Coordinate Geometry

(i) We know that distance between the two points is given by:

d = √[(x₁ – x₂)² + (y₁ – y₂)²]

Distance between A(2, 3) and B(4, 1) is:

f =√ [(4 - 2)² + (1 – 3)²]

= √[4 + 4]= √8

= 2√2

(ii) Distance between A(-5, 7) and B(-1, 3) is:

f = √[(-5 + 1)² + (7 – 3)²]

= √[16 + 16] = √32

= 4√2

(iii) Distance between A(a, b) and B(-a, -b) is:

f = √[(a+ a)² + (b+ b)²]

f = √[2a + 2b] = √(4a² + 4b²)

f = 2√(a² + b²)

To find: Distance between two points
Given: Points A(0, 0), B(36, 15)



For two points A(x₁, y₁) and B(x₂, y₂),

Distance is given by

f = [(x_{2 -} x₁)² + (y_{2 -} y₁)²]^1/2

Distance between (0, 0) and (36, 15) is:

f = [(36 – 0)² + (15 – 0)²]^1/2

= [1296 + 225]^1/2 = (1521)^1/2

= 39

Yes, we can find the distance between the given towns A and B. Let us take town A at origin point (0, 0)

Hence, town B will be at point (36, 15) with respect to town A

And, as calculated above, the distance between town A and B will be 39 km

Let the points (1, 5), (2, 3), and (- 2,-11) be representing the vertices A, B, and C of the given triangle respectively.
Let A = (1, 5), B = (2, 3) and C = (- 2,-11)
Case 1)


Since AB + BC ≠ CA
Case 2)
Now,




BA + AC ≠ BC
Case 3)
Now,




As BC+CA≠BA
As three of the cases are not satisfied.
Hence the points are not collinear.

Distance between two points A(x₁, y₁) and B(x₂, y₂) is given by

D = √(x₂ - x₁)² + (y₂ - y₁)²

Let us assume that points (5, - 2), (6, 4), and (7, - 2) are representing the vertices A, B, and C of the given triangle respectively as shown in the figure.

AB = [(5- 6)² + (-2- 4)²]^1/2

=

=

BC = [(6- 7)² + (4 + 2)²]^1/2

=

=

CA = [(5- 7)² + (-2+ 2)²]^1/2

=

= 2

Therefore, AB = BC

As two sides are equal in length, therefore, ABC is an isosceles triangle

It can be seen that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of 4 friends



distance between two points A(x_1, y₁) and B(x₂, y₂) is given by

D = √(x₂ - x₁)² + (y₂ - y₁)²

Hence,

AB = [(3- 6)² + (4 - 7)²]^1/2

=

=

BC = [(6 - 9)² + (7- 4)²]^1/2

=

=

CD = [(9- 6)² + (4- 1)²]^1/2

=

=

AD = [(3 - 6)² + (4 - 1)²]^1/2

=

=

Diagonal AC = [(3 - 9)² + (4 - 4)²]^1/2

=

= 6

Diagonal BD = [(6- 6)² + (7- 1)²]^1/2

=

= 6

It can be seen that all sides of quadrilateral ABCD are of the same length and diagonals are of the same length

Therefore, ABCD is a square and hence, Champa was correct

To Find: Type of quadrilateral formed

(i) Let the points ( - 1, - 2), (1, 0), ( - 1, 2), and ( - 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x₁, y₁) and (x₂, y₂).

AB = [(-1- 1)² + (-2- 0)²]^1/2

= =

= 2√2

BC = √[(1 + 1)² + (0- 2)²]

= =

= 2√2

CD = √[(-1 + 3)² + (2- 0)²]

= =

= 2√2

AD = √[(-1+ 3)² + (-2- 0)²]

= =

= 2√2

Diagonal AC = √[(-1 + 1)² + (-2 - 2)²]

=

= 4

Diagonal BD = √[(1 + 3)² + (0- 0)²]

=

= 4

It is clear that all sides of this quadrilateral are of the same length and the diagonals are of the same length. Therefore, the given points are the vertices of a square

(ii)Let the points (- 3, 5), (3, 1), (0, 3), and ( - 1, - 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x₁, y₁) and (x₂, y₂).

AB = √[(-3- 3)² + (5- 1)²]

= =

= 2√13

BC = √[(3- 0)² + (1- 3)²]

=

=

CD = √[(0 + 1)² + (3+ 4)²]

= =

= 5√2

AD = √[(-3+ 1)² + (5+ 4)²]

=

=

We can observe that all sides of this quadrilateral are of different lengths.

Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc

(iii)Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively

The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x₁, y₁) and (x₂, y₂).

AB = √[(4- 7)² + (5 - 6)²]

=

=

BC = √[(7- 4)² + (6 - 3)²]

=

=

CD = √[(4- 1)² + (3 - 2)²]

=

=

AD = √[(4- 1)² + (5 - 2)²]

=

=

Diagonal AC =√[(4 - 4)² + (5- 3)²]

=

= 2

Diagonal BD = √[(7 - 1)² + (6 - 2)²]

=

=

= 2√13

We can observe that opposite sides of this quadrilateral are of the same length.

However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram

We have to find a point on x-axis.

Hence, its y-coordinate will be 0

Let the point on x-axis be (x, 0)

By distance formula, Distance between two points A(x₁, y₁) and B(x₂, y₂) is

AB = √((x₂ - x₁)² + (y₂ - y₁)²)

Distance between (x, 0) and (2, -5) = √[(x- 2)² + (0+ 5)²]

= √[(x- 2)² + (5)²]

Distance between (x, 0) and (-2, 9) = √[(x + 2)² + (0 + 9)²]

= √[(x + 2)² + (9)²]

By the given condition, these distances are equal in measure

√[(x - 2)² + (5)²] = √[(x + 2)² + (9)²]
Squaring both sides we get,

(x – 2)² + 25 = (x + 2)² + 81

⇒ x² + 4 – 4x + 25 = x² + 4 + 4x + 81
⇒ ( x² - x² ) - 4x - 4x = 81 - 25
⇒ - 8 x = 56

⇒ 8x = -56

⇒ x = - 7

Therefore, the point is (- 7, 0)

Given: Distance between (2, - 3) and (10, y) is 10

To find: y

By distance formula, Distance between two points A(x₁, y₁) and B(x₂, y₂) is

AB = √((x₂ - x₁)² + (y₂ - y₁)²)

Therefore,

√[(2 - 10)² + (-3- y)²] = 10

√[(-8)² + (3+ y)²] = 10

⇒ 64 + (y + 3)² = 100

⇒ (y + 3)² = 100 - 64

⇒ (y + 3)² = 36

⇒ y + 3 = ±6

⇒ (y + 3) will give two values 6 and -6 as answer because both the values when squared will give 36 as answer.)

y + 3 = 6 or y + 3 = -6

Therefore, y = 3 or -9

We know, By distance formula distance between two coordinates A(x₁, y₁) and B(x₂, y₂) is

AB = √[(x₂ - x₁)² + (y₂ - y₁)² ]

Now, as

PQ = QR

√[(5 - 0)² + (- 3 - 1)²] = √[(0 - x)² + (1 - 6)²]

= √x² + 25

41 = x² + 25

x² = 16

x = ± 4

Hence, point R is (4, 6) or ( - 4, 6).

When point R is (4, 6)

PR = [(5 - 4)² + (-3 - 6)²]^1/2

=

=

QR = [(0 - 4)² + (1 - 6)²]^1/2

=

=

When point R is (- 4, 6),

PR = [(5 + 4)² + (-3 - 6)²]^1/2

=

= 9

QR = [(0+ 4)² + (1 - 6)²]^1/2

=

=

To Find: Relation between x and y
Given: (x, y) is equidistant from (3, 6) and (-3, 4)
From the figure it can be seen that Point (x, y) is equidistant from (3, 6) and (- 3, 4)
This means that the distance of (x, y) from (3, 6) will be equal to distance of (x, y) from (- 3, 4)
We know by distance formula that, distance between two points A(x_1, y₁), B(x₂, y₂) is given by

Therefore,

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

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

x² + 9 – 6x + y² + 36 – 12y = x² + 9 + 6x + y² + 16 – 8y

36 – 16 = 6x + 6x + 12y – 8y

20 = 12x + 4y

3x + y = 5

3x + y – 5 = 0 is the relation between x and y

If (x₁, y₁) and (x₂, y₂) are points that are divided in ratio m:n then,

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

Now,

Using the section formula, we get

x = 1

y =

y = 3

Therefore, the point is (1, 3)

The points of trisection means that the points which divide the line in three equal parts. From the figure C, and D are these two points.
Let C (x₁, y₁) and D (x₂, y₂) are the points of trisection of the line segment joining the given points i.e., BC = CD = DA
Let BC = CD = DA = k
Point C divides the BC and CA as:
BC = k
CA = CD + DA
= k + k
= 2k
Hence ratio between BC and CA is:

Therefore, point C divides BA internally in the ratio 1:2

The point D divides the line BA in the ratio 2:1
So now applying section formula again we get,

It can be observed that Niharika posted the green flag at of the distance AD i.e., 1/4 x 100 = 25 m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25)

Similarly, Preet posted red flag at of the distance AD i.e., 1/5 x 100 = 20 m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20)

Distance between these flags by using distance formula, D

= [(8 – 2)² + (25 – 20)²]^1/2

= (36 + 25)^1/2

= m

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x,y)

x = = 5

y = = 22.5

Hence, A (x, y) = (5, 22.5)

Therefore, Rashmi should post her blue flag at 22.5m on 5th line

Let the ratio in which the line segment joining (- 3, 10) and (6, - 8) is divided by point (- 1, 6) be k : 1

Using section formula

i.e. the coordinates of the points P(x, y) which divides the line segment joining the points A(x₁, y­₁) and B(x₂,y₂), internally in the ratio m : n are

Therefore,

-1 =

⇒ -k – 1 = 6k – 3

⇒ - k -6k = -3 + 1

⇒ 7k = 2

⇒ k =

Therefore, the required ratio is 2: 7.

The diagram for the question is

Let the ratio in which the line segment joining A (1, - 5) and B ( - 4, 5) is divided by x-axis be k: 1

Therefore, the coordinates of the point of division is (, )

We know that y-coordinate of any point on x-axis is 0

Therefore,

= 0

k = 1

Therefore, x-axis divides it in the ratio 1:1.

Division point = (, )

= ()

= (, 0)

Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD.

Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

If O is the mid-point of AC, then the coordinates of O are:

() = (, 4)

If O is the mid-point of BD, then the coordinates of O are:

(, ) = (, )

Since both the coordinates are of the same point O

x + 1 = 7

x = 6

And,

= 4

5 + y = 8

y = 3
Hence, x = 6 and y = 3.

For points A(x₁, y₁) and B(x₂, y₂), the midpoints (x, y) is given by

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

Mid-point of AB is (2, - 3), which is the center of circle and point B is (1, 4)
Therefore,

Therefore, the coordinates

To find: The coordinates of P
Given: Points A(-2, -2) and B(2, -4) and ratio AP:AB = 3:7

The coordinates of point A and B are ( - 2, - 2) and (2, - 4) respectively

AP = AB

Therefore, AP: PB = 3:4

Point P divides the line segment AB in the ratio 3:4

(-2/7, -20/7) is the point which divides line in the ratio of 3:4.

It can be observed from the figure that points P, Q, R are dividing the line segment in a ratio 1:3, 1:1, 3:1respectively

Coordinates of X = ()

= (-1, )

Coordinates of Y = ()

= (0, 5)

Coordinates of Z = ()

= (1, )

Let (3, 0), (4, 5), ( - 1, 4) and ( - 2, - 1) are the vertices A, B, C, D of a rhombus ABCD

Length of diagonal AC = [(3 + 1)² + (0 – 4)²]^1/2

=

=

Length of diagonal BD = [(4 + 2)² + (5+ 1)²]^1/2

=

= 6

Therefore,

Area of rhombus ABCD = 1/2 × × 6 = 1/2 × 48

= 24 square units

For a triangle with points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃)

(i) For triangle given: A(2, 3), B(–1, 0), C(2, – 4)

Area of the triangle = [2 (0+ 4) - 1 (-4 – 3) + 2 (3 – 0)

= (8 + 7 + 6)

= Square units

(ii) For triangle given: A(–5, –1), B(3, –5), C(5, 2)

Area of the triangle = [-5 (-5 – 2) + 3 (2+ 1) + 5 (-1+ 5)

= (35 + 9 + 20)

= 32 square units

For three points A(x₁, y₁), B(x₂, y₂), C(x₃, y₃)



(i)

Collinear points mean that the points lie on a straight line. So,

For collinear points, area of triangle formed by them is zero

Therefore, for points (7, - 2) (5, 1), and (3, k), area = 0

[7 (1 – k) + 5 (k+ 2) + 3 (-2 – 1) = 0

7 – 7 k + 5 k + 10 – 9 = 0

-2 k + 8 = 0

k = 4

(ii) For collinear points, area of triangle formed by them is zero.

Therefore, for points (8, 1), (k, - 4), and (2, - 5), area = 0

[8 (-4+ 5) + k (-5- 1) + 2 (1+ 4) = 0

8 – 6k + 10 = 0

6k = 18

k = 3

Let the vertices of the triangle be A (0, - 1), B (2, 1), C (0, 3)
We know, mid-points of a line joining points (x₁, y₁) and (x₂, y₂) is given by

Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

D = (, ) = (1, 0)

E = () = (0, 1)

F = () = (1, 2)

Area of the triangle (DEF) = [1 (2 – 1) + 1(1- 0) + 0 (0- 2)

= (1 + 1)

= 1 square unit

Area of the triangle (ABC) = = [0 (1 – 3) + 2(3 + 1) + 0 (-1 - 1)

= × 8

= 4 square units

Therefore, required ratio = 1 : 4

Let the vertices of the quadrilateral be A ( - 4, - 2), B ( - 3, - 5), C (3, - 2), and D (2, 3). Join AC to form two trianglesΔABC and ΔACD

Area of the triangle (ABC) = = [-4 (-5 + 2) -3 (-2+ 2) + 3 (-2+ 5)]

= (12 + 0 + 9)

= Square units

Area of the triangle (ACD) = [-4 (-2 – 3) + 3(3 + 2) + 2 (-2+ 2)]

= (20 + 15 + 0)

= Square units

Area of Quadrilateral ABCD = Area of the triangle (ABC) + Area of the triangle (ACD)

= +

= 28 Square units

Let the vertices of the triangle be A (4, - 6), B (3, - 2), and C (5, 2)

Let D be the mid-point of side BC of ΔABC Therefore, AD is the median in ΔABC

To prove: Area ΔABC = Area ΔABD + Area ΔADC

Midpoint of the line joining the points A(x_1, y₁) and B(x₂, y₂) is given by



Therefore, by applying the formula we can get the coordinate of D

Coordinates of point D = () = (4, 0)


We also know that if A(x_1, y₁), B(x₂, y₂) and C(x₃, y₃) are vertices of a triangle then


Area of triangle is given by,



For A (4, - 6), B (3, - 2), and C (5, 2)

Area of the Δ (ABC) = = [4 (-2 - 2) + 3 (0+ 8) + 5 (-6+ 2)]

= (-16 + 24 – 20)

= - 6 Square units

However, area cannot be negative. Therefore, area of ΔABC is 6 square units

Area of the Δ (ADC) = = [4 (0 - 2) + 4 (2 + 6) + 5 (-6 - 0)]

= (-8 + 32 – 30)

= - 3 Square units

However, area cannot be negative. Therefore, area of ΔADC is 3 square units

Now for for A(4, -6), B(3, -2) and D(4, 0)

Area of Δ ABD = [4(-2 - 0) + 3(0 + 6) + 4(-6 +2)]

Area = [-8 + 18 - 16]

Area = -3 square units

Now the Area cannot be negative so Area of ΔABD = 3 units

Area of Δ ABD + Area of Δ ADC = 3 + 3 square units

Area of Δ ABD + Area of Δ ADC = 6 = Area of Δ ABC

Clearly, median AD has divided ΔABC in two triangles of equal areas

Let the given line divide the line segment joining the points A(2, - 2) and B(3, 7) in a ratio k : 1

Coordinates of the point of division = ()

This point also lies on 2x + y - 4 = 0

Therefore,

2 () + () – 4 = 0

= 0

9k – 2 = 0

k =

Hence, the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the points A(2, - 2) and B(3, 7) is 2:9.

If the given points are collinear, then the area of triangle formed by these points will be 0

Area of the triangle = [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)

= [x (2 – 0) + 1 (0 – y) + 7 (y – 2)

0 = (2x – y + 7y – 14)

2x + 6y – 14 = 0

x + 3y – 7 = 0

This is the required relation between x and y

The figure for the question is:



Let O (x, y) be the centre of the circle. And let the points (6, - 6), (3, - 7), and (3, 3) be representing the points A, B,and C on the circumference of the circle

By distance formula, Distance between two points A(x₁, y₁) and B(x₂, y₂) is

AB = √((x₂ - x₁)² + (y₂ - y₁)²)

Then,

OA = √[(x- 6)² + (y+ 6)²]

OB = √[(x- 3)² + (y+ 7)²]

OC = √[(x- 3)² + (y- 3)²]

As OA, OB and OC are radii of same circle, OA = OB

[(x - 6)² + (y + 6)²]^1/2 = [(x - 3)² + (y + 7)²]^1/2

-6x – 2y + 14 = 0

3x + y = 7 (i)

Similarly, OA = OC

[(x - 6)² + (y + 6)²]^1/2 = [(x - 3)² + (y - 3)²]^1/2

-6x + 18y + 54 = 0

-3x + 9y = -27 (ii)

On adding equation (i) and (ii), we obtain

10y = - 20

y = - 2

From equation (i), we obtain

3x - 2 = 7

3x = 9

x = 3

Therefore, the centre of the circle is (3, - 2)

Let ABCD be a square having ( - 1, 2) and (3, 2) as vertices A and C respectively. Let (x, y), (x₁, y₁) be the coordinate of vertex B and D respectively

We know that the sides of a square are equal to each other

From the figure,
AB = BC
Therefore,

√[(x + 1)² + (y - 2)²] = √[(x - 3)² + (y - 2)²]

x² + 2x + 1 + y² – 4y + 4 = x² + 9 – 6x + y² + 4 – 4y

8x = 8

x = 1

We know that in a square, all interior angles are of 90°.

In ΔABC,

AB² + BC² = AC²

√[(x + 1)² + (y - 2)²] ² + √[(x - 3)² + (y - 2)²] ² = √[(3 + 1)² + (2 - 2)²] ²

4 + y² + 4 - 4y + 4 + y² - 4y + 4 =16

2y² + 16 - 8 y =16

2y² - 8 y = 0

y(y - 4) = 0

y = 0 or 4

We know that in a square, the diagonals are equal in length and bisect each other at 90°.

Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD

(i) Taking A as origin,

We will take AD as x-axis and AB as y-axis.

It can be observed that the coordinates of point P,Q, and R are (4, 6), (3, 2), and (6, 5) respectively

Area of the triangle PQR = [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)

= [4 (2 – 5) + 3 (5 – 6) + 6 (6 – 2)

= (-12 – 3 + 24)

= Square units

(ii) Taking C as origin, CB as x-axis, and CD as y-axis, the coordinates of vertices P, Q, and R are (12, 2), (13, 6),and (10, 3) respectively

Area of the triangle PQR = [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ - y₂)

= [12 (6 – 3) + 13 (3 – 2) + 10 (2 – 6)

= (36 + 13 – 40)

= Square units

It can be observed that the area of the triangle is same in both the cases

Given that,

= =

Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3

Coordinates of point D = ()

= (, )

Coordinates of point E = (, )

= (, )

Area of the triangle = [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ - y₂)

= [4 (– ) + ( - 6) + (6 – )]

= [3 – + ]

= []

= 15/32 square units

Area of triangle ABC = [4 (5 – 2) + 1 (2 - 6) + 7 (6 – 5)] /2

= (12 – 4 + 7)/ 2

= 15/2 square units

Clearly, the ratio between the areas of ΔADE and ΔABC is 1:16

We know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel tothe third side of the triangle.

These two triangles so formed (here ΔADE and ΔABC) will be similar to each other.

Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles

And, ratio between the areas of ΔADE and ΔABC = (1/4)²

= 1/16

(i) Median AD of the triangle will divide the side BC in two equal parts

Therefore, D is the mid-point of side BC

Coordinates of D = ()

= ()

(ii) Point P divides the side AD in a ratio 2:1

Coordinates of P = ()

= (

(iii) Median BE of the triangle will divide the side AC in two equal parts.

Therefore, E is the mid-point of side AC

Coordinates of E = (, )

= ()

Point Q divides the side BE in a ratio 2:1

Coordinates of Q = ()

= (

Median CF of the triangle will divide the side AB in two equal parts. Therefore, F is the mid-point of side AB

Coordinates of F = ()

= (5, )

Point R divides the side CF in a ratio 2:1

Coordinates of R = ()

= ()

(iv) It can be observed that the coordinates of point P, Q, R are the same.Therefore, all these are representing the same point on the plane i.e., the centroid of the triangle

(v)



Consider a triangle, ΔABC, having its vertices as A(x₁, y₁), B(x₂, y₂), and C(x₃,y₃)

Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-point of side BC

Coordinates of D = (, )

Let the centroid of this triangle be O. Point O divides the side AD in a ratio 2:1

By section formula if (x, y) divides line joining points A(x₁, y₁), B(x₂, y₂) in ratio m:n

then,

Coordinates of O = ()

Centroid of ABC = ()

P is the mid-point of AB

Coordinates of P = (-1, )

Similarly coordinates of Q, R and S are (2, 4), (5, ) and (2, -1) respectively

Length of PQ = [(-1 – 2)² + ()²]^1/2

=

Length of QR = [(2 – 5)² + ()²]^1/2

=

Length of RS = [(5 – 2)² + (+ 1)²]^1/2

=

Length of SP = [(2+ 1)² + (-1 – )²]^1/2

=

Length of PR = [(-1- 5)² + (– )²]^1/2

= 6

Length of QS = [(2 – 2)² + (4 + 1)²]^1/2

= 5

It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are ofdifferent lengths. Therefore, PQRS is a rhombus