**Coordinate geometry** is a branch of mathematics that uses algebra to study geometry. It involves representing geometric shapes using numbers (coordinates) on a plane.

**Key Concepts:**

**Cartesian Plane:**This is a plane formed by two perpendicular lines, the x-axis and the y-axis, intersecting at the origin (0, 0).**Coordinates of a Point:**Any point on the plane can be represented by an ordered pair (x, y), where x is the distance from the y-axis (abscissa) and y is the distance from the x-axis (ordinate).**Distance Formula:**Used to find the distance between two points on a plane given their coordinates.**Area of a Triangle:**Calculates the area of a triangle given the coordinates of its vertices.

**Applications:**

**Physics:**Representing motion, forces, and fields.**Engineering:**Designing structures and systems.**Computer Graphics:**Creating images and animations.**Navigation:**Determining locations and distances.

**Exercise 7.1**

**1. Find the distance between the following pairs of points:**

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

**(ii) (-5, 7), (-1, 3)**

**(iii) (a, b), (-a, -b)**

**Ans : **

**Distance Formula:**

= √[(x₂ – x₁)² + (y₂ – y₁)²]

**Let’s calculate the distance for each pair of points:**

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

- x₁ = 2, y₁ = 3,
- x₂ = 4, y₂ = 1
- D = √[(4 – 2)² + (1 – 3)²] = √[2² + (-2)²] = √8 = 2√2 units

(ii) Points (-5, 7) and (-1, 3)

- x₁ = -5, y₁ = 7, x₂ = -1, y₂ = 3
- D = √[(-1 + 5)² + (3 – 7)²] = √[4² + (-4)²] = √32 = 4√2 units

(iii) Points (a, b) and (-a, -b)

- x₁ = a, y₁ = b, x₂ = -a, y₂ = -b
- D = √[(-a – a)² + (-b – b)²] = √[(-2a)² + (-2b)²] = √(4a² + 4b²) = 2√(a² + b²) units

**2. Find the distance between the points (0, 0) and (36, 15).**

**Ans : **

**Given points:** (0, 0) and (36, 15)

**Formula:**

D = √((x₂ – x₁)² + (y₂ – y₁)²)

**Substituting the given points:** D = √((36 – 0)² + (15 – 0)²)

= √(36² + 15²)

= √(1296 + 225)

= √1521

= 39 Units

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

**Ans : **

**Given points:** (1, 5), (2, 3), and (-2, -11)

Let’s denote the points as A(1, 5), B(2, 3), and C(-2, -11).

**Calculate the distances:**

- AB = √[(2-1)² + (3-5)²] = √5
- BC = √[(-2-2)² + (-11-3)²] = √208
- AC = √[(-2-1)² + (-11-5)²] = √261

**Check for collinearity:** Since AB + BC ≠ AC (√5 + √208 ≠ √261), the points A, B, and C are **not collinear**.

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

**Ans : **

**Step 1: **We can use the distance formula: d = √((x₂ – x₁)² + (y₂ – y₁)²)

- Distance between (5, -2) and (6, 4): d₁ = √((6-5)² + (4-(-2))²) = √(1 + 36) = √37
- Distance between (6, 4) and (7, -2): d₂ = √((7-6)² + (-2-4)²) = √(1 + 36) = √37
- Distance between (5, -2) and (7, -2): d₃ = √((7-5)² + (-2-(-2))²) = √4 = 2

**Step 2: Compare the distances:**

We can see that d₁ = d₂ = √37.

**Conclusion:**

Since two sides of the triangle have equal length (d₁ = d₂), the triangle formed by the given points is an **isosceles triangle**

**5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in the given figure. 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.**

**Ans : **

**6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.**

**(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)**

**(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)**

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

**Ans : **

**7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).**

**Ans : **

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

**Ans : **

D = √((x₂ – x₁)² + (y₂ – y₁)²)

In this case:

- D = 10
- (x₁, y₁) = (2, -3)
- (x₂, y₂) = (10, y)

Substituting the values in the distance formula:

10 = √((10 – 2)² + (y – (-3))²)

10 = √(64 + (y + 3)²)

Squaring both sides: 100 = 64 + (y + 3)² (y + 3)² = 36

Taking the square root of both sides: y + 3 = ±6

**Case 1:** y + 3 = 6 y = 3

**Case 2:** y + 3 = -6 y = -9

**Therefore, the possible values of y are 3 and -9.**

**9. 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.**

**Ans : **

**Step 1: Finding the value of x**

Since Q is equidistant from P and R, PQ = QR.

Using the distance formula:

- PQ = √((5-0)² + (-3-1)²) = √(25 + 16) = √41
- QR = √((x-0)² + (6-1)²) = √(x² + 25)

Equating PQ and QR:

√41 = √(x² + 25)

Squaring both sides:

41 = x² + 25

x² = 16

x = ±4

**Step 2: Finding QR and PR**

We already found QR = √41.

For x = 4, R(4, 6)

- PR = √((4-5)² + (6-(-3))²) = √(1 + 81) = √82

For x = -4, R(-4, 6)

- PR = √((-4-5)² + (6-(-3))²) = √(81 + 81) = 9√2

**Therefore, the values of x are 4 and -4, QR = √41, and PR can be √82 or 9√2 depending on the value of x.**

**10. Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).**

**Ans : **

Let P(x, y) be any point on the locus. Given: PA = PB

Using the distance formula:

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

Since PA = PB, we can equate their squares:

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

Expanding and simplifying: x² – 6x + 9 + y² – 12y + 36 = x² + 6x + 9 + y² – 8y + 16 -12x – 4y + 20 = 0

Dividing by -4: 3x + y – 5 = 0

**Therefore, the relation between x and y is 3x + y – 5 = 0.**

**Exercise 7.2**

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

**Ans : **

**Given:**

- (x₁, y₁) = (-1, 7)
- (x₂, y₂) = (4, -3)
- m : n = 2 : 3

**Substituting the values in the section formula:**

x = (2*4 + 3*(-1)) / (2 + 3) = (8 – 3) / 5 = 1

y = (2*(-3) + 3*7) / (2 + 3) = (-6 + 21) / 5 = 3

**Therefore, the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3 are (1, 3).**

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

**Ans : **

**Given:**

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

**For the first point of trisection (dividing the line in the ratio 1:2):**

- m = 1, n = 2

Substituting the values in the section formula:

- x = (1*(-2) + 2*4) / (1 + 2) = 2
- y = (1*(-3) + 2*(-1)) / (1 + 2) = -5/3

**For the second point of trisection (dividing the line in the ratio 2:1):**

- m = 2, n = 1

Substituting the values in the section formula:

- x = (2*(-2) + 1*4) / (2 + 1) = 0
- y = (2*(-3) + 1*(-1)) / (2 + 1) = -7/3

**Therefore, the coordinates of the points of trisection are (2, -5/3) and (0, -7/3).**

**3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in given**

**figure below. Niharika runs ****1/4**** th the distance AD on the 2nd line and posts a green flag. Preet runs ****1/5**** th 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?**

**Ans : **

**Finding the Positions of the Flags**

**Niharika:**

Runs 1/4th the distance AD on the 2nd line.

- Distance along AD = (1/4) * 100 = 25 m
- Position of green flag = (2, 25) (2nd line, 25 m distance)
**Preet:**

Runs 1/5th the distance AD on the 8th line.

- Distance along AD = (1/5) * 100 = 20 m
- Position of red flag = (8, 20) (8th line, 20 m distance)

**Finding the Distance Between Flags**

Using the distance formula: d = √((x₂ – x₁)² + (y₂ – y₁)²)

- Distance between green and red flags = √((8-2)² + (20-25)²) = √(36 + 25) = √61 m

**Finding the Position for the Blue Flag**

The blue flag should be at the midpoint of the line joining the green and red flags.

- Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)
- Midpoint = ((2+8)/2, (25+20)/2) = (5, 22.5)

**Therefore, Rashmi should post the blue flag at 22.5 m on the 5th line.**

**4. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).**

**Ans : **

Let the ratio be k:1.

Using the section formula, we can find the coordinates of the point dividing the line segment in the ratio k:1.

x-coordinate of the point = (kx₂ + x₁) / (k + 1)

y-coordinate of the point = (ky₂ + y₁) / (k + 1)

Substituting the given values:

-1 = (k*6 + (-3)) / (k + 1) *

*6 = (k*(-8) + 10) / (k + 1)

Solving the first equation for k: -k – 1 = 6k – 3 7k = 2 k = 2/7

Therefore, the ratio in which the line segment is divided is **2:7**.

**5. Find the ratio in which 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.**

**Ans : **

**Let the ratio be k:1.**

Since the point of division lies on the x-axis, its y-coordinate is 0.

Using the section formula, we can find the y-coordinate of the point of division:

- y = (ky₂ + y₁) / (k + 1)

Where:

- (x₁, y₁) = (1, -5)
- (x₂, y₂) = (-4, 5)
- y = 0 (as the point lies on the x-axis)

Substituting the values:

- 0 = (k*5 + (-5)) / (k + 1)
- 5k – 5 = 0
- k = 1

**Therefore, the ratio in which the line segment is divided is 1:1.**

Now, let’s find the x-coordinate of the point of division using the section formula for the x-coordinate:

- x = (kx₂ + x₁) / (k + 1) = (1*(-4) + 1*1) / (1 + 1) = -3/2

**So, the coordinates of the point of division are (-3/2, 0).**

**In conclusion, the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis in the ratio 1:1, and the point of division is (-3/2, 0).**

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

**Ans : **

**Midpoint formula:**

- Midpoint of a line segment with endpoints (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2)

**Midpoint of AC (O):**

- ((1+x)/2, (2+6)/2) = ((1+x)/2, 4)

**Midpoint of BD (O):**

- ((4+3)/2, (y+5)/2) = (7/2, (y+5)/2)

Since O is the midpoint of both AC and BD, their coordinates are equal:

- (1+x)/2 = 7/2
- 1+x = 7
- x = 6
- 4 = (y+5)/2
- 8 = y+5
- y = 3

**Therefore, the coordinates of the fourth vertex C are (6, 6) and the value of y is 3.**

**7. 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).**

**Ans : **

**Let the coordinates of point A be (x, y).**

Since (2, -3) is the midpoint of AB, we can use the midpoint formula:

- Midpoint formula: ((x₁ + x₂)/2, (y₁ + y₂)/2)

Applying the midpoint formula to points A(x, y) and B(1, 4):

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

Equating corresponding coordinates:

- 2 = (x + 1)/2
- -3 = (y + 4)/2

Solving for x and y:

- x + 1 = 4
- x = 3
- y + 4 = -6
- y = -10

**Therefore, the coordinates of point A are (3, -10).**

**8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = ****3/7**** AB and P lies on the line segment AB.**

**Ans : **

Since P divides AB in the ratio 3:4 (as AP:PB = 3:7), we can use the section formula to find the coordinates of P.

If a point P(x, y) divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂) in the ratio m:n, then:

- x = (mx₂ + nx₁) / (m + n)
- y = (my₂ + ny₁) / (m + n)

**Substituting the given values:**

- x₁ = -2, y₁ = -2
- x₂ = 2, y₂ = -4
- m = 3, n = 4

**Calculating the coordinates of P:**

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

**Therefore, the coordinates of point P are (-2/7, -20/7).**

**9. Find the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts.**

**Ans : **

Since we need to divide the line segment into four equal parts, we are essentially finding the midpoints of different segments.

**Step 1: Find the midpoint of AB (let’s call it P).**

- Midpoint formula: ((x₁ + x₂)/2, (y₁ + y₂)/2)
- P = ((-2 + 2)/2, (2 + 8)/2) = (0, 5)

**Step 2: Find the midpoint of AP (let’s call it Q).**

- Q = ((-2 + 0)/2, (2 + 5)/2) = (-1, 7/2)

**Step 3: Find the midpoint of PB (let’s call it R).**

- R = ((0 + 2)/2, (5 + 8)/2) = (1, 13/2)

**Therefore, the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts are (-1, 7/2), (0, 5), and (1, 13/2).**

**10. 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)]**

**Ans : **

**Step 1: Find the coordinates of the diagonals**

- Diagonal 1: Join (3, 0) and (-1, 4)
- Diagonal 2: Join (4, 5) and (-2, -1)

**Step 2: Calculate the length of the diagonals**

- We can use the distance formula: d = √((x₂ – x₁)² + (y₂ – y₁)²)

**Diagonal 1:**

- Length = √((-1 – 3)² + (4 – 0)²) = √(16 + 16) = √32 = 4√2

**Diagonal 2:**

- Length = √((-2 – 4)² + (-1 – 5)²) = √(36 + 36) = √72 = 6√2

**Step 3: **

Area = (1/2) * (4√2) * (6√2) = (1/2) * 48 = 24 square units