Coordinates

[To solve such questions, following are the key points:

1. X coordinate of any point on Y axis is 0.

2. Y coordinate of any point on X axis is 0.

3. If a line is parallel to X axis, then its X coordinate varies, but Y coordinate remains constant.

4. If a line is parallel to Y axis, then its Y coordinate varies, but X coordinate remains constant.]

Since, the equation of X – axis is y = 0, y coordinate of any point on X – axis is 0(zero).

Since, the equation of Y axis is x = 0,coordinate of any point on Y axis is 0(zero).

Origin is the intersection of X axis and Y axis, the i.e. intersection of lines y = 0 and x = 0.

Therefore,

Coordinates of origin is (0,0).

Since the line is parallel to x – axis, its equation will be of the form y = c, where c is a constant.

Point (0,1) is on the line, therefore, it will satisfy the equation of the line.

Putting y = 1 in the equation of a line,

We get,

1 = c.

Therefore, the equation of line will be y = 1.

Hence, y coordinate of any point on the line through (0,1),parallel to the x axis is 1.

Since the line is parallel to y – axis , its equation will be of the form x = c, where c is a constant.

Point (1,0) is on the line, therefore, it will satisfy the equation of the line.

Putting x = 1 in the equation of a line,

We get,

1 = c.

Therefore, the equation of line will be x = 1.

Hence, y coordinate of any point on the line through (1,0), parallel to the y axis is 1.

Vertex O is the origin.

Therefore, its coordinates are (0,0).

Vertex A is on the Y – axis.

Therefore its X coordinate = 0.

Line joining vertex A and (4,3) in parallel to X – axis, therefore its y coordinate will remain constant. Since point A is on the line, its Y coordinate = 3

Therefore coordinates of vertex A are(0,3)

Vertex B is on X – axis.

Therefore it's Y coordinate = 0.

Line joining vertex B and (4,3) in parallel to Y – axis, therefore its x coordinate will remain constant. Since point A is on the line, its x coordinate = 4

Therefore coordinates of vertex A are(4,0)

Hence, the coordinate of the other three vertices of the given rectangle is (0,0),(0,3) and (4,0).

Since, O is the midpoint of the rectangle and sides of the rectangle are parallel to coordinate axis,

The rectangle is symmetrical about both X and Y axis.

Line AB is parallel to Y – axis,

∴ X coordinate will remain same

∴ X coordinate of B = 3

A is 2 units above X – axis , therefore by symmetry B will be 2 units below X – axis.

∴ Y coordinate of B = – 2

∴ coordinates of B = (3, – 2)

BC is parallel to the X axis

∴ Y coordinate will remain constant

∴ Y coordinate of C = – 2

B is 3 units right to the Y axis,

∴ by symmetry C will be 3 units to left of Y – axis.

∴ X coordinate of C = – 3

∴ coordinates of C = (– 3, – 2)

CD is parallel to Y – axis,

∴ X coordinate will remain constant

∴ X coordinate of D = – 3

The AD is parallel to X – axis,

∴ Y coordinate will remain constant

∴Y coordinate of D = 2

∴ coordinates of D = (– 3,2)

Hence, the coordinates of other three vertices are (3, – 2),(– 3, – 2)

And (– 3,2)

Vertex O is the origin.

Therefore, its coordinates (0,0).

Vertex A is on X axis, therefore, its y coordinate = 0.

Vertex A is 4 units away from y – axis in the direction of the positive X – axis.

Therefore, the x coordinate of A = 4

∴ coordinates of A = (4,0).

Consider triangle OBD,

OB = 4 units(All sides of an equilateral triangle are equal)

∠ BOD = 60°

Let coordinates of B be (q,p)

Then,

BD = p

OD = q

2√3 = p

Therefore, the y coordinate of B = 2

2 = q

Therefore, the x coordinate of B = 2

∴ coordinates of B = (2√3,2)

Hence, the coordinates of the vertices of the triangle are (0,0),(4,0) and(2√3,2)

Since, the four trapeziums are equal

OA = AB

∴ A bisects OB

OB = 8

∴ OA = 4

∴coordinates of A = (4,0)[A is on X axis]

And coordinates of B = (8,0)[B is on X axis]

BC is parallel to Y – axis,

∴ X coordinate will remain constant

∴ X coordinate of C = 8

Since all four trapeziums are equal,

BC = OA = 4 = Y coordinate of C

∴ Y coordinate of C = 4

∴coordinates of C = (8,4)

AF = GD[All trapeziums are equal]

AF + GD = BC = 4

∴ AF = GD = 2

AF = Y coordinate of F = 2

Line AF is parallel to Y – axis,

∴ X coordinate will remain constant.

∴ X coordinate of F = 4

∴coordinates of F = (4,2)

Line FG is parallel to X – axis,

∴ Y coordinate will remain constant

∴ Y coordinate of G = 2

FG = DC[All trapeziums are equal]

FG + DC = AB = 4

∴ FG = DC = 2

X coordinate of G = X coordinate of F + FG

= 4 + 2 = 6

∴coordinates of G = (6,2)

CD is parallel to X – axis,

∴ Y coordinate will remain constant

∴ Y coordinate of D = 4

GD is parallel to Y – axis,

∴ X coordinate will remain constant

∴ X coordinate of D = 6

∴coordinates of D = (6,4)

CI is parallel to X axis

∴ Y coordinate will remain constant

∴ Y coordinate of I = 4

AI us parallel to Y – axis,

∴ X coordinate will remain constant

∴ X coordinate of I = 4

∴coordinates of I = (4,4)

OH and HI will be equal in length and will be at equal angles with X axis[All trapeziums are equal]

∴ H will be the midpoint of OI

∴ H will be the mid point of(0,0) and (4,4)

∴coordinates of H = (2,2)

Coordinates of A = (– 2,3)

Coordinates of C = (2,4)

Line AB is parallel to X – axis ,

∴ Y coordinate will remain constant.

∴ Y – coordinate of B = 3

Line BC is parallel to Y – axis ,

∴ X coordinate will remain constant.

∴ X coordinate of B = 2

∴coordinates of B = (2,3)

Line AD is parallel to Y – axis ,

∴ X coordinate will remain constant.

∴ X coordinate of D = – 2

Line CD is parallel to X – axis ,

∴ Y coordinate will remain constant.

∴ Y coordinate of D = 4

∴coordinates of D = (– 2,4)

Hence, the remaining vertices of the rectangle are (2,3),(– 2,4).

Coordinates of A = (2, – 4)

Coordinates of C = (– 1, – 2)

Line AB is parallel to X – axis ,

∴ Y coordinate will remain constant.

∴ Y – coordinate of B = – 4

Line BC is parallel to Y – axis ,

∴ X coordinate will remain constant.

∴ X coordinate of B = – 1

∴coordinates of B = (– 1, – 4)

Line AD is parallel to Y – axis ,

∴ X coordinate will remain constant.

∴ X coordinate of D = 2

Line CD is parallel to X – axis ,

∴ Y coordinate will remain constant.

∴ Y coordinate of D = – 2

∴coordinates of D = (2, – 2)

Hence, the remaining vertices of the rectangle are (– 1, – 4) and (2, – 2).

Coordinates of A = (– 1,3)

Coordinates of C = (2,6)

Line AB is parallel to X – axis ,

∴ Y coordinate will remain constant.

∴ Y – coordinate of B = 3

Line BC is parallel to Y – axis ,

∴ X coordinate will remain constant.

∴ X coordinate of B = 2

∴coordinates of B = (2,3)

Line AD is parallel to Y – axis ,

∴ X coordinate will remain constant.

∴ X coordinate of D = – 1

Line CD is parallel to X – axis ,

∴ Y coordinate will remain constant.

∴ Y coordinate of D = 6

∴coordinates of D = (– 1,6)

Hence, the remaining vertices of the rectangle are (2,3) and (– 1,6).

Steps for marking the given points:

1) Mark Point (3,5) anywhere.

2) Move (7 – 3) = 4 units right.

3) Move (8 – 5) = 3 units up.

4) Mark the point at the current position.

Since sides of the rectangle are parallel to coordinate axes,

Their equations will be

x = 3

x = 7

y = 5

y = 8

Point of intersection of x = 3 and y = 5 is (3,5) [Given point]

Point of intersection of x = 3 and y = 8 is (3,8)

Point of intersection of x = 7 and y = 5 is (7,5)

Point of intersection of x = 7 and y = 8 is (7,8) [Given point]

Hence, the coordinates of the other vertices of the rectangle are (3,8) and (7,5)

Steps for marking the given points:

1) Mark Point (6,2) anywhere.

2) Move (5 – 6) = – 1 unit right, i.e. 1 unit left.

3) Move (4 – 2) = 2 units up.

4) Mark the point at the current position.

Since sides of the rectangle are parallel to coordinate axes,

Their equations will be

x = 6

x = 5

y = 2

y = 4

Point of intersection of x = 6 and y = 2 is (6,2) [Given point]

Point of intersection of x = 5 and y = 2 is (5,2)

Point of intersection of x = 6 and y = 4 is (6,4)

Point of intersection of x = 5 and y = 4 is (5,4) [Given point]

Hence, the coordinates of the other vertices of the rectangle are (5,2) and (6,4)

Steps for marking the given points:

1) Mark Point (– 3,5) anywhere.

2) Move (– 7 – (– 3)) = – 4 units right i.e. 4 units left.

3) Move (1 – 5) = – 4 units up i.e. 4 units down.

4) Mark the point at the current position.

Since sides of the rectangle are parallel to coordinate axes,

Their equations will be

x = – 3

x = – 7

y = 5

y = 1

Point of intersection of x = – 3 and y = 5 is (3,5) [Given point]

Point of intersection of x = – 7 and y = 5 is (– 7,5)

Point of intersection of x = – 3 and y = 1 is (– 3,1)

Point of intersection of x = – 7 and y = 1 is (– 7,1) [Given point]

Hence, the coordinates of the other vertices of the rectangle are (– 3,1) and (– 7,5)

Steps for marking the given points:

1) Mark Point (– 1, – 2) anywhere.

2) Move (– 5 – (– 1)) = – 4 units right i.e. 4 units left.

3) Move (– 4 – (– 2)) = – 2 units up i.e. 2 units down.

4) Mark the point at the current position.

Since sides of the rectangle are parallel to coordinate axes,

Their equations will be

x = – 1

x = – 5

y = – 2

y = – 4

Point of intersection of x = – 1 and y = – 2 is (– 1, – 2) [Given point]

Point of intersection of x = – 1 and y = – 4 is (– 1, – 4)

Point of intersection of x = – 5 and y = – 2 is (– 5, – 2)

Point of intersection of x = – 5 and y = – 4 is (– 5, – 4) [Given point]

Hence, the coordinates of the other vertices of the rectangle are (– 1, – 4) and (– 5, – 2)

Let A = (0,0)

B = (1, – 2)

C = (– 3, – 2)

D = (– 3,1)

Here, AB, BC, CD, DA are the sides of the quadrilateral and AC and BD are the diagonals of the quadrilateral.

Length of Side

AB = distance between point A and B =

Here x₂ = 1,y₂ = – 2,x₁ = 0,y₁ = 0

∴ AB =

∴ AB =

∴ AB =

∴AB = √5unit

BC = distance between point B and C =

Here x₂ = – 3,y₂ = – 2,x₁ = 1,y₁ = – 2

∴ BC =

∴ BC =

∴ BC =

∴BC = 2 units

CD = distance between point C and D =

Here x₂ = – 3,y₂ = 1,x₁ = – 3,y₁ = – 2

∴ CD =

∴ CD =

∴ CD =

∴CD = 3 units

DA = distance between point D and A =

Here x₂ = 0,y₂ = 0,x₁ = – 3,y₁ = 1

∴ DA =

∴ DA =

∴ DA =

∴DA = √10 units

Length of diagonal

AC = distance between point A and C =

Here x₂ = – 3,y₂ = – 2,x₁ = 0,y₁ = 0

∴ AC =

∴ AC =

∴ AC =

∴AC = √13 units

Length of diagonal

BD = distance between point B and D =

Here x₂ = – 3,y₂ = 1,x₁ = 1,y₁ = – 2

∴ BD =

∴ BD =

∴ BD =

∴BD = 5 units

Hence, length of the sides of the quadrilateral is √5,2,3,√10 units.

Length of the diagonals of the quadrilateral are √13 and 5 units.

Let A = (2,1)

B = (3,4)

C = (– 3,6)

Length of Side

AB = distance between point A and B =

Here x₂ = 3,y₂ = 4,x₁ = 2,y₁ = 1

∴ AB =

∴ AB =

∴ AB =

∴AB = √10 units

BC = distance between point B and C =

Here x₂ = – 3,y₂ = 6,x₁ = 3,y₁ = 4

∴ BC =

∴ BC =

∴ BC =

∴BC = √40units

CA = distance between point C and A =

Here x₂ = 2,y₂ = 1,x₁ = – 3,y₁ = 6

∴ CA =

∴ CA =

∴ CA =

∴CA = √50 units

Here CA is the largest side.

For ΔABC to be right angled triangle

Here,

And

50 = 50

⇒

∴ the given triangle is a right – angled triangle.

Hence Proved.

(i) [If the distance between the centre and the point is greater than Radius, then the point is outside the circle.

If the distance between the centre and the point is smaller than Radius, then the point is inside the circle.

If the distance between the centre and the point is equal to Radius, then the point is on the circle.]

Centre of the circle = (0,0)(given)

Radius = 10

The distance between origin and (6,9) = = √117>10

⇒ Point (6,9) is outside the circle.

The distance between origin and (5,9) = = √106>10

⇒ Point (5,9) is outside the circle.

The distance between origin and (6,8) = = √100 = 10

⇒ Point (6,8) is outside the circle.

(ii) Let the coordinates of the point on the circle be (x,y)

Then, distance of it from the origin(center) = 10

⇒ = 10

⇒ = 10

⇒

All the possible solutions of the above equation will be on the circle.

Such 8 points are,(6,8),

(√10,√90),

(√20,√80),

(√30,√70),

(√40,√60),

(√50,√50),

(√60,√40),

(√70,√30).

Let the coordinates of the required point is (x, y).

Since the point is on the circle,

Its distance from the centre(1,1) = Radius =

⇒ = √2

⇒ (x – 1)² + (y – 1)² = 2

If the point is on X axis,

y = 0

⇒ (x – 1)² + (0 – 1)² = 2

⇒ (x – 1)² + 1 = 2

⇒ (x – 1)² = 1

⇒ x – 1 = 1 or x – 1 = – 1

⇒ x = 2 or x = 0

Hence, coordinates of the points where a circle of radius √2, centred on the point with coordinates (1, 1) cut the axis are (2,0)and (0,0)

If the point is on Y axis,

x = 0

⇒ (0 – 1)² + (y – 1)² = 2

⇒ (– 1)² + (y – 1)² = 2

⇒ (y – 1)² = 1

⇒ y – 1 = 1 or y – 1 = – 1

⇒ y = 2 or y = 0

Hence, coordinates of the points where a circle of radius √2, centred on the point with coordinates (1, 1) cut the axis are (0,2)and (0,0)

Hence, the coordinates of the points where a circle of radius_√2, centred on the point with coordinates (1, 1) cut the axes are (0,0),(2,0) and (0,2).

Let the centre of the circumcircle be (x,y)

Since, centre of circumcircle is a point which is equidistant from all the points,

(x – 1)² + (y – 2)² = (x – 2)² + (y – 3)² = (x – 3)² + (y – 1)²

⇒ x² + y² – 2x – 4y + 5 = x² + y² – 4x – 6y + 13 = x² + y² – 6x – 2y + 10

⇒ – 2x – 4y + 5 = – 4x – 6y + 13 = – 6x – 2y + 10

– 2x – 4y + 5 = – 4x – 6y + 13

⇒ 2x + 2y = 8

⇒ x + y = 4 …….(1)

– 4x – 6y + 13 = – 6x – 2y + 10

⇒ 2x – 4y = – 3…..(2)

Substituting x = 4 – y from eq(1)

2(4 – y) – 4y = – 3

⇒ 8 – 2y – 4y = – 3

⇒ 8 – 6y = – 3

⇒ – 6y = – 11

⇒

And

x = 4 – y

⇒

Hence coordinates of the centre of the circumcircle of the triangle =)

And its radius =

=

=units.

Let coordinates of A be (x₁,y₁) and coordinates of B be(x₂,y₂).

Drop a perpendicular from A on X – axis intersecting X axis at P.

Now,

In Δ OPA

And

⇒ x₁ = √3

∴ coordinates of A = (√3)

Drop a perpendicular from B on X – axis intersecting X axis at Q.

Now,

In Δ OQB

And

∴ coordinates of B = (– 1,√3)

Hence, length of the chord AB =

The distance between point A and B =

= √8 = √2