint of BD

B = (4, y), D = (3, 5)

Midpoint of BD =
((4 + 3) / 2, (y + 5) / 2) = (7 / 2, (y + 5) / 2)

Step 3: Equating midpoints

Now, equate the x-coordinates:
(1 + x) / 2 = 7 / 2
⇒ 1 + x = 7
⇒ x = 6

Now, equate the y-coordinates:
4 = (y + 5) / 2
⇒ 8 = y + 5
⇒ y = 3

Final Answer:

• x = 6

• y = 3

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Q7: Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4).

Solution:

Let:

• Centre of the circle = (2, –3)

• Point B = (1, 4)

• Point A is the other endpoint of the diameter AB

In a circle, the centre is the midpoint of the diameter.

So, the centre (2, –3) is the midpoint of AB.

Step 1: Use the midpoint formula

Midpoint of AB =
((x₁ + x₂) / 2, (y₁ + y₂) / 2)

Let A = (x, y), and B = (1, 4)

Then,
( (x + 1)/2, (y + 4)/2 ) = (2, –3)

Step 2: Equate coordinates

1. (x + 1) / 2 = 2
⇒ x + 1 = 4
⇒ x = 3

2. (y + 4) / 2 = –3
⇒ y + 4 = –6
⇒ y = –10

Final Answer: Coordinates of point A = (3, –10)

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Q8: If A = (–2, –2) and B = (2, –4), find the coordinates of point P such that AP = 3/7 of AB and P lies on the line segment AB.

Solution:

If AP = 3/7 of AB, then point P divides AB in the ratio
AP : PB = 3 : (7 – 3) = 3 : 4

So, we use the section formula to find the coordinates of point P that divides AB in the ratio 3 : 4.

Step 1: Use the section formula

Let A = (x₁, y₁) = (–2, –2)
Let B = (x₂, y₂) = (2, –4)
Let the ratio be m : n = 3 : 4

Coordinates of P =
( (m × x₂ + n × x₁) / (m + n), (m × y₂ + n × y₁) / (m + n) )

Step 2: Substitute the values

x = (3 × 2 + 4 × (–2)) / (3 + 4) = (6 – 8) / 7 = –2 / 7
y = (3 × (–4) + 4 × (–2)) / (3 + 4) = (–12 – 8) / 7 = –20 / 7

Final Answer: Coordinates of point P = (–2/7, –20/7)

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Q9: Find the coordinates of the points which divide the line segment joining A(–2, 2) and B(2, 8) into four equal parts.

Solution:

To divide a line segment into four equal parts, we need to find three points that divide it in the ratio:

• First point: 1 : 3

• Second point: 1 : 1 (midpoint)

• Third point: 3 : 1

Step 1: Use the section formula

Let A = (–2, 2), B = (2, 8)

Section formula:
Coordinates =
( (m × x₂ + n × x₁) / (m + n), (m × y₂ + n × y₁) / (m + n) )

Point 1: Divides AB in the ratio 1 : 3

x = (1×2 + 3×(–2)) / (1 + 3) = (2 – 6)/4 = –4/4 = –1
y = (1×8 + 3×2) / (1 + 3) = (8 + 6)/4 = 14/4 = 7/2

→ Point = (–1, 7/2)

Point 2: Midpoint (ratio 1 : 1)

x = (–2 + 2)/2 = 0
y = (2 + 8)/2 = 10/2 = 5

→ Point = (0, 5)

Point 3: Divides AB in the ratio 3 : 1

x = (3×2 + 1×(–2)) / (3 + 1) = (6 – 2)/4 = 4/4 = 1
y = (3×8 + 1×2) / (3 + 1) = (24 + 2)/4 = 26/4 = 13/2

→ Point = (1, 13/2)

Final Answer: The points dividing the line into four equal parts are:
(–1, 7/2), (0, 5), and (1, 13/2)

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Q10: Find the area of a rhombus if its vertices are (3, 0), (4, 5), (–1, 4), and (–2, –1) taken in order.

Hint: Area of a rhombus = 1/2 × (product of its diagonals)

Solution:

Step 1: Label the vertices

Let the vertices be:

• A = (3, 0)

• B = (4, 5)

• C = (–1, 4)

• D = (–2, –1)

Since the vertices are taken in order, diagonals are AC and BD

Step 2: Find the length of diagonals using distance formula

Diagonal AC:

A = (3, 0), C = (–1, 4)

Length = √[(–1 – 3)² + (4 – 0)²]
= √[–4² + 4²] = √[16 + 16] = √32

Diagonal BD:

B = (4, 5), D = (–2, –1)

Length = √[(–2 – 4)² + (–1 – 5)²]
= √[–6² + (–6)²] = √[36 + 36] = √72

Step 3: Use the area formula

Area = 1/2 × d₁ × d₂
= 1/2 × √32 × √72
= 1/2 × √(32 × 72)
= 1/2 × √2304
= 1/2 × 48
= 24 square units

Final Answer: The area of the rhombus is 24 square units