Coordinate Geometry

i. (5,7) is point in the IV quadrant.

False

Reason: X –coordinate(abscissa) and y –coordinate (ordinate) both are positive. When both are positives, then they lie in the I quadrant.

ii. (–2, –7) is point in the III quadrant.

True

Reason: X–coordinate (Abscissa) and y –coordinate (ordinate) both are negative. When both are negatives, then they lie in the III quadrant.

iii. (8, −7) lies below the x–axis.

True

Reason: x – coordinate (Abscissa) is positive and y – coordinate (ordinate) is negative. Hence, this point lies in the IV quadrant. IV quadrant is the area below the x–axis.

iv. (5, 2) and (–7, 2) are points on the line parallel to y–axis.

False

Reason: (5, 2) and (–7, 2) are the line parallel to x–axis. Because, for any points to lie on line parallel to y–axis, the x–coordinates should be same. Hence, these points cannot lie on the line parallel to y–axis.

v. (–5, 2) lies to the left of y–axis.

True

Reason: x – coordinate (Abscissa) is negative and y – coordinate (ordinate) is positive. Hence, this point lies in the II quadrant. II quadrant is the area left of y–axis.

vi. (0, 3) is point on x–axis.

False

Reason: For any point on x–axis, the value of y–coordinate(ordinate) is 0. Hence, this point does not lie on x–axis.

vii. (–2, 3) lies in the II quadrant.

True

Reason: X – coordinate (Abscissa) is negative and y – coordinate (ordinate) is positive. Hence, this point lies in the II quadrant.

viii. (–10, 0) is point on x–axis.

True

Reason: For any point on the x–axis, the value of y–coordinate is zero. Hence, this point lies on the x–axis.

ix. (–2, –4) lies above x–axis

False

Reason: When both coordinates, i.e., x–coordinate and y–coordinate are negative, the point lies in the III quadrant. Therefore (–2, –4) lies in the III quadrant, which is below the axis.

x. For any point on the x–axis its y–coordinate is zero.

True

i (5, 2) – I quadrant

ii (–1, –1) – III quadrant

iii (7, 0) – on X–axis

iv (–8, 1) – II quadrant

v (0, –5) – on down y–axis

vi (0, 3) – on y – axis

vii (4, –5) IV quadrant

viii (0, 0) – on origin

ix (1, 4) – I quadrant

x (–5, 7) – II quadrant

Abscissa is the x–coordinate of any point A (x, y)

i. (–7, 2)

Abscissa of point (–7, 2) is –7

ii. (3, 5)

Abscissa of point (3, 5) is 3

iii. (8, –7)

Abscissa of point (8, –7) is 8

iv. (–5, –3)

Abscissa of point (–5, –3) is –5

Ordinate is the y–coordinate of any point A (x, y)

i. (7, 5)

Ordinate of point (7, 5) is 5

ii. (2, 9)

Ordinate of point (2, 9) is 9

iii. (–5, 8)

Ordinate of point (–5, 8) is 8

iv. (–5, –3)

Ordinate of point (–5, –3) is –3

Let (4, 2) be A, (4, –5) be B, (4,0) be C and (4, –2) be D.

The line joining the coordinates A, B, C and D is parallel to the y–axis.

Let the coordinates of two points i.e. A and B be (2, –6) and (–3, –6) respectively.

As we can see that, the line joining the point A and B is parallel to x–axis.

Let the coordinate of two points i.e. A and B are (0, 3) and (0, –3) respectively.

As we can see that, the line joining the point A and B lies on the y–axis.

To plot A (2, 4) move 2 units in positive x direction and 4 units in positive y direction.
To plot B (−3, 4) move 3 units in negative x direction and 4 units in positive y direction.
To plot C (−3, −1)move 3 units in negative x direction and 1 unit in negative y direction.
To plot D (2, −1)move 2 units in positive x direction and 1 unit in negative y direction.

Now use distance formula to find the lengths of each side,

For AB,

For AD,

For CD,

For BC,

Now AC,

For BD,

As AB = AC = BC = CD
Also AC = BD
Hence the given points make a square.

For OABC to be square, the coordinate should be in a line where point B is and where it meets the y–axis. Therefore, the point C should be (0, 4).

To obtain the coordinate C, extend a line from D towards right and extend a line from the coordinate B. the intersection point is the point C.

Hence, the coordinates of point C is (4, 3).

Formula used:

(7, 8) and (–2, –3)

x₁ = 7 and x₂ = –2

y₁ = 8 and y₂ = –3

⇒ D = √ ((–2 – 7)² + (–3 – 8)²)

⇒ D = √ ((–9)² + (–11)²)

⇒ D = √ (81 + 121)

⇒ D = √ 202

Formula used:

(6, 0) and (–2, 4)

x₁ = 6 and x₂ = –2

y₁ = 0 and y₂ = 4

⇒ D = √ ((–2 – 6)² + (4 – 0)²)

⇒ D = √ ((–8)² + (4)²)

⇒ D = √ (64 + 16)

⇒ D = √ 80

⇒ D = √ (5 × 4 × 4)

⇒ D = 4√ 5

Formula used:

(–3, 2) and (2, 0)

x₁ = –3 and x₂ = 2

y₁ = 2 and y₂ = 0

⇒ D = √ ((2 – (–3)² + (0 – 2)²)

⇒ D = √ ((2 + 3)² + (0 – 2)²)

⇒ D = √ ((5)² + (–2)²)

⇒ D = √ (25 + 4)

⇒ D = √ 29

Formula used:

(–2, –8) and (–4, –6)

x₁ = –2 and x₂ = –4

y₁ = –8 and y₂ = –6

⇒ D = √ ((–4 – (–2))² + (–6 – (–8))²)

⇒ D = √ ((–4 + 2)² + (–6 + 8)²)

⇒ D = √ ((–2)² + (2)²)

⇒ D = √ (4 + 4)

⇒ D = √ 8

⇒ D = √ (2 × 2 × 2)

⇒ D = 2√ 2

Formula used:

(–2, –3) and (3, 2)

x₁ = –2 and x₂ = 3

y₁ = –3 and y₂ = 2

⇒ D = √ ((3 – (–2))² + (2 – (–3))²)

⇒ D = √ ((3 + 2)² + (2 + 3)²)

⇒ D = √ ((5)² + (5)²)

⇒ D = √ (25 + 25)

⇒ D = √ 50

⇒ D = √ (5 × 5 × 2)

⇒ D = 5√ 2

Formula used:

(2, 2) and (3, 2)

x₁ = 2 and x₂ = 3

y₁ = 2 and y₂ = 2

⇒ D = √ ((3 – 2)² + (2 – 2)²)

⇒ D = √ ((1)² + (0)²)

⇒ D = √ (1 + 0)

⇒ D = √ 1

⇒ D = 1

Formula used:

(–2, 2) and (3, 2)

x₁ = –2 and x₂ = 3

y₁ = 2 and y₂ = 2

⇒ D = √ ((3 – (–2))² + (2 – 2)²)

⇒ D = √ ((5)² + (0)²)

⇒ D = √ (25 + 0)

⇒ D = √ 25

⇒ D = √ (5 × 5)

⇒ D = 5

Formula used:

(7, 0) and (–8, 0)

x₁ = 7 and x₂ = –8

y₁ = 0 and y₂ = 0

⇒ D = √ ((–8 – 7)² + (0 – 0)²)

⇒ D = √ ((–15)² + (0)²)

⇒ D = √ (225 + 0)

⇒ D = √ 225

⇒ D = √ (5 × 3 × 5 × 5)

⇒ D = 5 × 3

⇒ D = 15

Formula used:

(0, 17) and (0, –1)

x₁ = 0 and x₂ = 0

y₁ = 17 and y₂ = –1

⇒ D = √ ((0 – 0)² + (–1 – 17)²)

⇒ D = √ ((0)² + (–18)²)

⇒ D = √ (0 + 324)

⇒ D = √ 324

⇒ D = √ (18 × 18)

⇒ D = 18

Formula used:

(5, 7) and (0, 0)

x₁ = 5 and x₂ = 0

y₁ = 7 and y₂ = 0

⇒ D = √ ((0 – 5)² + (0 – 7)²)

⇒ D = √ ((–5)² + (–7)²)

⇒ D = √ (25 + 49)

⇒ D = √ 74

Formula used:

(3, 7), (6, 5) and (15, –1)

Let the points be A (15, –1), B (6, 5) and C (3, 7)

Distance of AB

⇒ AB = √ (6 – 15)² + (5 – (–1))²

⇒ AB = √ (–9)² + (6)²

⇒ AB = √ (81 + 36)

⇒ AB = √ 117 = √ 3 × 3 × 13

⇒ AB = 3√13

Distance of BC

⇒ BC = √ (3 – 6)² + (7 – 5)²

⇒ BC= √ (3)² + (2)²

⇒ BC = √ (9 + 4)

⇒ BC= √ 13

Distance of AC

⇒ AC = √ (3 – 15)² + (7 – (–1))²

⇒ AC = √ (3 – 15)² + (7 + 1)²

⇒ AC= √ (–12)² + (8)²

⇒ AC = √ (144 + 64)

⇒ AC= √ 208 = √ 4 × 4 × 13

⇒ AC = 4√13

i.e. AB + BC = AC

⇒ 3√13 + √13 = 4√13

∴ A, B and C are collinear

Formula used:

(3, 2), (–2, 8) and (0, 4)

Let A (–2, 8), B (0, 4) and C (3, 2)

Distance of AB

⇒ AB = √ ((0 – (–2))² + (4 – 8)²)

⇒ AB = √ (2)² + (–4)²

⇒ AB = √ (4 + 16)

⇒ AB = √20

Distance of BC

⇒ BC = √ ((3 – 0)² + (2 – 4)²)

⇒ BC = √ (3)² + (–2)²

⇒ BC = √ (9 + 4)

⇒ BC = √13

Distance of AC

⇒ AC = √ ((3 – (–2))² + (2 – 8)²)

⇒ AC = √ (5)² + (–6)²

⇒ AC = √ (25 + 36)

⇒ AC = √ 61

Formula used:

(1, 4), (3, –2) and (–1, 10)

Let A (–1, 10), B (1, 4) and C (3, –2)

Distance of AB

⇒ AB =√ ((1 – (–1))² + (4 – 10)²)

⇒ AB = √ ((1 + 1)² + (4 – 10)²)

⇒ AB = √ (2)² + (–6)²

⇒ AB = √ (4 + 36)

⇒ AB = √ 40

Distance of BC

⇒ BC =√ ((3 – 1)² + (–2 – 4)²)

⇒ BC = √ (2)² + (–6)²

⇒ BC = √ (4 + 36)

⇒ BC = √ 40

Distance of AC

⇒ AC =√ ((3 – (–1))² + (–2 – 10)²)

⇒ AC = √ ((3 + 1)² + (2 – 10)²)

⇒ AC = √ (4)² + (–8)²

⇒ AC = √ (16 + 64)

⇒ AC = √ 80

i.e. AB + BC = AC

⇒ √40 + √40 = √80

∴ A, B and C are collinear.

Formula used:

(6, 2), (2, –3) and (–2, –8)

Let A (6, 2), B (2, –3) and C (–2, –8)

Distance of AB

⇒ AB =√ ((2 – (6))² + (–3 – 2)²)

⇒ AB = √ (4)² + (–5)²

⇒ AB = √ (16 + 25)

⇒ AB = √ 41

Distance of BC

⇒ BC =√ ((–2 – 2)² + (–8 – (–3))²)

⇒ BC = √ ((–2 – 2)² + (–8 + 3)²)

⇒ BC = √ (–4)² + (–5)²

⇒ BC = √ (16 + 25)

⇒ BC = √ 41

Distance of AC

⇒ AC =√ ((–2 – 6)² + (–8 – 2)²)

⇒ AC = √ (–8)² + (–10)²

⇒ AC = √ (64 + 100)

⇒ AC = √ 164 = √ 2 × 2 × 41

⇒ AC =2√ 41

i.e. AB + BC = AC

⇒ √41 + √41 = 2√41

∴ A, B and C are collinear.

Formula used:

(4, 1), (5, –2) and (6, –5)

Let A (4, 1), B (5, –2) and C (6, –5)

Distance of AB

⇒ AB =√ ((5 – 4)² + (–2 – 1)²)

⇒ AB = √ (1)² + (–3)²

⇒ AB = √ (1 + 9)

⇒ AB = √10

Distance of BC

⇒ BC =√ ((6 – 5)² + (–5 – (–2))²)

⇒ BC = √ (6 – 5)² + (–5 + 2)²)

⇒ BC = √ (1)² + (–3)²

⇒ BC = √ (1 + 9)

⇒ BC = √ 10

Distance of AC

⇒ AC =√ ((6 – 4)² + (–5 – 1)²)

⇒ AC = √ (2)² + (–6)²

⇒ AC = √ (4 + 36)

⇒ AC = √20 =

i.e. AB + BC = AC

⇒ √10 + √10 = √20

Squaring both sides

⇒ (√10)² + (√10)² = (√20)²

⇒ 10 + 10 = 20

∴ A, B and C are collinear.

Formula used:

(–2, 0), (4,0) and (1, 3)

Let the point be A (1, 3) B (–2, 0) and C (4, 0)

Distance of AB

⇒ AB = √ ((–2 – 1)² + (0 – 3)²)

⇒ AB = √ ((–3)² + (–3)²)

⇒ AB = √ (9 + 9)

⇒ AB = √ 18 = 3√ 2

Distance of AC

⇒ AC = √ ((4 – 1)² + (0 – 3)²)

⇒ AC = √ ((3)² + (–3)²)

⇒ AC = √ (9 + 9)

⇒ AC = √ 18 = 3√2

Distance of BC

⇒ BC = √ ((4 – (–2))² + (0 – 0)²)

⇒ BC = √ ((6)² + (0)²)

⇒ BC = √ (36 + 0)

⇒ BC = √ 36 = 6

We notice that AB = AC =3√2

∴ Points A, B and C are coordinates of an isosceles triangle.

Formula used:

(1, −2), (−5, 1) and (1, 4)

Let the point be A (–5, 1) B (1, –2) and C (1, 4)

Distance of AB

⇒ AB = √ (1 – (–5))² + (–2 – 1)²)

⇒ AB = √ (1 + 5)² + (–2 – 1)²

⇒ AB = √ ((6)² + (–3)²)

⇒ AB = √ (36 + 9)

⇒ AB = √ 45 = 3√5

Distance of AC

⇒ AC = √ ((1 – (–5))² + (4 – 1)²)

⇒ AC = √ ((1 + 5)² + (4 – 1)²)

⇒ AC = √ ((6)² + (3)²)

⇒ AC = √ (36 + 9)

⇒ AC = √45 = 3√5

Distance of BC

⇒ BC = √ ((1 – 1)² + (4 – (–2)²)

⇒ BC = √ ((1 – 1)² + (4 + 2²)

⇒ BC = √ ((0)² + (6)²)

⇒ BC = √ (0 + 36)

⇒ BC = √ 36 = 6

We notice that AB = AC =3√5

∴ Points A, B and C are coordinates of an isosceles triangle.

Formula used:

(−1, −3), (2, −1) and (−1, 1)

Let the point be A (2, –1) B (–1, –3) and C (–1, 1)

Distance of AB

⇒ AB = √ ((–1 – 2)² + (–3 – (–1))²)

⇒ AB = √ ((–1 – 2)² + (–3 + 1)²)

⇒ AB = √ ((–3)² + (–2)²)

⇒ AB = √ (9 + 4)

⇒ AB = √ 13

Distance of AC

⇒ AC = √ ((–1 – 2)² + (1 – (–1))²)

⇒ AC = √ ((–1 – 2)² + (1 + 1)²)

⇒ AC = √ ((–3)² + (2)²)

⇒ AC = √ (9 + 4)

⇒ AC = √ 13

Distance of BC

⇒ BC = √ ((–1 – (–1))² + (1 – (–3))²)

⇒ BC = √ ((–1 + 1))² + (1 + 3)²)

⇒ BC = √ ((0)² + (4)²)

⇒ BC = √ (0 + 16)

⇒ BC = √ 16

We notice that AB = AC = √13

∴ Points A, B and C are coordinates of an isosceles triangle.

Formula used:

(1, 3), (–3, –5) and (–3, 0)

Let the point be A (–3, 0) B (1, 3) and C (–3, –5)

Distance of AB

⇒ AB = √ ((1 – (–3))² + (3 – 0)²)

⇒ AB = √ ((1 + 3)² + (3 – 0)²)

⇒ AB = √ ((4)² + (3)²)

⇒ AB = √ (16 + 9)

⇒ AB = √25 = 5

Distance of AC

⇒ AC = √ ((–3 – (–3))² + (–5 – 0)²)

⇒ AC = √ ((–3 + 3)² + (–5 + 0)²)

⇒ AC = √ ((0)² + (–5)²)

⇒ AC = √ (0 + 25)

⇒ AC = √25 = 5

Distance of BC

⇒ BC = √ ((–3 – 1)² + (–5 – 3)²)

⇒ BC = √ ((–4)² + (–8)²)

⇒ BC = √ (16 + 64)

⇒ BC = √ 80

We notice that AB = AC = 5

∴ Points A, B and C are coordinates of an isosceles triangle.

Formula used:

(2, 3), (5, 7) and (1, 4)

Let the point be A (5, 7) B (2, 3) and C (1, 4)

Distance of AB

⇒ AB = √ (2 – 5)² + (3 – 7)²)

⇒ AB = √ ((–3)² + (–4)²)

⇒ AB = √ (9 + 16)

⇒ AB = √ 25 = 5

Distance of AC

⇒ AC = √ ((1 – 5)² + (4 – 7)²)

⇒ AC = √ ((–4)² + (–3)²)

⇒ AC = √ (16 + 9)

⇒ AC = √ 25 = 5

Distance of BC

⇒ BC = √ ((1 – 2)² + (4 – 3)²)

⇒ BC = √ ((–1)² + (1)²)

⇒ BC = √ (1 + 1)

⇒ BC = √ 2

We notice that AB = AC = 5

∴ Points A, B and C are coordinates of an isosceles triangle.

Formula used:

(2, –3), (–6, –7) and (–8, –3)

Let the points be A (2, –3), B (–6, –7) and C (–8, –3)

Distance of AB

⇒ AB = √ ((–6 – 2)² + (–7 – (–3))²)

⇒ AB = √ ((–6 – 2)² + (–7 + 3)²)

⇒ AB = √ ((–8)² + (–4)²)

⇒ AB = √ (64 + 16)

⇒ AB = √ 80

Distance of BC

⇒ B C= √ ((–8 – (–6))² + (–3 – (–7))²)

⇒ BC = √ ((–8 + 6)² + (–3 + 7)²)

⇒ BC = √ ((–2)² + (4)²)

⇒ BC = √ (4 + 16)

⇒ BC = √ 20

Distance of AC

⇒ AC = √ ((–8 – 2)² + (–3 – (–3))²)

⇒ AC = √ ((–8 – 2)² + (–3 + 3)²)

⇒ AC = √ ((–10)² + (0)²)

⇒ AC = √ (100 + 0)

⇒ AC = √ 100

i.e. AB² + BC²

= (√80)² + (√20)²

= 80 + 20

= 100 = (AC)²

Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.

Formula used:

(–11, 13), (–3, –1) and (4, 3)

Let the points be A (–11, 13), B (–3, –1) and C (4, 3)

Distance of AB

⇒ AB = √ ((–3 – (–11))² + (–1 – 13)²)

⇒ AB = √ ((–3 + 11)² + (–1 – 13)²)

⇒ AB = √ ((8)² + (–14)²)

⇒ AB = √ (64 + 196)

⇒ AB = √260

Distance of BC

⇒ B C= √ ((4 – (–3))² + (3 – (–1))²)

⇒ BC = √ ((4 + 3)² + (3 + 1)²)

⇒ BC = √ ((7)² + (4)²)

⇒ BC = √ (49 + 16)

⇒ BC = √ 65

Distance of AC

⇒ AC = √ ((4 – (–11))² + (3 – 13))²)

⇒ AC = √ ((4 + 11)² + (3 – 13)²)

⇒ AC = √ ((15)² + (–10)²)

⇒ AC = √ (225 + 100)

⇒ AC = √ 325

i.e. AB² + BC²

= (√260)² + (√65)²

= 260 + 65

= 325 = (AC)²

Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.

Formula used:

(0, 0), (a, 0) and (0, b)

Let the points be A (0, 0), B (a, 0) and C (0, b)

Distance of AB

⇒ AB = √ ((a – 0)² + (0 – 0)²)

⇒ AB = √ ((a)² + (0)²

⇒ AB = √ a²

Distance of BC

⇒ BC = √ ((0 – a)² + (b – 0)²)

⇒ BC = √ ((–a)² + (b)²

⇒ BC = √ a² + b²

Distance of AC

⇒ AC = √ ((0 – 0)² + (b – 0)²)

⇒ AC = √ ((0)² + (b)²

⇒ AC = √ b²

i.e. AB² + AC²

= (√a²)² + (√b²)²

= a² + b² = BC²

Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.

Formula used:

(10, 0), (18, 0) and (10, 15)

Let the points be A (10, 15), B (10, 0) and C (18, 0)

Distance of AB

⇒ AB = √ ((10 – 10))² + (0 – 15)²)

⇒ AB = √ ((0)² + (–15)²)

⇒ AB = √ (0 + 225)

⇒ AB = √225

Distance of BC

⇒ B C= √ ((18 – 10)² + (0 – 0)²)

⇒ BC = √ ((8)² + (0)²)

⇒ BC = √ (64 + 0)

⇒ BC = √ 64

Distance of AC

⇒ AC = √ ((18 – 10)² + (0 – 15))²)

⇒ AC = √ ((8)² + (–15)²)

⇒ AC = √ (64 + 225)

⇒ AC = √289

i.e. AB² + BC²

= (√225)² + (√64)²

= 225 + 64

= 289 = (AC)²

Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.

Formula used:

(5, 9), (5, 16) and (29, 9)

Let the points be A (5, 16), B (5, 9) and C (29, 9)

Distance of AB

⇒ AB = √ ((5 – 5)² + (9 – 16)²)

⇒ AB = √ ((0)² + (–7)²)

⇒ AB = √ (0 + 49)

⇒ AB = √49

Distance of BC

⇒ B C= √ ((29 – 5)² + (9 – 9)²)

⇒ BC = √ ((24)² + (0)²)

⇒ BC = √ (576 + 0)

⇒ BC = √576

Distance of AC

⇒ AC = √ ((29 – 5)² + (9 – 16))²)

⇒ AC = √ ((24)² + (–7)²)

⇒ AC = √ (576 + 49)

⇒ AC = √ 625

i.e. AB² + BC²

= (√49)² + (√576)²

= 49 + 576

= 625 = (AC)²

Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.

Formula used:

(0, 0), (10, 0) and (5, 5√3)

Let the points be A (0, 0), B (10, 0) and C (5, 5√3)

Distance of AB

⇒ AB = √ ((10 – 0)² + (0 – 0)²)

⇒ AB = √ ((10)² + (0)²)

⇒ AB = √ (100 + 0)

⇒ AB = √100

⇒ AB = 10

Distance of BC

⇒ B C= √ ((5 – 10)² + (5√3 – 0)²)

⇒ BC = √ ((–5)² + (5√3)²)

⇒ BC = √ (25 + 75)

⇒ BC = √100

⇒ BC = 10

Distance of AC

⇒ AC = √ ((5 – 0)² + (5√3 – 0))²)

⇒ AC = √ ((5)² + (5√3)²)

⇒ AC = √ (25 + 75)

⇒ AC = √ 100

⇒ AC = 10

∴ AB = BC = AC = 10

Since, all the sides are equal the points form an equilateral triangle.

Formula used:

(a, 0), (–a, 0) and (0, a√3)

Let the points be A (a, 0), B (–a, 0) and C (0, a√3)

Distance of AB

⇒ AB = √ ((–a – a)² + (0 – 0)²)

⇒ AB = √ ((–2a)² + (0)²)

⇒ AB = √ (4a² + 0)

⇒ AB = √4a²

⇒ AB = 2a

Distance of BC

⇒ B C= √ ((0 – a)² + (a√3 – 0)²)

⇒ BC = √ ((–a)² + (a√3)²)

⇒ BC = √ (a² + 3a²)

⇒ BC = √4a²

⇒ BC = 2a

Distance of AC

⇒ AC = √ ((0 – a)² + (a√3 – 0))²)

⇒ AC = √ ((–a)² + (a√3)²)

⇒ AC = √ (a² + 3a²)

⇒ AC = √ 4a²

⇒ AC = 2a

∴ AB = BC = AC = 2a

Since, all the sides are equal the points form an equilateral triangle.

Formula used:

(2, 2), (–2, –2) and (–2√3, 2√3)

Let the points be A (2, 2), B (–2, –2) and C (–2√3, 2√3)

Distance of AB

⇒ AB = √ ((–2 – 2)² + (–2 – 2)²)

⇒ AB = √ ((–4)² + (–4)²)

⇒ AB = √ (16 + 16)

⇒ AB = √32

⇒ AB = 4√2

Distance of BC

⇒ B C= √ ((–2√3 – (–2))² + (2√3 – (–2))²)

⇒ B C= √ ((–2√3 + 2))² + (2√3 + 2)²)

⇒ BC = √ (((–2√3)² + 2 (–2√3) (2) + (2)²) + ((2√3)² + 2 (2√3) (2) + (2)²))

⇒ BC = √ (12 – 8√3 + 4 + 12 + 8√3 + 4)

⇒ BC = √ (12 + 4 + 12 + 4

⇒ BC = √ 32

⇒ BC = 4√2

Distance of AC

⇒ AC= √ ((–2√3 – 2))² + (2√3 – 2)²)

⇒ AC = √ (((–2√3)² + 2 (–2√3) (–2) + (2)²) + ((2√3)² + 2 (2√3) (–2) + (–2)²))

⇒ AC = √ (12 + 8√3 + 4 + 12 – 8√3 + 4)

⇒ AC = √ (12 + 4 + 12 + 4)

⇒ AC = √ 32

⇒ AC = 4√2

∴ AB = BC = AC = 4√2

Since, all the sides are equal the points form an equilateral triangle.

Formula used:

(√3, 2), (0, 1) and (0, 3)

Let the points be A (√3, 2), B (0, 1) and C (0, 3)

Distance of AB

⇒ AB = √ ((0 – √3)² + (1 – 2)²)

⇒ AB = √ ((√3)² + (–1)²)

⇒ AB = √ (3 + 1)

⇒ AB = √4

⇒ AB = 2

Distance of BC

⇒ B C= √ ((0 – 0)² + (3 – 1)²)

⇒ BC = √ ((0)² + (2)²)

⇒ BC = √ (0 + 4)

⇒ BC = √4

⇒ BC = 2

Distance of AC

⇒ AC = √ ((0 – √3)² + (3 – 2))²)

⇒ AC = √ ((√3)² + (1)²)

⇒ AC = √ (3 + 1)

⇒ AC = √ 4

⇒ AC = 2

∴ AB = BC = AC = 2

Since, all the sides are equal the points form an equilateral triangle.

Formula used:

(–√3, 1), (2√3, –2) and (2√3, 4)

Let the points be A (–√3, 1), B (2√3, –2) and C (2√3, 4)

Distance of AB

⇒ AB = √ ((2√3 – (–√3))² + (–2 – 1)²)

⇒ AB = √ ((2√3 + √3))² + (–2 – 1)²)

⇒ AB = √ ((12 + 12 + 3)² + (–3)²)

⇒ AB = √ (27 + 9)

⇒ AB = √36

⇒ AB = 6

Distance of BC

⇒ B C= √ ((2√3 – 2√3)² + (4 – (–2))²)

⇒ B C= √ ((2√3 – 2√3)² + (4 + 2)²)

⇒ BC = √ ((0)² + (6)²)

⇒ BC = √ (0 + 36)

⇒ BC = √36

⇒ BC = 6

Distance of AC

⇒ AC = √ ((2√3 – (–√3))² + (4 – 1))²)

⇒ AC = √ ((2√3 + √3))² + (4 – 1))²)

⇒ AC = √ ((3√3)² + (3)²)

⇒ AC = √ (27 + 9)

⇒ AC = √ 36

⇒ AC = 6

∴ AB = BC = AC = 6

Since, all the sides are equal the points form an equilateral triangle.

Formula used:

(–7, –5), (–4, 3), (5, 6) and (2, –2)

Let A, B, C and D represent the points (–7, –5), (–4, 3), (5, 6) and (2, –2)

Distance of AB

⇒ AB = √ ((–4 – (–7)))² + (3 – (–5))²)

⇒ AB = √ ((–4 + 7))² + (3 + 5)²)

⇒ AB = √ ((3)² + (8)²)

⇒ AB = √ (9 + 64)

⇒ AB = √73

Distance of BC

⇒ BC= √ ((5 – (–4))² + (6 – 3)²)

⇒ BC= √ ((5 + 4))² + (6 – 3)²)

⇒ BC = √ ((9)² + (3)²)

⇒ BC = √ (81 + 9)

⇒ BC = √ 90

Distance of CD

⇒ CD = √ ((2 – 5)² + (–2 – 6)²)

⇒ CD = √ ((–3)² + (–8)²)

⇒ CD = √ (9 + 64)

⇒ CD = √73

Distance of AD

⇒ AD = √ ((2 – (–7)))² + (–2 – (–5))²)

⇒ AD = √ ((2 + 7))² + (–2 + 5)²)

⇒ AD = √ ((9)² + (3)²)

⇒ AD = √ (81 + 9)

⇒ AD = √ 90

So, AB = CD = √73 and BC = AD = √90

i.e., The opposite sides are equal. Hence ABCD is a parallelogram.

Formula used:

(9,5), (6, 0), (–2, –3) and (1, 2)

Let A, B, C and D represent the points (9, 5), (6, 0), (–2, –3) and (1, 2)

Distance of AB

⇒ AB = √ ((6 – 9))² + (0 – 5)²)

⇒ AB = √ ((–3)² + (5)²)

⇒ AB = √ (9 + 25)

⇒ AB = √ 34

Distance of BC

⇒ BC= √ ((–2 – 6)² + (–3 – 0)²)

⇒ BC = √ ((–8)² + (–3)²)

⇒ BC = √ (64 + 9)

⇒ BC = √73

Distance of CD

⇒ CD = √ ((1 – (–2))² + (2 – (–3))²)

⇒ CD = √ ((1 + 2)² + (2 + 3))²)

⇒ CD = √ ((3)² + (5)²)

⇒ CD = √ (9 + 25)

⇒ CD = √36

Distance of AD

⇒ AD = √ ((1 – 9))² + (2 – 5)²)

⇒ AD = √ ((–8)² + (–3)²)

⇒ AD = √ (64 + 9)

⇒ AD = √ 73

So, AB = CD = √36 and BC = AD = √73

i.e., The opposite sides are equal. Hence ABCD is a parallelogram.

Formula used:

(0,0) (7, 3), (10, 6) and (3, 3)

Let A, B, C and D represent the points (0, 0), (7, 3), (10, 6) and (3, 3)

Distance of AB

⇒ AB = √ ((7 – 0))² + (3 – 0)²)

⇒ AB = √ ((7)² + (3)²)

⇒ AB = √ (49 + 9)

⇒ AB = √ 58

Distance of BC

⇒ BC= √ ((10 – 7)² + (6 – 3)²)

⇒ BC = √ ((3)² + (3)²)

⇒ BC = √ (9 + 9)

⇒ BC = √18

Distance of CD

⇒ CD = √ ((3 – 10)² + (3 – 6)²)

⇒ CD = √ ((–7)² + (–3)²)

⇒ CD = √ (49 + 9)

⇒ CD = √58

Distance of AD

⇒ AD = √ ((3 – 0))² + (3 – 0)²)

⇒ AD = √ ((3)² + (3)²)

⇒ AD = √ (9 + 9)

⇒ AD = √18

So, AB = CD = √58 and BC = AD = √18

i.e., The opposite sides are equal. Hence ABCD is a parallelogram.

Formula used:

(–2, 5), (7, 1), (–2, –4) and (7, 0)

Let A, B, C and D represent the points (–2, 5), (7, 1), (–2, –4) and (7, 0)

Distance of AB

⇒ AB = √ ((7 – (–2)))² + (1 – 5)²)

⇒ AB = √ ((7 + 2))² + (1 – 5)²)

⇒ AB = √ ((9)² + (–4)²)

⇒ AB = √ (81 + 16)

⇒ AB = √ 97

Distance of BC

⇒ BC= √ ((–2 – 7)² + (–4 – 1)²)

⇒ BC = √ ((–9)² + (–5)²)

⇒ BC = √ (81 + 25)

⇒ BC = √106

Distance of CD

⇒ CD = √ ((7 – (–2))² + (0 – (–4))²)

⇒ CD = √ ((7 + 2)² + (0 + 4))²)

⇒ CD = √ ((9)² + (4)²)

⇒ CD = √ (81 + 16)

⇒ CD = √97

Distance of AD

⇒ AD = √ ((7 – (–2))² + (0 – 5)²)

⇒ AD = √ ((7 + 2)² + (0 – 5)²)

⇒ AD = √ ((9)² + (–5)²)

⇒ AD = √ (81 + 25)

⇒ AD = √ 106

So, AB = CD = √97 and BC = AD = √106

i.e., The opposite sides are equal. Hence ABCD is a parallelogram.

Formula used:

(3, –5), (–5, –4), (7, 10) and (15, 9)

Let A, B, C and D represent the points (3, –5), (–5, –4), (7, 10) and (15, 9)

Distance of AB

⇒ AB = √ ((–5 – 3)² + ((–4 – (–5))²)

⇒ AB = √ ((–5 – 3))² + (–4 + 5)²)

⇒ AB = √ ((–8)² + (1)²)

⇒ AB = √ (64 + 1)

⇒ AB = √ 65

Distance of BC

⇒ BC= √ ((7 – (–5))² + (10 – (–4))²)

⇒ BC= √ ((7 + 5)² + (10 + 4)²)

⇒ BC = √ ((12)² + (14)²)

⇒ BC = √ (144 + 196)

⇒ BC = √ 340

Distance of CD

⇒ CD = √ ((15 – 7)² + (9 – 10)²)

⇒ CD = √ ((8)² + (–1)²)

⇒ CD = √ (64 + 1)

⇒ CD = √65

Distance of AD

⇒ AD = √ ((15 – 3)² + (9 – (–5))²)

⇒ AD = √ ((15 – 3)² + (9 + 5)²)

⇒ AD = √ ((12)² + (14)²)

⇒ AD = √ (144 + 196)

⇒ AD = √ 340

So, AB = CD = √65 and BC = AD = √340

i.e., The opposite sides are equal. Hence ABCD is a parallelogram.

Formula used:

(0, 0), (3, 4), (0, 8) and (–3, 4)

Let the vertices be taken as A (0, 0), B (3, 4), C (0, 8) and D (–3, 4).

Distance of AB

⇒ AB = √ ((3 – 0)² + ((4 – 0)²)

⇒ AB = √ ((3)² + (4)²)

⇒ AB = √ (9 + 16)

⇒ AB = √ 25

⇒ AB = 5

Distance of BC

⇒ BC= √ ((0 – 3)² + (8 – 4)²)

⇒ BC = √ ((–3)² + (4)²)

⇒ BC = √ (9 + 16)

⇒ BC = √ 25

⇒ BC = 5

Distance of CD

⇒ CD = √ ((–3 – 0)² + (4 – 8)²)

⇒ CD = √ ((–3)² + (–4)²)

⇒ CD = √ (9 + 16)

⇒ CD = √25

⇒ CD = 5

Distance of AD

⇒ AD = √ ((–3 – 0)² + (4 – 0)²)

⇒ AD = √ ((–3)² + (4)²)

⇒ AD = √ (9 + 16)

⇒ AD = √ 25

⇒ AD = 5

Distance of AC

⇒ AC = √ ((0 – 0)² + (8 – 0)²)

⇒ AC = √ ((0)² + (8)²)

⇒ AC = √ (64)

⇒ AC = 8

Distance of BD

⇒ BD = √ ((–3 – 3)² + (4 – 4)²)

⇒ BD = √ ((–6)² + (0)²)

⇒ BD = √ (36 +0)

⇒ BD = √ 36

⇒ BD = 6

AB = BC = CD = DA = 5 (That is, all the sides are equal.)

AC ≠ BD (That is, the diagonals are not equal.)

Hence the points A, B, C and D form a rhombus.

Formula used:

(–4, –7), (–1, 2), (8, 5) and (5, –4)

Let the vertices be taken as A (–4,–7), B (–1, 2), C (8, 5) and D (5, –4).

Distance of AB

⇒ AB = √ ((–1 – (–4))² + (2 – (–7)²))

⇒ AB = √ ((–1+4)² + (2+7)²)

⇒ AB = √ ((3)² + (9)²)

⇒ AB = √ (9 + 81)

⇒ AB = √ 100

⇒ AB = 10

Distance of BC

⇒ BC= √ ((8 – (–1))² + (5 – 2)²)

⇒ BC = √ ((8+1)² + (3)²)

⇒ BC = √ ((9)²+ 9)

⇒ BC = √ (81 + 9)

⇒ BC = √ 100

⇒ BC = 10

Distance of CD

⇒ CD = √ ((5 – 8)² + (–4 –5 )²)

⇒ CD = √ ((3)² + (–9)²)

⇒ CD = √ (9 + 81)

⇒ CD = √100

⇒ CD = 10

Distance of AD

⇒ AD = √ ((5 – (–4))² + (–4 –(–7) )²)

⇒ AD = √ ((5+4)² + (–4+7)²)

⇒ AD = √ ((9)² +(3)²)

⇒ AD = √ (81+9)

⇒ AD = √ 100

⇒ AD = 10

Distance of AC

⇒ AC = √ ((8 – (–4))² + (5 – (–7))²)

⇒ AC = √ ((8+4)² + (5+7)²)

⇒ AC = √ ((12)² +(12)²)

⇒ AC = √ (144 + 144)

⇒ AC = √ (288)

Distance of BD

⇒ BD = √ ((5 – (–1))² + (–4 – 2)²)

⇒ BD = √ ((5 + 1))² + (–4 – 2)²)

⇒ BD = √ ((6)² + (–6)²)

⇒ BD = √ (36 + 36)

⇒ BD = √ 72

AB = BC = CD = DA = 10 (That is, all the sides are equal.)

AC ≠ BD (That is, the diagonals are not equal.)

Hence the points A, B, C and D form a rhombus.

Formula used:

(1, 0), (5, 3), (2, 7) and (–2, 4)

Let the vertices be taken as A (1, 0), B (5, 3), C (2, 7) and D (–2, 4).

Distance of AB

⇒ AB = √ ((5 – 1)² + (3 – 0)²)

⇒ AB = √ ((4)² + (3)²)

⇒ AB = √ (16 + 9)

⇒ AB = √ 25

⇒ AB = 5

Distance of BC

⇒ BC= √ ((2 – 5)² + (7 – 3)²)

⇒ BC = √ ((3)² + (4)²)

⇒ BC = √ (9 + 16)

⇒ BC = √ 25

⇒ BC = 5

Distance of CD

⇒ CD = √ ((–2 – 2)² + (4 – 7)²)

⇒ CD = √ ((–4)² + (–3)²)

⇒ CD = √ (16 + 9)

⇒ CD = √25

⇒ CD = 5

Distance of AD

⇒ AD = √ ((–2 – 1)² + (4 – 0)²)

⇒ AD = √ ((–3)² +(4)²)

⇒ AD = √ (9 + 16)

⇒ AD = √ 25

⇒ AD = 5

Distance of AC

⇒ AC = √ ((2 – 1)² + (7 – 0)²)

⇒ AC = √ ((1)² + (7)²)

⇒ AC = √ (1 + 49)

⇒ AC = √ 50

Distance of BD

⇒ BD = √ ((–2 – 5)² + (4 – 3)²)

⇒ BD = √ ((–7)² + (1)²)

⇒ BD = √ (49 + 1)

⇒ BD = √ 50

AB = BC = CD = DA = 10 (That is, all the sides are equal.)

AC ≠ BD (That is, the diagonals are not equal.)

Hence the points A, B, C and D form a rhombus.

Formula used:

(2, –3), (6, 5), (–2, 1) and (–6, –7)

Let the vertices be taken as A (2, –3), B (6, 5), C (–2, 1) and D (–6, –7).

Distance of AB

⇒ AB = √ ((6 – 2)² + (5 – (–3)²))

⇒ AB = √ ((6 – 2)² + (5 + 3)²)

⇒ AB = √ ((4)² + (8)²)

⇒ AB = √ (16 + 64)

⇒ AB = √ 80

Distance of BC

⇒ BC= √ ((–2 – 6)² + (1 – 5)²)

⇒ BC = √ ((–8)² + (–4)²)

⇒ BC = √ (64 + 16)

⇒ BC = √ 80

Distance of CD

⇒ CD = √ ((–6 – (–2))² + (–7 – 1)²)

⇒ CD = √ ((–6 + 2)² + (–7 – 1)²)

⇒ CD = √ ((–4)² + (–8)²)

⇒ CD = √ (16 + 64)

⇒ CD = √80

Distance of AD

⇒ AD = √ ((–6 – (2))² + (–7 – (–3))²)

⇒ AD = √ ((–6 – 2)² + (–7 + 3)²)

⇒ AD = √ ((–8)² +(–4)²)

⇒ AD = √ (64 + 16)

⇒ AD = √ 80

Distance of AC

⇒ AC = √ ((–2 – 2)² + (1 – (–3))²)

⇒ AC = √ ((–2 – 2)² + (1 + 3)²)

⇒ AC = √ ((–4)² +(4)²)

⇒ AC = √ (16 + 16)

⇒ AC = √ 32

Distance of BD

⇒ BD = √ ((–6 – 6)² + (–7 – 5)²)

⇒ BD = √ ((–6 – 6))² + (–7 – 5)²)

⇒ BD = √ ((–12)² + (–12)²)

⇒ BD = √ (144 + 144)

⇒ BD = √ 288

AB = BC = CD = DA = √80 (That is, all the sides are equal.)

AC ≠ BD (That is, the diagonals are not equal.)

Hence the points A, B, C and D form a rhombus.

Formula used:

(15, 20), (–3, 12), (–11, –6) and (7, 2)

Let the vertices be taken as A (15, 20), B (–3, 12), C (–11, –6) and D (7, 2).

Distance of AB

⇒ AB = √ ((–3 – 15)² + (12 – 20)²)

⇒ AB = √ ((–18)² + (–8)²)

⇒ AB = √ (324 + 64)

⇒ AB = √ 388

Distance of BC

⇒ BC= √ ((–11 –(–3))² + (–6 – 12)²)

⇒ BC = √ (–11 + 3)² + (–6 – 12)²)

⇒ BC = √ ((–8)² + (–18)²)

⇒ BC = √ (64 + 324)

⇒ BC = √ 388

Distance of CD

⇒ CD = √ ((7 – (–11))² + (2 – (–6))²)

⇒ CD = √ ((7 + 11)² + (2 + 6)²)

⇒ CD = √ ((18)² + (8)²)

⇒ CD = √ (324 + 64)

⇒ CD = √388

Distance of AD

⇒ AD = √ ((7 – 15))² + (2 – 20)²)

⇒ AD = √ ((–8)² +(–18)²)

⇒ AD = √ (64 + 324)

⇒ AD = √ 388

Distance of AC

⇒ AC = √ ((–11 – 15)² + (–6 – 20)²)

⇒ AC = √ ((–26)² +(–26)²)

⇒ AC = √ (676 + 676)

⇒ AC = √ 1352

Distance of BD

⇒ BD = √ ((7 – (–3))² + (2 – 12)²)

⇒ BD = √ ((7 + 3))² + (2 – 12)²)

⇒ BD = √ ((10)² + (–10)²)

⇒ BD = √ (100 + 100)

⇒ BD = √ 200

AB = BC = CD = DA = √388 (That is, all the sides are equal.)

AC ≠ BD (That is, the diagonals are not equal.)

Hence the points A, B, C and D form a rhombus.

Formula used:

(0, –1), (2, 1), (0, 3) and (–2, 1)

Let the vertices be taken as A (0, –1), B (2, 1), C (0, 3) and D (–2, 1).

Distance of AB

⇒ AB = √ ((2 – 0)² + ((1 – (–1))²)

⇒ AB = √ ((2 – 0))² + (1 + 1)²)

⇒ AB = √ ((2)² + (2)²)

⇒ AB = √ (4 + 4)

⇒ AB = √ 8

Distance of BC

⇒ BC= √ ((0 – 2)² + (3 – 1)²)

⇒ BC = √ ((–2)² + (2)²)

⇒ BC = √ (4 + 4)

⇒ BC = √ 8

Distance of CD

⇒ CD = √ ((–2 – 0)² + (1 – 3)²)

⇒ CD = √ ((–2)² + (–2)²)

⇒ CD = √ (4 + 4)

⇒ CD = √8

Distance of AD

⇒ AD = √ ((–2 – 0)² + (1 – (–1))²)

⇒ AD = √ ((–2 – 0)² + (1 + 1)²)

⇒ AD = √ ((–2)² + (2)²)

⇒ AD = √ (4 + 4)

⇒ AD = √ 8

Distance of AC

⇒ AC = √ ((0 – 0)² + (3 – (–1))²)

⇒ AC = √ ((0 – 0)² + (3 + 1)²)

⇒ AC = √ ((0)² + (4)²)

⇒ AC = √ (0 + 16)

⇒ AC = √ 16

⇒ AC = 4

Distance of BD

⇒ AC = √ ((–2 – 2)² + (1 – 1)²)

⇒ AC = √ ((–4)² + (0)²)

⇒ AC = √ (16 + 0)

⇒ AC = √ 16

⇒ AC = 4

AB = BC = CD = DA = √8 (That is, all the sides are equal.)

AC = BD = 4. (That is, the diagonals are equal.)

Hence the points A, B, C and D form a square.

Formula used:

(5, 2), (1, 5), (–2, 1) and (2, –2)

Let the vertices be taken as A (5, 2), B (1, 5), C (–2, 1) and D (2, –2).

Distance of AB

⇒ AB = √ ((1 – 5)² + ((5 – 2)²)

⇒ AB = √ ((–4)² + (3)²)

⇒ AB = √ (16 + 9)

⇒ AB = √25

⇒ AB = 5

Distance of BC

⇒ BC= √ ((–2 – 1)² + (1 – 5)²)

⇒ BC = √ ((–3)² + (–4)²)

⇒ BC = √ (9 + 16)

⇒ BC = √ 25

⇒ BC = 5

Distance of CD

⇒ CD = √ ((2 – (–2))² + (–2 – 1)²)

⇒ CD = √ ((2 + 2)² + (–2 – 1)²)

⇒ CD = √ ((4)² + (–3)²)

⇒ CD = √ (16 + 9)

⇒ CD = √25

⇒ CD = 5

Distance of AD

⇒ AD = √ ((2 – 5)² + (–2 – 2)²)

⇒ AD = √ ((–3)² + (–4)²)

⇒ AD = √ (9 + 16)

⇒ AD = √ 25

⇒ AD = 5

Distance of AC

⇒ AC = √ ((–2 – 5)² + (1 – 2)²)

⇒ AC = √ ((–7)² + (–1)²)

⇒ AC = √ (49 + 1)

⇒ AC = √50

⇒ AC = 5√2

Distance of BD

⇒ BD = √ ((2 – 1)² + (–2 – 5)²)

⇒ BD = √ ((1)² + (–7)²)

⇒ BD = √ (1 + 49)

⇒ BD = √ 50

⇒ BD = 5√2

AB = BC = CD = DA = 5 (That is, all the sides are equal.)

AC = BD = 5√2. (That is, the diagonals are equal.)

Hence the points A, B, C and D form a square.

Formula used:

(3, 2), (0, 5), (–3, 2) and (0, –1)

Let the vertices be taken as A (3, 2), B (0, 5), C (–3, 2) and D (0, –1).

Distance of AB

⇒ AB = √ ((0 – 3)² + ((5 – 2)²)

⇒ AB = √ ((–3)² + (3)²)

⇒ AB = √ (9 + 9)

⇒ AB = √18

Distance of BC

⇒ BC= √ ((–3 – 0)² + (2 – 5)²)

⇒ BC = √ ((–3)² + (–3)²)

⇒ BC = √ (9 + 9)

⇒ BC = √ 18

Distance of CD

⇒ CD = √ ((0 – (–3))² + (–1 – 2)²)

⇒ CD = √ ((0 + 3)² + (–1 – 2)²)

⇒ CD = √ ((3)² + (–3)²)

⇒ CD = √ (9 + 9)

⇒ CD = √18

Distance of AD

⇒ AD = √ ((0 – 3)² + (–1 – 2)²)

⇒ AD = √ ((–3)² + (–3)²)

⇒ AD = √ (9 + 9)

⇒ AD = √ 18

Distance of AC

⇒ AC = √ ((–3 – 3)² + (2 – 2)²)

⇒ AC = √ ((–6)² + (0)²)

⇒ AC = √ (36 + 0)

⇒ AC = √36

⇒ AC = 6

Distance of BD

⇒ BD = √ ((0 – 0)² + (–1 – 5)²)

⇒ BD = √ ((0)² + (–6)²)

⇒ BD = √ (0 + 36)

⇒ BD = √ 36

⇒ BD = 6

AB = BC = CD = DA = √18. (That is, all the sides are equal.)

AC = BD = 6. (That is, the diagonals are equal.)

Hence the points A, B, C and D form a square.

Formula used:

(12, 9), (20, –6), (5, –14) and (–3, 1)

Let the vertices be taken as A (12, 9), B (20, –6), C (5, –14) and D (–3, 1).

Distance of AB

⇒ AB = √ ((20 – 12)² + ((–6 – 9)²)

⇒ AB = √ ((8)² + (–15)²)

⇒ AB = √ (64 + 225)

⇒ AB = √289

Distance of BC

⇒ BC= √ ((5 – 20)² + (–14 – (–6))²)

⇒ BC= √ ((5 – 20)² + (–14 + 6)²)

⇒ BC = √ ((–15)² + (–8)²)

⇒ BC = √ (225 + 64)

⇒ BC = √ 289

Distance of CD

⇒ CD = √ ((–3 – 5)² + (1 – (–14))²)

⇒ CD = √ ((–3 – 5)² + (1 + 14)²)

⇒ CD = √ ((–8)² + (15)²)

⇒ CD = √ (64 + 225)

⇒ CD = √289

Distance of AD

⇒ AD = √ ((–3 – 12)² + (1 – 9)²)

⇒ AD = √ ((–15)² + (–8)²)

⇒ AD = √ (225 + 64)

⇒ AD = √ 289

Distance of AC

⇒ AC = √ ((5 – 12)² + (–14 – 9)²)

⇒ AC = √ ((–7)² + (–23)²)

⇒ AC = √ (49 + 529)

⇒ AC = √578

Distance of BD

⇒ BD = √ ((–3 – 20)² + (1 – (–6))²)

⇒ BD = √ ((–3 – 20)² + (1 + 6)²)

⇒ BD = √ ((–23)² + (7)²)

⇒ BD = √ (529 + 49)

⇒ BD = √ 578

AB = BC = CD = DA = √ 289 (That is, all the sides are equal.)

AC = BD = √578. (That is, the diagonals are equal.)

Hence the points A, B, C and D form a square.

Formula used:

(–1, 2), (1, 0), (3, 2) and (1, 4)

Let the vertices be taken as A (–1, 2), B (1, 0), C (3, 2) and D (1, 4).

Distance of AB

⇒ AB = √ ((1 – (–1))² + ((0 – 2)²)

⇒ AB = √ ((1 + 1)² + (0 – 2)²)

⇒ AB = √ ((2)² + (–2)²)

⇒ AB = √ (4 + 4)

⇒ AB = √8

Distance of BC

⇒ BC= √ ((3 – 1)² + (2 – 0)²)

⇒ BC = √ ((2)² + (2)²)

⇒ BC = √ (4 + 4)

⇒ BC = √ 8

Distance of CD

⇒ CD = √ ((1 – 3)² + (4 – 2))²)

⇒ CD = √ ((–2)² + (2)²)

⇒ CD = √ (4 + 4)

⇒ CD = √8

Distance of AD

⇒ AD = √ ((1 – (–1))² + (4 – 2)²)

⇒ AD = √ ((1 + 1)² + (4 – 2)²)

⇒ AD = √ ((2)² + (2)²)

⇒ AD = √ (4 + 4)

⇒ AD = √ 8

Distance of AC

⇒ AC = √ ((3 – (–1))² + (2 – 2)²)

⇒ AC = √ ((3 + 1) + (2 – 2)²)

⇒ AC = √ ((4)² + (0)²)

⇒ AC = √ (16 + 0)

⇒ AC = √16

⇒ AC = 4

Distance of BD

⇒ BD = √ ((1 – 1)² + (4 – 0)²)

⇒ BD = √ ((0)² + (4)²)

⇒ BD = √ (0 + 16)

⇒ BD = √16

⇒ BD = 4

AB = BC = CD = DA = √8 (That is, all the sides are equal.)

AC = BD = 4. (That is, the diagonals are equal.)

Hence the points A, B, C and D form a square.

Formula used:

(8, 3), (0, –1), (–2, 3) and (6, 7)

Let the vertices be taken as A (8, 3), B (0, –1), C (–2, 3) and D (6, 7).

Distance of AB

⇒ AB = √ ((0 – 8)² + ((–1 – 3)²)

⇒ AB = √ ((–8)² + (–4)²)

⇒ AB = √ (64 + 16)

⇒ AB = √ 80

Distance of BC

⇒ BC= √ ((–2 – 0)² + (3 – (–1))²)

⇒ BC = √ ((–2 – 0)² + (3 + 1)²)

⇒ BC = √ ((–2)² + (4)²)

⇒ BC = √ (4 + 16)

⇒ BC = √ 20

Distance of CD

⇒ CD = √ ((6 – (–2))² + (7 – 3)²)

⇒ CD = √ ((6 + 2)² + (7 – 3)²)

⇒ CD = √ ((8)² + (4)²)

⇒ CD = √ (64 + 16)

⇒ CD = √80

Distance of AD

⇒ AD = √ ((6 – 8)² + (7 – 3)²)

⇒ AD = √ ((–2)² + (4)²)

⇒ AD = √ (4 + 16)

⇒ AD = √ 20

Distance of AC

⇒ AC = √ ((–2 – 8)² + (3 – 3)²)

⇒ AC = √ ((–10)² + (0)²)

⇒ AC = √ (100 + 0)

⇒ AC = √ 100

⇒ AC = 10

Distance of BD

⇒ BD = √ ((6 – 0 )² + (7 – (–1))²)

⇒ BD = √ ((6 – 0)² + (7 + 1)²)

⇒ BD = √ ((6)² + (8)²)

⇒ BD = √ (36 + 64)

⇒ BD = √ 100

⇒ BD = 10

AB = CD = √80 and BC = AD = √ 20 (opposite sides of rectangle are equal).

AC = BD = 10 (Diagonals of rectangle are equal)

Hence the points A, B, C and D form a square.

Formula used:

(–1, 1), (0, 0), (3, 3) and (2, 4)

Let the vertices be taken as A (–1, 1), B (0, 0), C (3, 3) and D (2, 4).

Distance of AB

⇒ AB = √ ((0 – (–1))² + (0 – 1)²)

⇒ AB = √ ((0 + 1)² + (0 – 1)²)

⇒ AB = √ ((1)² + (–1)²)

⇒ AB = √ (1 + 1)

⇒ AB = √ 2

Distance of BC

⇒ BC= √ ((3 – 0)² + (3 – 0)²)

⇒ BC = √ ((3)² + (3)²)

⇒ BC = √ (9 + 9)

⇒ BC = √ 18

Distance of CD

⇒ CD = √ ((2 – 3)² + (4 – 3)²)

⇒ CD = √ ((1)² + (1)²)

⇒ CD = √ (1 + 1)

⇒ CD = √2

Distance of AD

⇒ AD = √ ((2 – (–1))² + (4 – 1)²)

⇒ AD = √ ((2 + 1)² + (4 – 1)²)

⇒ AD = √ ((3)² + (3)²)

⇒ AD = √ (9 + 9)

⇒ AD = √ 18

Distance of AC

⇒ AC = √ ((3 – (–1))² + (3 – 1)²)

⇒ AC = √ ((3 + 1)² + (3 – 1)²)

⇒ AC = √ ((4)² + (2)²)

⇒ AC = √ (16 + 4)

⇒ AC = √ 20

Distance of BD

⇒ BD = √ ((2 – 0)² + (4 – 0)²)

⇒ BD = √ ((2)² + (4)²)

⇒ BD = √ (4 + 16)

⇒ BD = √ 20

AB = CD = √2 and BC = AD = √ 18 (opposite sides of rectangle are equal).

AC = BD = √ 20 (Diagonals of rectangle are equal)

Hence the points A, B, C and D form a square.

Formula used:

(–3, 0), (1, –2), (5, 6) and (1, 8)

Let the vertices be taken as A (–3, 0), B (1, –2), C (5, 6) and D (1, 8).

Distance of AB

⇒ AB = √ ((1 – (–3))² + ((–2 – 0)²)

⇒ AB = √ ((1 + 3)² + (–2 – 0)²)

⇒ AB = √ ((4)² + (–2)²)

⇒ AB = √ (16 + 4)

⇒ AB = √ 20

Distance of BC

⇒ BC= √ ((5 – 1)² + (6 – (–2))²)

⇒ BC = √ ((5 – 1)² + (6 + 2)²)

⇒ BC = √ ((4)² + (8)²)

⇒ BC = √ (16 + 64)

⇒ BC = √ 80

Distance of CD

⇒ CD = √ ((1 – 5)² + (8 – 6)²)

⇒ CD = √ ((–4)² + (2)²)

⇒ CD = √ (16 + 4)

⇒ CD = √ 20

Distance of AD

⇒ AD = √ ((1 – (–3))² + (8 – 0)²)

⇒ AD = √ ((1 + 3)² + (8 – 0)²)

⇒ AD = √ ((4)² + (8)²)

⇒ AD = √ (16 + 64)

⇒ AD = √ 80

Distance of AC

⇒ AC = √ ((5 – (–3))² + (6 – 0)²)

⇒ AC = √ ((5 + 3)² + (6 – 0)²)

⇒ AC = √ ((8)² + (6)²)

⇒ AC = √ (64 + 36)

⇒ AC = √ 100

⇒ AC = 10

Distance of BD

⇒ BD = √ ((1 – 1)² + (8 – (–2))²)

⇒ BD = √ ((1 – 1)² + (8 + 2)²)

⇒ BD = √ ((0)² + (10)²)

⇒ BD = √ (0 + 100)

⇒ BD = √ 100

⇒ BD = 10

AB = CD = √20 and BC = AD = √ 80 (opposite sides of rectangle are equal).

AC = BD = 10 (Diagonals of rectangle are equal)

Hence the points A, B, C and D form a square.

Formula used:

Given: Distance = 10 and coordinates of two points is A (x, 7) and B (1, 15)

AB = √ (x₂ – x₁)² + (y₂ – y₁)²

⇒ 10 = √ (1 – x)² + (15 – 7)²

⇒ 10 = √ (1 – x)² + 8²

Squaring both sides

⇒ 10² = (1 – x)² + 8²

⇒ 100 = 1 – 2x + x² + 64

⇒ 100 = x² – 2x + 65

⇒ x² – 2x + 65 – 100 = 0

⇒ x² – 2x – 35 = 0

⇒ x² – 7x + 5x – 35 = 0

⇒ x (x – 7) + 5(x – 7) = 0

⇒ (x – 7) (x + 5) = 0

x – 7 = 0 or x + 5 = 0

x = 7 or x = –5

Formula used:

Let the points be A (4, 1), B (–10, 6) and C (9, –13)

Distance of AB

⇒ AB = √ ((–10 – 4)² + (6 – 1)²)

⇒ AB = √ ((–14)² + (5)²)

⇒ AB = √ (196 + 25

⇒ AB = √ 221

Distance of BC

⇒ BC = √ ((9 – 4)² + (–13 – 1)²)

⇒ BC = √ ((5)² + (–14)²)

⇒ BC = √ (25 + 196

⇒ BC = √ 221

∴ AB = BC = √ 221

Formula used:

Let the points be A (x, y), B (2, 3) and C (–6, –5)

Distance of AB

⇒ AB = √ ((2 – x)² + (3 – y)²)

⇒ AB = √ ((4 – 4x + x²) + (9 – 6y + y²))

⇒ AB = √ (4 – 4x + x² + 9 – 6y + y²)

⇒ AB = √ x² + y² – 4x – 6y + 13

Distance of BC

⇒ BC = √ ((–6 – x)² + (–5 – y)²)

⇒ BC = √ ((36 + x² + 12x) + (25 + y² + 10y))

⇒ BC = √ (36 + x² + 12x + 25 + y² + 10y)

⇒ BC = √ (x² + y² + 12x + 10y + 61)

i.e. AB = BC (∵ Given)

⇒ √x² + y² – 4x – 6y + 13 = √ x² + y² + 12x + 10y + 61

Squaring both sides

⇒ x² + y² – 4x – 6y + 13 = x² + y² + 12x + 10y + 61

⇒ x² + y² – 4x – 6y + 13 – x² – y² – 12x – 10y – 61 = 0

⇒ –16x – 16 y – 48 = 0

⇒ –4(x + y + 3) = 0

⇒ x + y + 3 = 0

Hence proved.

Formula used:

Given: Distance = 4 and coordinates of two points is A (2, –6) and B (2, y)

AB = √ (x₂ – x₁)² + (y₂ – y₁)²

⇒ 4 = √ (2 – 2)² + (y – (–6))²

⇒ 4 = √ (0) + (y + 6)²

Squaring both sides

⇒ 4² = (y + 6)²

⇒ 16 = y² + 12y + 36

⇒ y² + 12y + 36 – 16 = 0

⇒ y² + 12y + 20 = 0

⇒ y² + 10y + 2y + 20 = 0

⇒ y (y + 10) + 2(y + 10) = 0

⇒ (y + 2) (y + 10) = 0

y + 2 = 0 or y + 10 = 0

y = –2 or y = –10

∴ y = –2 or –10

Formula used:

i). (0, 8), (6, 0) and (0, 0)

Let the points be A (0, 8), B (6, 0) and C (0, 0)

Distance of AB

⇒ AB = √ ((6 – 0)² + (0 – 8)²)

⇒ AB = √ ((6)² + (–8)²)

⇒ AB = √ (36 + 64)

⇒ AB = √ 100

⇒ AB = 10

Distance of BC

⇒ BC = √ ((0 – 6)² + (0 – 0)²)

⇒ BC = √ ((–6)² + (0)²)

⇒ BC = √ (36 + 0)

⇒ BC = √ 36

⇒ BC = 6

Distance of AC

⇒ AC = √ ((0 – 0)² + (0 – 8)²)

⇒ AC = √ ((0)² + (–8)²)

⇒ AC = √ (0 + 64)

⇒ AC = √ 64

⇒ AC = 8

Perimeter of ΔABC = AB + BC + AC

= 10 + 6 + 8

= 24

ii). (9, 3), (1, –3) and (0, 0)

Let the points be A (9, 3), B (1, –3) and C (0, 0)

Distance of AB

⇒ AB = √ ((1 – 9)² + (–3 – 3)²)

⇒ AB = √ ((–8)² + (–6)²)

⇒ AB = √ (64 + 36)

⇒ AB = √ 100

⇒ AB = 10

Distance of BC

⇒ BC = √ ((0 – 1)² + (0 – (–3))²)

⇒ BC = √ ((0 – 1)² + (0 + 3)²)

⇒ BC = √ ((–1)² + (3)²)

⇒ BC = √ (1 + 9)

⇒ BC = √10

Distance of AC

⇒ AC = √ ((0 – 9)² + (0 – 3)²)

⇒ AC = √ ((–9)² + (–3)²)

⇒ AC = √ (81 + 8)

⇒ AC = √ 90

⇒ AC = 3√10

Perimeter of ΔABC = AB + BC + AC

= 10 + √ 10 + 3√ 10

= 10 + 4√10

Formula used:

Let the point A (–5, 2), B (9, –2) and C be the point on y–axis i.e. (0, y)

Distance of AC

⇒ AC = √ ((0 – (–5))² + (y – 2)²)

⇒ AC = √ ((0 + 5)² + (y – 2)²)

⇒ AC = √ ((5)² + (y – 2)²)

⇒ AC = √ (25 + y² – 4y + 4)

⇒ AC = √ y² – 4y + 29

Distance of BC

⇒ BC = √ ((0 – 9)² + (y – (–2)²)

⇒ BC = √ ((0 – 9)² + (y + 2)²)

⇒ BC = √ ((9)² + (y + 2)²)

⇒ BC = √ (81 + y² + 4y + 4)

⇒ BC = √ y² + 4y + 85

i.e. AC = BC (∵ Given)

⇒ √ y² – 4y + 29 = √ y² + 4y + 85

Squaring both sides

⇒ y² – 4y + 29 = y² + 4y + 8

⇒ y² – 4y + 29 – y² – 4y – 85 = 0

⇒ –8y – 56 = 0

⇒ –8 (y + 7) = 0

⇒ y + 7 = 0

y = –7

∴ the point on y–axis is (0, –7).

Formula used:

Let the point be A (–5, 6) and O (3, 2)

Distance of OA

⇒ OA = √ ((–5 – 3)² + (6 – 2)²)

⇒ OA = √ ((–8)² + (4)²)

⇒ OA = √ (64 + 16)

⇒ OA = √ 80

⇒ OA = 4√5

Formula used:

Let the point A (0, –5), B (4, 3) and C (–4, –3) lie on the circle with center O (0, 0)

Distance of AO

⇒ AO = √ ((0 – 0)² + (0 – (–5))²)

⇒ AO = √ ((0 – 0)² + (0 + 5)²)

⇒ AO = √ ((0)² + (5)²)

⇒ AO = √ (0 + 25)

⇒ AO = √ 25

⇒ AO = 5

Distance of BO

⇒ BO = √ ((0 – 4)² + (0 – 3)²)

⇒ BO = √ ((–4)² + (–3)²)

⇒ BO = √ (16 + 9)

⇒ BO = √ 25

⇒ BO = 5

Distance of CO

⇒ CO = √ ((0 – (–4))² + (0 – (–3))²)

⇒ CO = √ ((0 + 4)² + (0 + 3)²)

⇒ CO = √ ((4)² + (3)²)

⇒ CO = √ (16 + 9)

⇒ CO = √ 25

⇒ CO = 5

∴ AO = BO = CO = 5 = Radius

Hence, point A, B and C lie on the circle.

Formula used:

Let the point P (4, 0)

PB is perpendicular segment from point A to B

∴ let B be (4, –y)

Distance of PA

⇒ PA = √ ((4 – 4)² + (3 – 0)²)

⇒ PA = √ ((0)² + (3)²)

⇒ PA = √ (0 + 9)

⇒ PA= √ 9

⇒ PA = 3

Distance of PB

⇒ PB = √ ((4 – 4)² + (–y – 0)²)

⇒ PB = √ ((4 – 4)² + (–y)²)

⇒ PB = √ ((0)² + (–y)²)

⇒ PB = √ 0 + y²

⇒ PB = y²

i.e. AP = BP

⇒ 3 = √ y²

Squaring both sides

⇒ 9 = y²

⇒ y = √9

⇒ y = 3

∴ Point B is (4, –3)

Formula used:

Coordinates of rhombus are A (2, 0), B (5, –5), C (8, 0) and D (5, 5)

Area of rhombus =

Distance of AC(d₁)

⇒ AC = √ ((8 – 2)² + (0 – 0)²)

⇒ AC = √ ((6)² + (0)²)

⇒ AC = √ (36 + 0)

⇒ AC = √ 36

⇒ AC = 6

Distance of BD(d₂)

⇒ BD = √ ((5 – 5)² + (5 – (–5))²)

⇒ BD = √ ((5 – 5)² + (5 + 5)²)

⇒ BD = √ ((0)² + (10)²)

⇒ BD = √ (0 + 100)

⇒ BD = √ 100

⇒ BD = 10

∴ Area of rhombus =

⇒ Area

⇒ Area = 3 × 10

⇒ Area = 30 units sq.

Formula used:

Let the points A (1, 5) B (5, 8) and C (13, 14)

Distance of AB

⇒ AB = √ ((5 – 1)² + (8 – 5)²)

⇒ AB = √ ((4)² + (3)²)

⇒ AB = √ (16 + 9)

⇒ AB = √ 25

⇒ AB = 5

Distance of BC

⇒ BC = √ ((13 – 5)² + (14 – 8)²)

⇒ BC = √ ((8)² + (6)²)

⇒ BC = √ (64 + 36)

⇒ BC = √ 100

⇒ BC = 10

Distance of AC

⇒ AC = √ ((13 – 1)² + (14 – 5)²)

⇒ AC = √ ((12)² + (9)²)

⇒ AC = √ (144 + 81)

⇒ AC = √ 225

⇒ AC = 15

Now, we can see that AB + BC = AC.

∴ A, B and C are collinear. Hence, we cannot draw triangle using these coordinates.

Formula used:

Let the point be A (x, y)

Center is at origin (0, 0)

Distance of OA

⇒ OA = √((x – 0)² + (y – 0)²)

⇒ OA = √ ((x)² + (y)²)

⇒ OA = √ x² + y²

Squaring both sides

⇒ (0A)² = x² + y²

⇒ (17)² = x² + y²

Using Pythagorean triplet

x and y can 8 and 5 or vice–a–versa.

∴ x = ± 8 or ±15

y = ± 8 or ±15

Hence, coordinate on circle other than coordinates on axis are

(8, 15), (–8, –15), (–8, 15) and (8, –15)

Formula used:

Let the points be A (3, 1), B (2, 2), C (1, 1) and S(2, 1)

Distance of SA

⇒ SA = √ ((3 – 2)² + (1 – 1)²)

⇒ SA = √ ((1)² + (0)²

⇒ SA = √ (1 + 0)

⇒ SA = √ 1 = 1

Distance of SB

⇒ SB = √ ((2 – 2)² + (2 – 1)²)

⇒ SB = √ ((0)² + (1)²

⇒ SB = √ (0 + 1)

⇒ SB = √ 1 = 1

Distance of SC

⇒ SC = √ ((1 – 2)² + (1 – 1)²)

⇒ SC = √ ((–1)² + (0)²

⇒ SC = √ (1 + 0)

⇒ SC = √ 1 = 1

It is known that the circum–centre is equidistant from all the vertices of a triangle.

Since S is equidistant from all the three vertices, it is the circum–centre of the triangle ABC.

Formula used:

Let the points be A (1, 0), B (0, –1), C and S (0, 0)

Distance of SA

⇒ SA = √ ((1 – 0)² + (0 – 0)²)

⇒ SA = √ ((1)² + (0)²

⇒ SA = √ (1 + 0)

⇒ SA = √ 1 = 1

Distance of SB

⇒ SB = √ ((0 – 0)² + (–1 – 0)²)

⇒ SB = √ ((0)² + (–1)²

⇒ SB = √ (0 + 1)

⇒ SB = √ 1 = 1

Distance of SC

⇒ SC

⇒ SC

⇒ SC

⇒ SC

⇒ SC =√ 1 = 1

It is known that the circum–centre is equidistant from all the vertices of a triangle.

Since S is equidistant from all the three vertices, it is the circum–centre of the triangle ABC.

Formula used:

Let A, B, C and D represent the points (6, 1), (8, 2), (9, 4) and (p, 3)

Distance of AB

⇒ AB = √ ((8 – 6))² + (2 – 1)²)

⇒ AB = √ ((2)² + (1)²)

⇒ AB = √ (4 + 1)

⇒ AB = √ 5

Distance of CD

⇒ CD = √ ((p – 9)² + (3 – 4)²)

⇒ CD = √ ((p – 9)² + (1)²)

⇒ CD = √ (p² + 81 – 18p + 1)

⇒ CD = √ p² – 18p + 82

i.e., The opposite sides are equal.

∴ AB = CD

⇒ √5 = √ p² – 18p + 82

Squaring both sides

⇒ 5 = p² – 18p + 82

⇒ p² – 18p + 82 – 5 =0

⇒ p² – 18p + 77 = 0

⇒ p² – 11p – 7p + 77 = 0

⇒ p(p – 11) – 7(p – 11)= 0

⇒ (p – 11)(p – 7) = 0

p – 11 = 0 or p – 7 = 0

p = 11 or p = 7

Formula used:

Let the point be A (x, 0) and B (0, y)

Given center O (0, 0) and radius = 10

Distance of OA

⇒ 5 = √ ((x – 0)² + (0 – 0)²)

⇒ 5 = √ ((x)² + (0)²)

⇒ 5= √ (x² + 0)

⇒ 5 = √ x²

⇒ 5 = x

∴ point A is (5, 0)

Distance of OB

⇒ 5 = √ ((0 – 0)² + (y – 0)²)

⇒ 5 = √ ((0)² + (y)²)

⇒ 5 = √ (0 + y²)

⇒ 5 = √ y²

⇒ 5 = y

∴ point B is (0, 5)

Now,

Distance AB = √ ((0 – 5)² + (5 – 0)²)

= √ ((–5)² + (5)²)

= √ (25 + 25)

= √ (50)

= 5√2

Option A: value to lie in I quadrant, both should be positive. Hence, this is not correct.

Option B: value to lie in II quadrant, x–coordinate should be negative and y–coordinate should be positive. Hence, this is correct.

Option C: value to lie in III quadrant, both should be negative. Hence, this is not correct.

Option D: value to lie in I quadrant, x–coordinate should be positive and y– coordinate should be negative. Hence, this is not correct.

Option A: point on OX, x–coordinate will be greater than 0 i.e. x > 0.

Option B: point on OY, y–coordinate will be greater than 0 i.e. y > 0.

Option C: point on OX’, x–coordinate will be lesser than 0 i.e. x < 0.

Option D: point on OY’, y–coordinate will be lesser than 0 i.e. y < 0.

Option A: point with a > 0, b < 0, lies in the IV quadrant.

Option B: point with a < 0, b < 0, lies in the III quadrant.

Option C: point with a > 0, b > 0, lies in the I quadrant.

Option D: point with a < 0, b > 0, lies in the II quadrant.

Formula used:

Let the points be A (1, 0), B (0, 1), C (–1, 0) and D (0, –1)

Distance of diagonal AC

⇒ AC = √ ((–1 – 1)² + ( 0 – 0)²)

⇒ AC = √ ((–2)² + (0)²)

⇒ AC =√ (4 + 0)

⇒ AC = √ 4

⇒ AC = 2

∴ Option A is correct.

Formula used:

Let the point be A (–5, 0) B (5, 0) and C (0, 6)

Distance of AB

⇒ AB = √ (5 – (–5))² + (0 – 0)²)

⇒ AB = √ (5 + 5)² + (0 – 0)²)

⇒ AB = √ ((10)² + (0)²)

⇒ AB = √ (100 + 0)

⇒ AB = √ 100 = 10

Distance of AC

⇒ AC = √ ((0 – (–5))² + (6 – 0)²)

⇒ AC = √ ((5)² + (6)²)

⇒ AC = √ (25 + 36)

⇒ AC = √ 61

Distance of BC

⇒ BC = √ ((0 – 5)² + (6 – 0)²)

⇒ BC = √ ((–5)² + (6)²)

⇒ BC = √ (25 + 36)

⇒ BC = √ 61

We notice that BC = AC = √ 61

∴ Points A, B and C are coordinates of an isosceles triangle.

Formula used:

Let the point be A (0, 8) and B (0, –2)

Distance of AB

⇒ AB = √ (0 – 0)² + (–2 – 8)²)

⇒ AB = √ ((0)² + (–10)²)

⇒ AB = √ (0 + 100)

⇒ AB = √ 100

⇒ AB = 10

∴ , Option B is correct.

Formula used:

Let the point be A (a, b) and B (–a, –b)

Distance of AB

⇒ AB = √ (–a – a)² + (–b – b)²)

⇒ AB = √ ((–2a)² + (–2b)²)

⇒ AB = √ (4a² + 4b²)

⇒ AB = √ 4(a² + b²)

⇒ AB = 2√ (a² + b²)

∴ Option D is correct.

For any point on y–axis, x–coordinate is 0.

∴ the point is (0, –5).

Formula used:

Let the point be A (p, q), B (–4, 0) and C (4, 0)

Distance of AB

⇒ AB = √ (–4 – p)² + (0 – q)²)

⇒ AB = √ ((–4 – p)² + (–q)²)

⇒ AB = √ (16 + p² + 8p + q²)

Distance of AC

⇒ AC = √ (4 – p)² + (0 – q)²)

⇒ AC = √ ((4 – p)² + (–q)²)

⇒ AC = √ (16 + p² – 8p + q²)

i.e. AB = AC (Given)

⇒ 16 + p² + 8p + q² = 16 + p² – 8p + q²

Squaring both sides

⇒ 16 + p² + 8p + q² = 16 + p² – 8p + q²

⇒ 16 + p² + 8p + q² – 16 – p² + 8p – q² = 0 …

⇒ 16 p = 0

⇒ p = 0

∴ Option A is correct.