Co-ordinate Geometry

The coordinates of a point in represented as the ordered pair (abscissa, ordinate).

where abscissa is the X-coordinate or distance of that point from Y-axis

and ordinate is the Y-coordinate or distance of that point from X-axis.

Abscissa of point P = 5

Ordinate of point P = 3

⇒ Coordinates of point P = (5,3)

Abscissa of point Q = –4

Ordinate of point Q = 6

⇒ Coordinates of point Q = (–4,6)

Abscissa of point R = –3

Ordinate of point R = –2

⇒ Coordinates of point R = (–3,–2)

Abscissa of point S = –1

Ordinate of point S = –5

⇒ Coordinates of point S = (–1,–5)

Draw the rectangular coordinate axes.

The given points (x,y) are marked by measuring the distance x from Y-axis and distance y from X-axis to obtain the following:

Mark the points O(0,0), P(3,0) and R(0,4) on the rectangular coordinate axes.

Since OPQR is a rectangle, OP = RQ = 3 units

OR = PQ = 4 units

So, abscissa of point Q = 3

ordinate of point Q = 4

The coordinates of point Q are (3,4).

On plotting the given points, we observe that a pentagon is obtained.

(i) The vertices are plotted on the rectangular axes.

Since only a pair of the opposite sides is parallel, the obtained quadrilateral is trapezium.

(ii) The vertices are plotted on the rectangular axes.

Since both pairs of the opposite sides are parallel and equal, the obtained quadrilateral is rhombus.

We know that distance between two points (x₁, y₁) and (x₂, y₂) is given by

(i) Let the points (–6, 7) and (–1, –5) be A and B respectively, so the distance between them

(ii) Let the points (–1, –1) and (8, –2) be A and B respectively, so the distance between them

(iii) Let the points (at₁², 2at₁) and (at₂², 2at₂) be A and B respectively, so the distance between them

We have A → (2, –2)

B → (–2, 1)

C → (5, 2)

Using the distance formula,

We find that (AB)² + (CA)² = (BC)², or simply (5, 5, √50) forms a pythagorean triplet and hence the points A, B and C form a right angled triangle.

We have A → (1, –2)

B → (3, 0)

C → (1, 2)

D → (–1, 0)

Using the distance formula,

We find that AB = BC = CD = DA.

⇒ ABCD is a square.

We have A → (a, a)

B → (–a, –a)

C → ()

Using the distance formula,

We know that (a – b)² + (a + b)² = (a² + b² – 2 ab) + (a² + b² + 2 ab) = 2(a² + b² )

We find that AB = BC = CA.

⇒ Points A, B and C form an equilateral triangle.

We have A → (1, 1)

B → (–2, 7)

C → (3, –3)

Using the distance formula,

We find that,

AB + BC = 3√5 + 2√5 = 5√5 = CA

⇒ The points A, B and C are collinear.

We have A → (–2, –5)

B → (x, 0) equidistant from A and C

(y = 0 since B lies on X-axis)

C → (2, –3)

Using the distance formula,

Since B is equidistant from A and C

∴ AB = BC

Squaring both sides

8x + 16 = 0

x = –2

The required point is B (–2, 0).

We have A → (–5, –2)

B → (0, y) equidistant from A and C

(x = 0 since B lies on Y-axis)

C → (3, 2)

Using the distance formula,

Since B is equidistant from A and C

∴ AB = BC

Squaring both sides

8y + 16 = 0

y = –2

The required point is B (0, –2).

We have A → (3, k)

B → (0, 2)

C → (k, 5)

Using the distance formula,

Since B is equidistant from A and C

∴ AB = BC

Squaring both sides

4k = 4

∴ k = 1

We have P → (a cos θ, b sin θ)

Q → (–a sin θ, b cos θ)

We know that distance of a point A (x,y) from origin O (0, 0) is given as OA =

Using the above formula,

OP =

=

OP² =

OQ =

=

OQ² =

Now, OP² + OQ² = a² cos²θ + b² sin²θ + a² sin²θ + b² cos²θ

= a² (cos²θ +sin²θ) + b² (sin²θ + cos²θ)

We know that, cos²θ +sin²θ =1

∴ OP² + OQ² = a² + b²

Given two vertices of an equilateral traingle ABC

A → (0, 0)

B → (3, )

Let the third vertex be C → (x, y)

We know that distance of a point A (x,y) from origin O (0, 0) is given as OA =

So,

Using distance formula,

Since ∆ABC is an equilateral, AB = BC = CA.

Using CA = AB,

Squaring both sides, we get

x² + y² = 12 …(i)

Now, using AB = BC

(Using eq (i))

Squaring both sides, we get

– 6x – 2√3y + 24 = 12

⇒ 6x + 2√3y = 12

⇒ 3x + √3y = 6

Substituting the value of x in eq (i), we get

⇒ 12y² – 12√3y +36 = 108

⇒ 12y² – 12√3y – 72 = 0
⇒ y² – √3y – 6 = 0

⇒ y² – 2√3y + √3y – 6 = 0

⇒ y (y – 2√3) + √3 (y – 2√3) = 0

⇒ (y – 2√3) (y + √3) = 0
⇒ y = 2√3 or – √3
If y = 2√3, then
If y = –√3,
So, the third vertex of the equilateral triangle = (0, 2√3) or (3, –√3).

We know that

Coordinates of a point P(x,y) dividing the line segment joining A (x₁, y₁) and B (x₂, y₂) in the ratio m:n internally are

We have A → (3, 5)

and B → (7, 9).

Coordinates of point P(x,y) dividing AB in the ratio 2:3 internally are

is the required point.

We know that

Coordinates of a point P(x,y) dividing the line segment joining A (x₁, y₁) and B (x₂, y₂) in the ratio m:n externally are

We have A → (5, –2)

and B → or B

Coordinates of point P (x,y) dividing AB in the ratio 7:9 externally are

is the required point.

Internal divison formula:

Coordinates of a point P(x,y) dividing the line segment joining A (x₁, y₁) and B (x₂, y₂) in the ratio m:n internally are

Let O divide the line segment joining the given points in the ratio λ:1. Then by internal division formula,

3λ = 1

Hence proved that, the origin O divides the line segment joining the points, A(1, –3) and B(–3, 9) in the ratio 1 : 3 internally.

External divison formula:

Coordinates of a point P(x,y) dividing the line segment joining A (x₁, y₁) and B (x₂, y₂) in the ratio m:n externally are

Using external division formula,

x =

y =

P (3,–9) is the required point.

We know that, coordinates of the mid–point of of the line segment joining the points A (x₁, y₁) and B (x₂, y₂) are given by

So, the coordinates of the mid–point of the line joining the points (22, 20) and (0, 16) are

=(11,18)

Let the line segment joining (5, 3) and (–3, –2) be divided by X-axis in the ratio λ : 1 by point P.

Since P lies on X-axis, so its ordinate = 0

By internal division formula,

y =

⇒ 2λ = 3

⇒ λ =

∴ Required ratio is =3:2

Let the line segment joining (2, –3) and (5, 6) be divided by Y-axis in the ratio λ : 1 by point P.

Since P lies on Y-axis, so its abscissa = 0

By internal division formula are

x =

⇒ 5λ = –2

The negative sign implies external division.

⇒ λ = (externally)

∴ Required ratio is =2:5 externally.

Let the line segment joining (15, 5) and (9, 20) be divided by P(11, 15) in the ratio λ : 1.

By internal division formula,

11 =

⇒ 11λ + 11 = 9λ + 15

⇒ 2λ = 4

⇒ λ = 2

∴ Required ratio is =2:1

Note: Alternatively, the procedure could have been carried out by y-coordinate.

Let the coordinates of B be (x, y).

Using internal division formula, we have

⇒

⇒ 4x – 14 = 33

⇒ 4x = 47

⇒ x =

and 5 =

⇒

⇒ 4y + 21 = 55

⇒ 4y = 34

⇒ y =

B is the required point.

Let the line segment joining the points A (1, 2) and B (11, 9) be trisected at points P (x₁, y₁) and Q (x₂, y₂).

Clearly, P divides the line segment AB internally in the ratio1:2.

∴ By internal division formula,

x₁ =

y₁ =

Again, Q divides the line segment AB internally in the ratio 2:1.

∴ By internal division formula,

x₂ =

y₂ =

∴ The required points are P and Q .

Let the line segment joining the points A (–4, 0) and B (0, 6) be divided into 4 parts by points P (x₁, y₁), Q (x₂, y₂) and R(x₃, y₃).

Clearly, P divides the line segment AB internally in the ratio1:3.

∴ By internal division formula,

x₁ = = –3

y₁ =

Again, Q divides the line segment AB internally in the ratio 1:1.

∴ By mid–point formula,

x₂ = = –2

y₂ = =3

Also, R divides the line segment AB internally in the ratio3:1.

∴ By internal division formula,

x₃ = = –1

y₃ =

∴ The required points are P , Q (–2, 3)and R .

Let the line segment joining (1, 3) and (2, 7) be divided by the given line in the ratio λ : 1 at point P (x, y).

By internal division formula

x =

y =

Since P lies on the line 3x + y = 9, it must satisfy its equation.

Hence, 3× = 9

6λ +3 + 7λ +3 = 9 (λ + 1)

13λ + 6 = 9λ + 9

4λ = 3

⇒ λ =

∴ Required ratio is =3:4.

Let the line segment joining (–5, –4) and (–2, 3) be divided by the point (–3, p) in the ratio λ : 1.

By internal division formula

⇒ –2λ – 5 = –3λ – 3

⇒ λ = 2

∴ Required ratio is =2:1.

Now,

∴ p

Distance of point (3,4) from Y-axis = abscissa of the point = 3

∴The correct option is D.

Distance of point (5, –2) from X-axis = ordinate of the point = |–2| = 2

∴The correct option is B.

We have A → (0, 3)

B → (–2, 0)

Using the distance formula,

AB =

=

=

∴ The correct option is C.

We have A → (–2, 1)

B → (2, –2)

C → (5, 2)

Using the distance formula,

AB =

=

=

= = 5

BC =

=

=

=

CA =

=

=

=

We find that (AB)² + (CA)² = (BC)², or simply (5, 5, √50) forms a pythagorean triplet and hence the points A, B and C form a right angled triangle.

∴ The correct option is A.

We have A → (–1, 1)

B → (0, –3)

C → (5, 2)

D → (4, 6)

Using the distance formula,

AB =

=

=

=

BC =

=

=

=

CD=

=

=

=

DA =

=

=

=s

We find that AB = CDand BC = DA.

The oppposite pairs of sides are equal for the given quadrilateral.

Consider the diagonals, AC and BD.

AC =

=

=

=

BD =

=

=

=

Since the diagonals are not equal, the given quadrilateral is not a rectangle.

(This is also visible from the figure since angles are not 90°.)

⇒ ABCD is a parallelogram.

∴ The correct option is D.

We have A → (0, 0)

B → (2, 0)

C → (0, 2)

Let D (x,y) be equidistant from A, B and C.

We know that distance of a point A (x,y) from origin O (0, 0) is given as OA =

∴ AD = …(i)

Using the distance formula,

BD =

=

= –––(ii)

CD =

= –––(iii)

Since D is equidistant from A, B and C

Equating eq (i) and (ii)

Squaring both sides

4y – 4 = 0

y = 1

Equating eq (i) and (iii)

Squaring both sides

4x – 4 = 0

x = 1

The required point is D (1, 1).

∴ The correct option is D.

Coordinates of point P (x,y) dividing the line segment joining the points (5, 0) and (0, 4) in the ratio 2 : 3 internally are:

x =

∴ x = 3

And y =

P is the required point.

∴ The correct option is A.

Given, points A (1, 2), B (–1, x) and C (2, 3) are collinear.

⇒ Slope of AB = Slope of AC

∴ x – 2 = –2 (1)

∴ x = 0.

∴ The correct option is B.

Using the distance formula, the distance between the points (3, a) and (4, 1)

√10 =

=

=

=

Squaring both sides, we get

a² – 2a +2 = 10

⇒ a² – 2a – 8 = 0

⇒ a² – 4a + 2a – 8 = 0

⇒ a (a – 4)+ 2 (a – 4) = 0

⇒ (a – 4) (a + 2) = 0

∴ Either a = 4 or a = –2.

∴ The correct option is C.

We have A → (2, 1)

B → (1, –2)

P → (x, y)

Using the distance formula,

AP =

=

=

BP =

=

=

=

Since P is equidistant from A and B

∴ AP = BP

=

Squaring both sides, we get

x² +y² – 4x – 2y + 5 = x² +y² – 2x + 4y + 5

2x + 6y = 0

or x + 3y = 0

∴ The correct option is A.

We have A → (1, 4)

B → (–5, 4)

C → (–5, –3)

D → (1, –3)

Using the distance formula,

AB =

=

=

=6

BC =

=

=

= 7

CD =

=

=

=

= 6

DA =

=

=

= 7

The opposite sides are equal in length.

Consider the diagonals, AC and BD.

AC =

=

=

=

BD =

=

=

=

Since the diagonals are also equal, the given quadrilateral is a rectangle.

On plotting and joining the points (–2, 0), (2, 0), (2, 2), (0, 4), (–2, 2) in order on the rectangular axes, we obtain the following figure:

which corresponds to a pentagon.

Let the line segment joining (1, 2) and (6, 7) be divided by the point (3, 4) in the ratio λ : 1.

By internal division formula,

3 =

3λ +3 = 6λ + 1

⇒ 3λ = 2

⇒ λ =

∴ Required ratio is =2:3.

Using the distance formula, length of the diagonal

=

=

=

= = 10

∴ The length of the diagonal of given square = 10 units.

Let the coordinates of the other end of the line segment be (x, y).

Using the mid–point formula, we get

4 =

⇒ 4 + x = 8

⇒ x = 4

and

1 =

⇒ y = 2

∴ The coordinates of the other end of the line segment are (4, 2).

Using mid–point formula:

Coordinates of the mid–point of the line segment joining the points (6, 8) and (2, 4) are

= (4, 6)

Using the distance formula,

Distance between the points (1, 2) and (4, 6) =

=

=

= 5

We have P → (2,–1)

Q → (3,4)

R → (–2,3)

S → (–3,–2)

Using the distance formula,

PQ =

=

=

=

QR =

=

=

=

RS =

=

=

=

SP =

=

=

=

All four sides of quadrilateral PQRS are equal.

Consider the diagonals, AC and BD.

PR =

=

=

=

QS =

=

=

=

Since the diagonals are not equal, the given quadrilateral is not a square.

⇒ PQRS is a rhombus.

Consider a right angled ∆AOB, such that C is the mid–point of hypotenuse AB.

We have O → (0, 0)

Since A lies on y-axis, A→ (0, y)

and B lies on x-axis, B→ (x, 0)

Using mid–point formula, coordinates of C are

Using distance formula,

AC =

=

=

=

=

BC =

=

=

=

=

We know that distance of a point P (x,y) from origin O (0, 0) is given as OP =

∴ OC =

=

=

We can observe that OA = OB = OC.

∴ C (mid–point of hypotenuse AB) is equidistant from all the three vertices of the right angled ∆AOB.

We have the vertices of ∆ABC, A (1, –1), B (0, 4) and C (–5, 3).

Let D, E and F be the mid points of the sides BC, CA and AB respectively.

Using the mid–point formula,

Coordinates of D are

Coordinates of E are = (–2, 1)

Coordinates of F are

Using distance formula,

Length of median AD =

=

=

=

=

=

Length of median BE =

=

=

=

Length of median CF =

=

=

=

=

=

∴ The lengths of medians of the triangle whose vertices are (1, –1), (0, 4) and (–5, 3) are and √13.

Using mid–point formula:

Coordinates of the mid–point of the line segment joining the points (5, 7) and (3, 9) are

= (4, 8)

and

Coordinates of the mid–point of the line segment joining the points (8, 6) and (0, 10) are

= (4, 8)

Hence proved that the mid–point of the line segment joining the points (5, 7) and (3, 9) is the same as the mid–point the line segment joining the points (8, 6) and (0, 10).

Let A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) be the vertices of the given triangle

and D (1, 2), E (2, –1) and F (0, –1) be the mid points of the sides BC, CA and AB respectively.

Using the mid–point formula,

=(0, –1)

⇒ x₁ +x₂ = 0 and y₁ +y₂ = –2 ……(i)

=(2, –1)

⇒ x₁ +x₃ = 4 and y₁ +y₃ = –2 ……(ii)

=(1, 2)

⇒ x₂ +x₃ = 2 and y₂ +y₃ = 4 ……(iii)

Adding all the equations obtained for x coordinates and y coordinates respectively, we get

2(x₁ + x₂ + x₃) = 6

and 2(y₁ + y₂ + y₃) = 0

⇒ x₁ + x₂ + x₃ = 3 and y₁ + y₂ + y₃ = 0 –––(iv)

From eq (i) and (iv), we get x₃ = 3 , y₃ = 2 , i.e. C (3, 2)

From eq (ii) and (iv), we get x₂ = –1 , y₂ = 2, i.e. B (–1, 2)

From eq (iii) and (iv), we get x₁ = 1 , y₃ = –4, i.e. A (1, –4)

∴ The coordinates of the vertices of the triangle are A (1, –4), B (–1, 2) and C (3, 2).