State whether the following statements are true / false.
i. (5, 7) is a point in the IV quadrant.
ii. (−2, −7) is a point in the III quadrant.
iii. (8, −7) lies below the x–axis.
iv. (5, 2) and (−7, 2) are points on the line parallel to y–axis.
v. (−5, 2) lies to the left of y–axis.
vi. (0, 3) is a point on x–axis.
vii. (−2, 3) lies in the II quadrant.
viii. (−10, 0) is a point on x–axis.
ix. (−2, −4) lies above x–axis.
x. For any point on the x–axis its y–coordinate is zero.
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
Plot the following points in the coordinate system and specify their quadrant.
i. (5, 2) ii. (−1, −1)
iii. (7, 0) iv. (−8, −1)
v. (0, −5) vi. (0, 3)
vii. (4, −5) viii. (0, 0)
ix. (1, 4) x. (−5, 7)
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
Write down the abscissa for the following points.
i. (−7, 2) ii. (3, 5)
iii. (8, −7) iv. (−5, −3)
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
Write down the ordinate of the following points.
i. (7, 5) ii. (2, 9)
iii. (−5, 8) iv. (−7, −3)
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
Plot the following points in the coordinate plane.
i. (4, 2) ii. (4, −5)
iii. (4, 0) iv. (4, −2)
How is the line joining them situated?
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.
The ordinates of two points are each −6. How is the line joining them related with reference to x–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.
The abscissa of two points is 0. How is the line joining situated?
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.
Mark the points A (2, 4), B (−3, 4),C (−3, −1) and D (2, −1) in the cartesian plane. State the figure obtained by joining A and B, B and C, C and D and D and A.
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.
With rectangular axes plot the points O (0, 0), A (5, 0), B (5, 4). Find the coordinate of point C such that OABC forms a rectangle.
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).
In a rectangle ABCD, the coordinates of A, B and D are (0, 0) (4, 0) (0, 3). What are the coordinates of C?
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).
Find the distance between the following pairs of points.
(7, 8) and (−2, −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
Find the distance between the following pairs of points.
(6, 0) and (−2, 4)
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
Find the distance between the following pairs of points.
(−3, 2) and (2, 0)
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
Find the distance between the following pairs of points.
(−2, −8) and (−4, −6)
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
Find the distance between the following pairs of points.
(−2, −3) and (3, 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
Find the distance between the following pairs of points.
(2, 2) and (3, 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
Find the distance between the following pairs of points.
(−2, 2) and (3, 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 = √ ((5)2 + (0)2)
⇒ D = √ (25 + 0)
⇒ D = √ 25
⇒ D = √ (5 × 5)
⇒ D = 5
Find the distance between the following pairs of points.
(7, 0) and (8, 0)
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
Find the distance between the following pairs of points.
(0, 17) and (0, −1)
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
Find the distance between the following pairs of points.
(5, 7) and the origin
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
Show that the following points are collinear.
(3, 7), (6, 5) and (15, −1)
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
Show that the following points are collinear.
(3, −2), (−2, 8) and (0, 4)
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
Show that the following points are collinear.
(1, 4), (3, −2) and (−1, 10)
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.
Show that the following points are collinear.
(6, 2), (2, −3) and (−2, −8)
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.
Show that the following points are collinear.
(4, 1), (5, −2) and (6, −5)
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.
Show that the following points form an isosceles triangle.
(−2, 0), (4, 0) and (1, 3)
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.
Show that the following points form an isosceles triangle.
(1, −2), (−5, 1) and (1, 4)
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.
Show that the following points form an isosceles triangle.
(−1, −3), (2, −1) and (−1, 1)
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.
Show that the following points form an isosceles triangle.
(1, 3), (−3, –5) and (−3, 0)
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.
Show that the following points form an isosceles triangle.
(2, 3), (5, 7) and (1, 4)
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.
Show that the following points form a right–angled triangle.
(2, −3), (−6, −7) and (−8, −3)
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.
Show that the following points form a right–angled triangle.
(−11, 13), (−3, −1) and (4, 3)
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.
Show that the following points form a right–angled triangle.
(0, 0), (a, 0) and (0, b)
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.
Show that the following points form a right–angled triangle.
(10, 0), (18, 0) and (10, 15)
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.
Show that the following points form a right–angled triangle.
(5, 9), (5, 16) and (29, 9)
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.
Show that the following points form an equilateral triangle.
(0, 0), (10, 0) and (5, 5√3)
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.
Show that the following points form an equilateral triangle.
(a, 0), (−a, 0) and (0, a√3)
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.
Show that the following points form an equilateral triangle.
(2, 2), (−2, −2) and (−2√3, 2√3)
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.
Show that the following points form an equilateral triangle.
(√3, 2), (0,1) and (0, 3)
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.
Show that the following points form an equilateral triangle.
(−√3, 1), (2√3, −2) and (2√3, 4)
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.
Show that the following points taken in order form the vertices of a parallelogram.
(−7, –5), (−4, 3), (5, 6) and (2, −2)
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.
Show that the following points taken in order form the vertices of a parallelogram.
(9, 5), (6, 0), (−2, −3) and (1, 2)
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.
Show that the following points taken in order form the vertices of a parallelogram.
(0, 0), (7, 3), (10, 6) and (3, 3)
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.
Show that the following points taken in order form the vertices of a parallelogram.
(−2, 5), (7, 1), (−2, −4) and (7, 0)
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.
Show that the following points taken in order form the vertices of a parallelogram.
(3, −5), (−5, −4), (7, 10) and (15, 9)
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.
Show that the following points taken in order form the vertices of a rhombus.
(0, 0), (3, 4), (0, 8) and (−3, 4)
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.
Show that the following points taken in order form the vertices of a rhombus.
(−4, −7), (−1, 2), (8, 5) and (5, −4)
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.
Show that the following points taken in order form the vertices of a rhombus.
(1, 0), (5, 3), (2, 7) and (−2, 4)
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.
Show that the following points taken in order form the vertices of a rhombus.
(2, −3), (6, 5), (−2, 1) and (−6, −7)
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.
Show that the following points taken in order form the vertices of a rhombus.
(15, 20), (−3, 12), (−11, −6) and (7, 2)
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.
Examine whether the following points taken in order form a square.
(0, −1), (2, 1), (0, 3) and (−2, 1)
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.
Examine whether the following points taken in order form a square.
(5, 2), (1, 5), (−2, 1) and (2, −2)
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.
Examine whether the following points taken in order form a square.
(3, 2), (0, 5), (−3, 2) and (0, −1)
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.
Examine whether the following points taken in order form a square.
(12, 9), (20, −6), (5, −14) and (−3, 1)
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.
Examine whether the following points taken in order form a square.
(−1, 2), (1, 0), (3, 2) and (1, 4)
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.
Examine whether the following points taken in order form a rectangle.
(8, 3), (0, −1), (−2, 3) and (6, 7)
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.
Examine whether the following points taken in order form a rectangle.
(−1, 1), (0, 0), (3, 3) and (2, 4)
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.
Examine whether the following points taken in order form a rectangle.
(−3, 0), (1, −2), (5, 6) and (1, 8)
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.
If the distance between two points (x,7) and (1, 15) is 10, find x.
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
Show that (4, 1) is equidistant from the points (−10, 6) and (9, −13).
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
If two points (2, 3) and (−6, −5) are equidistant from the point (x, y), show that x + y + 3 = 0.
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.
If the length of the line segment with end points (2, −6) and (2, y) is 4, find y.
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
Find the perimeter of the triangle with vertices (i) (0, 8), (6, 0) and origin; (ii) (9, 3), (1, −3) and origin.
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
Find the point on the y–axis equidistant from (−5, 2) and (9, −2) (Hint: A point on the y–axis will have its x–coordinate as zero).
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).
Find the radius of the circle whose center is (3, 2) and passes through (−5, 6).
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
Prove that the points (0, −5), (4, 3) and (−4, −3) lie on the circle centered at the origin with radius 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.
In the Fig. 5.20, PB is perpendicular segment from the point A (4, 3). If PA = PB then find the coordinates of B.
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)
Find the area of the rhombus ABCD with vertices A (2, 0), B (5, –5), C (8, 0) and D (5, 5). [Hint: Area of the rhombus ABCD = 1/2d1 d2]
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.
Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14)? Give reason.
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.
If origin is the center of a circle with radius 17 units, find the coordinates of any four points on the circle which are not on the axes. (Use the Pythagorean triplets)
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)
Show that (2, 1) is the circum–center of the triangle formed by the vertices (3, 1), (2, 2) and (1, 1).
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.
Show that the origin is the circum–center of the triangle formed by the vertices (1, 0), (0, −1) and .
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.
If the points A (6, 1), B (8, 2), C (9, 4) and D (p, 3) taken in order are the vertices of a parallelogram, find the value of p using distance formula.
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
The radius of the circle with center at the origin is 10 units. Write the coordinates of the point where the circle intersects the axes. Find the distance between any two of such points.
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
The point (–2, 7) lies is the quadrant
A. I
B. II
C. III
D. IV
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.
The point (x, 0) where x < 0 lies on
A. OX
B. OY
C. OX’
D. OY’
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.
For a point A (a, b) lying in quadrant III
A. a > 0, b < 0
B. a < 0, b < 0
C. a > 0, b > 0
D. a < 0, b > 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.
The diagonal of a square formed by the points (1, 0) (0, 1) (–1, 0) and (0, –1) is
A. 2
B. 4
C. √2
D. 8
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.
The triangle obtained by joining the points A (–5, 0) B (5, 0) and C (0, 6) is
A. an isosceles triangle
B. right triangle
C. scalene triangle
D. an equilateral triangle
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.
The distance between the points (0, 8) and (0, –2) is
A. 6
B. 100
C. 36
D. 10
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.
(4, 1), (–2, 1), (7, 1) and (10, 1) are points
A. on x–axis
B. on a line parallel to x–axis
C. on a line parallel to y–axis
D. on y–axis
The distance between the points (a, b) and (–a, –b) is
A. 2a
B. 2b
C. 2a + 2b
D.
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.
The point which is on y–axis with ordinate –5 is
A. (0, −5)
B. (−5, 0)
C. (5, 0)
D. (0, 5)
For any point on y–axis, x–coordinate is 0.
∴ the point is (0, –5).
The relation between p and q such that the point (p, q) is equidistant from (–4, 0) and (4, 0) is
A. p = 0
B. q = 0
C. p + q = 0
D. p + q = 8
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.