Co-ordinate Geometry: Internal And External Division Of Straight Line Segment

Let the co-ordinates of the point which divides the line segment be (x, y) –

And, we know, by section formula

(Where, m and n are the ratios, (x₁, y₁) and (x₂, y₂) are the coordinates of the line segment.)

⇒ x = 0

And,

⇒ y = 2

∴ the co-ordinates of the point which divides the line segment joining two point s in the given ratio is (0, 2).

Let the co-ordinates of the point which divides the line segment be (x, y) –

And, we know, By section formula

(where, m and n are the ratios, (x₁, y₁) and (x₂, y₂) are the coordinates of the line segment.)

And,

⇒ y = 1

∴ The co-ordinates of the point which divides the line segment joining two point s in the given ratio is .

Let the co-ordinates of the point which divides the line segment be (x, y) –

And, we know, By section formula

(where, m and n are the ratios, (x₁, y₁) and (x₂, y₂) are the coordinates of the line segment.)

⇒ x = 14

And,

⇒ y = – 19

∴ The co-ordinates of the point which divides the line segment joining two point s in the given ratio is (14, – 19).

Let the co-ordinates of the point which divides the line segment be (x, y) –

And, we know, By section formula

(Where, m and n are the ratios, (x₁, y₁) and (x₂, y₂) are the coordinates of the line segment.)

⇒ x = 10

And,

⇒ y = 8

∴ the co-ordinates of the point which divides the line segment joining two point s in the given ratio is (10, 8).

Let the co-ordinates of mid-point be (x, y).

And, we know mid-point formula, i.e. the coordinates of mid-point of line joining (x₁, y₁) and (x₂, y₂) is

⇒x = 4 and y = 0

⇒ Co – ordinates of mid-point = (4, 0)

(ii) Let the co-ordinates of mid-point be (x, y).

And, since it is a mid-point –

⇒ co-ordinates of mid-point =

We know, By section formula

(where, m and n are the ratios(externally), (x₁, y₁) and (x₂, y₂) are the coordinates of the line segment, and (x, y) are coordinates of dividing point .)

Now taking in consideration the equation for x one only –

⇒ m – n = 3m – 4n

⇒ 3n = 2m

⇒ (1, 3) divides the line segment joining the point s (4, 6) and (3, 5) in the ratio 3: 2 externally.

Since it is divided by the y – axis, ∴ the coordinate of that point will be –(0, y).

We know, By section formula

(Where, m and n are the ratios (internally), (x₁, y₁) and (x₂, y₂) are the coordinates of the line segment, and (0, y) are coordinates of dividing point .)

Now taking in consideration the equation for x one only –

⇒ m + n = – 9m + 7n

⇒ 10m = 6n

⇒ (1, 3) divides the line segment joining the point s (4, 6) and (3, 5) in the ratio 3: 5 internally.

We know that a quadrilateral is a parallelogram if the co-ordinates of mid-point s of its both the diagonals are same.

Therefore, we’ll find the mid-point s of diagonal AC and BD.

Let the co-ordinates of mid-point of AC be (x₃, y3).

And, we know mid-point formula, i.e. the coordinates of mid-point of line joining (x₁, y₁) and (x₂, y₂) is

(Where, (x₁, y₁) and (x₂, y₂) are the coordinates of A and C

⇒ Co – ordinates of mid-point of AC =

Now, Let the co-ordinates of mid-point of BD be (x₄, y₄).

And, since it is a mid-point –

(Where, (x₅, y₅) and (x₆, y₆) are the coordinates of B and D.

⇒ Co – ordinates of mid-point of BD =

And, since these are equal –

⇒ ABCD is a parallelogram.

Let A (3, 2), B (6, 3), C (x, y) and D (6, 5) We know that a quadrilateral is a parallelogram if the co-ordinates of mid-point s of its both the diagonals are same.

Therefore, we’ll use the mid-point s of diagonal AC and BD.

Let the co-ordinates of mid-point of AC be (x₃, y₃).

And, we know mid-point formula, i.e. the coordinates of mid-point of line joining (x₁, y₁) and (x₂, y₂) is

(Where, (x₁, y₁) and (x₂, y₂) are the coordinates of A and C

Now, Let the co-ordinates of mid-point of BD be (x₄, y₄).

And, since it is a mid-point –

(Where, (x₅, y₅) and (x₆, y₆) are the coordinates of B and D.

⇒x = 6 and y = 4

⇒ Co – ordinates of mid-point of BD =

And, since it is a parallelogram –

⇒ x₄ = x₃ and y₄ = y₃

⇒ x = 9 and y = 6.

A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) and D(x₄, y₄)

We know that a quadrilateral is a parallelogram if the co-ordinates of mid-point s of its both the diagonals are same.

Therefore, we’ll find the mid-point s of diagonal AC and BD.

Let the co-ordinates of mid-point of AC be (x₀, y₀).

And, we know mid-point formula, i.e. the coordinates of mid-point of line joining (x₁, y₁) and (x₂, y₂) is

Similarly, let the co-ordinates of mid-point of AC be (x₅, y₅).

And, since it is a mid-point –

Now, since ABCD is a parallelogram –

⇒ (x₀, y₀) = (x₅, y₅)

⇒ x₀ = x₅ and y₀ = y₅

⇒x₁ + x₃ = x₂ + x₄ and y₁ + y₃ = y₂ + y₄

To calculate, the length of Median AD, first we’ll calculated the coordinates of mid-point of BC.

Let the coordinates of that mid-point be (x, y) –

And, we know mid-point formula, i.e. the coordinates of mid-point of line joining (x₁, y₁) and (x₂, y₂) is

⇒ x = 3 and y = 0

⇒ the coordinates of one end of median(x₁, y₁) = (– 1, 3) and of another end(x₂, y₂) = (3, 0).

Now, we know the length = √((x₂ – x₁)² + (y₂ – y₁)²)

⇒ Length of median = √ ((3 –(– 1))² + (0 – 3)²)

⇒ Length of median = √ (16 + 9)

⇒ Length of median = √ 25

⇒ Length of median = 5

The co-ordinates of vertices of a triangle ABC are A(2, – 4), B(6, –2) and C (– 4, 2) respectively.

To calculate, the length of Median AD, first we’ll calculated the coordinates of mid-point (d) of BC.

Let the coordinates of that mid-point be (x, y) –

And, we know mid-point formula, i.e. the coordinates of mid-point of line joining (x₁, y₁) and (x₂, y₂) is

⇒ x = 1 and y = 0

⇒ the coordinates of one end of median(x₁, y₁) = (2, – 4) and of another end(x₂, y₂) = (1, 0).

Now, we know the length = √ ((x₂ – x₁)² + (y₂ – y₁)²)

⇒ Length of median = √ ((1 –(2))² + (0 –(– 4))²)

⇒ Length of median = √ (1 + 16)

⇒ Length of median = √ 17

Now, to calculate, the length of Median BE, first we’ll calculated the coordinates of mid-point (E) of AC.

Let the coordinates of that mid-point be (x, y) –

⇒ x = – 1 and y = – 1

⇒ The coordinates of one end of median(x₁, y₁) = (6, – 2) and of another end(x₂, y₂) = (– 1, – 1).

Now, we know the length = √ ((x₂ – x₁)² + (y₂ – y₁)²)

⇒ Length of median = √ ((– 1 –(6))² + (– 1 –(– 2))²)

⇒ Length of median = √ (49 + 1)

⇒ Length of median = √ 50

And, now To calculate, the length of Median CG, first we’ll calculated the coordinates of mid-point (G) of AB.

Let the coordinates of that mid-point be (x, y) –

⇒ x = 4 and y = – 3

⇒ The coordinates of one end of median(x₁, y₁) = (– 4, 2) and of another end(x₂, y₂) = (4, – 3).

Now, we know the length)

⇒ Length of median

⇒ Length of median

⇒ Length of median

Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) be the vertices of triangle and D(4, 3) be the mid-point of AB, E(–2, 7) be the mid-point of BC and F(0, 11) be the mid-point of AC.

And, we know mid-point formula, i.e. the coordinates of mid-point of line joining (x₁, y₁) and (x₂, y₂) is

⇒ x₁ + x₂ = 8, ………(1)

y₁ + y₂ = 6, ………(2)

x₂ + x₃ = – 4, ……..(3)

y₂ + y₃ = 14, …….(4)

x₁ + x₃ = 0, ……..(5)

y₁ + y₃ = 22 ………(6)

Now, adding (1) and (3), we get –

x₁ + 2x₂ + x₃ = 4…..(7)

Now, subtracting (5) from (7), we get –

2x₂ = 4

⇒ x₂ = 2

Now, putting this in (3), we get –

2 + x₃ = – 4

⇒ x₃ = – 6

Now, putting this in (5), we get –

x₁ – 6 = 0

x₁ = 6

And now, adding (2) and (4), we get –

y₁ + 2y₂ + y₃ = 20…..(8)

Now, subtracting (6) from (8), we get –

2y₂ = – 2

⇒ y₂ = – 2

Now, putting this in (4), we get –

– 2 + y₃ = 14

⇒ y₃ = 16

Now, putting this in (6), we get –

y₁ + 16 = 22

y₁ = 6

∴, A(x₁, y₁) = (6, 7), B(x₂, y₂) = (2, – 1), C(x₃, y₃) = (– 6, 15)

Let the coordinates of that mid-point be (x, y) –

And, we know mid-point formula, i.e. the coordinates of mid-point of line joining (x₁, y₁) and (x₂, y₂) is

Let the co-ordinates of the point which divides the line segment be (x, y) –

We know, abscissa is x – coordinate of any point .

We know, By section formula

⇒ x = – 1

Since, centre of circle will be the mid-point of diameter.

And, we know mid-point formula, i.e. the coordinates of mid-point of line joining (x₁, y₁) and (x₂, y₂) is

⇒x = 3 and y = 3

Let the co-ordinates of the point which divides the line segment be (x, y) –

We know, ordinate is y – coordinate of any point .

And, we know, By section formula

⇒ y = 7

P(1, 2), Q(4, 6), R(5, 7) and S(x, y) We know that a quadrilateral is a parallelogram if the co-ordinates of mid-point s of its both the diagonals are same.

Therefore, we’ll use the mid-point s of diagonal PR and QS.

Let the co-ordinates of mid-point of PR be (x₃, y₃).

And, we know mid-point formula, i.e. the coordinates of mid-point of line joining (x₁, y₁) and (x₂, y₂) is

(where, (x₁, y₁) and (x₂, y₂) are the coordinates of P and R

Now, Let the co-ordinates of mid-point of QS be (x₄, y₄).

And, since it is a mid-point –

(where, (x₅, y₅) and (x₆, y₆) are the coordinates of Q and S.

And, since it is a parallelogram –

⇒ x₄ = x₃ and y₄ = y₃

⇒ x = 2 and y = 3.

Since, centre of circle will be the mid-point of diameter.

Now, Let the coordinates of point B be (x, y) –

⇒x = 10 – 6, and y = – 4 + 7

⇒ x = 4, and y = 3.

∴ coordinates of point B = (4, 3)

The coordinates of point P will be –(6, 4), because it lies in 1^st quadrant. And point Q will be (– 6, – 4), because it’s in 3^rd quadrant.

Let the co-ordinates of mid-point of PR be (x, y).

And, since it is a mid-point –

⇒ x = 0 and y = 0

⇒ The mid-point is actually the origin(0, 0).

The coordinates of point A will be –(– 8, 6), because it lies in 2^nd quadrant. And point B will be (8, – 6), because it’s in 4^th quadrant.

Let the co-ordinates of mid-point of PR be (x, y).

And, since it is a mid-point –

⇒ x = 0 and y = 0

⇒ The mid-point is actually the origin(0, 0).

Since, AP = BP ⇒ P is a mid-point .

And, since it is a mid-point –

⇒ x = – 1 and y = – 1

⇒ co-ordinates of P = (– 1, – 1).

Since, the sides are parallel to the co-ordinates axes –

⇒ the next coordinate can be determined by adding the distance between the two point s.

Or, the other co-ordinates can easily be found by interchanging abscissa of one with other and ordinate of one with another, which will make it –

C(2, 3), since its abscissa should be same as D and ordinate as B.

Similarly, A(7, 6), since its abscissa should be same as B and ordinate as D.