Coordinate Geometry

In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x₁,x₂) and Q(y₁,y₂) then

PQ = √(x₂ – x₁)² + (y₂ – y₁)²

AB = √{(15 – 9)² + (11 – 3)²}

= √{(6)² + (8)²}

= √{36 + 64}

= √100

∴ AB = 10 units.

In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x₁,x₂) and Q(y₁,y₂) then

PQ = √(x₂ – x₁)² + (y₂ – y₁)²

AB = √{(– 5 – 7)² + (1 – (– 4))²}

= √{(– 12)² + (5)²}

= √{144 + 25}

= √169

∴ AB = 13 units

In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x₁,x₂) and Q(y₁,y₂) then

PQ = √(x₂ – x₁)² + (y₂ – y₁)²

AB = √{(9 – (– 6))² + (– 12 – (– 4))²}

= √{(15)² + (– 8)²}

= √{225 + 64}

= √289

∴ AB = 17 units

In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x₁,x₂) and Q(y₁,y₂) then

PQ = √(x₂ – x₁)² + (y₂ – y₁)²

AB = √{(4 – 1)² + (– 6 – (– 3))²}

= √{(3)² + (– 3)²}

= √{9 + 9}

= √18

∴ AB = 3√2 units

In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x₁,x₂) and Q(y₁,y₂) then

PQ = √(x₂ – x₁)² + (y₂ – y₁)²

AB = √{((a – b) – (a + b))² + ((a + b) – (a – b))²}

= √{(– 2b)² + (2b)²}

= √{4b² + 4b²}

= √8b²

∴ AB = 2√2b units

In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x₁,x₂) and Q(y₁,y₂) then

PQ = √(x₂ – x₁)² + (y₂ – y₁)²

PQ = √{(a cos a – a sin a)² – (– a sin a – a cos a)²}

= √{(a² cos² a + a² sin² a – 2a² sina.cosa + a² cos² a + a² sin² a + 2a² sina.cosa }

= √{ a² (cos² a + sin² a) + (a² (cos² a + sin² a))}

= √a²(1) + a² (1)

= √a²(1 + 1)

∴ PQ = a√2 units

Since it is given that the distance is to be found from origin so in this question we have to use the distance formula keeping one – point fix i.e. O (0,0), as shown below:

OA = √{(5 – 0)² + (– 12 – 0)²}

= √{(5)² + (– 12)²}

= √{25 + 144}

= √169

∴ OA = 13 units

Since it is given that the distance is to be found from origin so in this question we have to use the distance formula keeping one – point fix i.e. O (0,0), as shown below:

OB = √{(– 5 – 0)² + (5 – 0)²}

= √{(– 5)² + (5)²}

= √{25 + 25}

= √50

∴ OB = 5√2 units

Since it is given that the distance is to be found from origin so in this question we have to use the distance formula keeping one – point fix i.e. O (0,0), as shown below:

OC = √{(– 4 – 0)² + (– 6 – 0)²}

= √{(– 4)² + (– 6)²}

= √{16 + 36}

∴ OC = √52 units

Given:

Distance AB = 5 units

By distance formula, as shown below:

AB = √{(5 – x)² + (3 – (– 1))²}

5 = √{(5 – x)² + (4)²}

5 = √{25 + x² – 10x + 16}

5 = √{41 + x² – 10x}

Squaring both sides we get

25 = 41 + x² – 10x

⇒ 16 + x² – 10x = 0

⇒ (x – 8)(x – 2) = 0

⇒ × = 8 or × = 2

∴ The values of × can be 8 or 2

Given, the distance AB = 10 units

By distance formula, as shown below:

AB = √{(10 – 2)² + (y – (– 3))²}

10 = √{(8)² + (y + 3)²}

10 = √{64 + y² + 6y + 9}

10 = √{73 + y² + 6y}

Squaring both sides we get

100 = 73 + y² + 6y

On solving the equation, 100 = 73 + y² + 6y

⇒ 27 + y² + 6y = 0

⇒ y² + 6y + 27 = 0

⇒(y – 3)(y + 9) = 0

⇒ y = 3 or y = – 9

∴ The values of y can be 3 or – 9

Given the distance PQ = 10 units

By distance formula, as shown below:

PQ = √{(9 – x)² + (10 – 4)²}

10 = √{(9 – x)² + (6)²}

10 = √{81 + x² – 18x + 36}

10 = √{117 + x² – 18x}

Squaring both sides we get

⇒ 100 = 117 + x² – 18x

⇒ x² – 18x + 17x = 0

⇒ (x – 1)(x – 17)

⇒ × = 1 or × = 17

Given that point A is equidistant from points B and C , so AB = AC

By distance formula, as shown below:

AB = √{(8 – x)² + (– 2 – 2)²}

= √{(8 – x)² + (– 4)²}

= √{64 + x² – 16x + 16}

= √{80 + x² – 16x}

AC = √{(2 – x)² + (– 2 – 2)²}

= √{(2 – x)² + (4)²}

= √{4 + x² – 4x + 16}

= √{20 + x² – 4x}

Now, AB = AC

Squaring both sides, we get,

(80 + x² – 16x) = (20 + x² – 4x)

60 = 12x

x = 5

⇒ AB = √{80 + x² – 16x}

⇒ AB = √(80 + 5² – 16× 5)

= 5 units

Given that point A is equidistant from points B and C, so AB = AC

By distance formula, as shown below:

AB = √{(3 – 0)² + (p – 2)²}

= √{(3)² + (p – 2)²}

= √{9 + p² – 4p + 4}

⇒ AB = √{13 + p² – 4p}

AC = √{(p – 0)² + (5 – 2)²}

= √{(p)² + (3)²}

⇒ AB = √{9 + p²}

Now, AB = AC

Squaring both sides, we get,

(13 + p² – 4p) = (9 + p²)

⇒ 4 = 4p

⇒ p = 1

Now, AB = √{13 + p² – 4p}

⇒ AB = √(13 + 1 – 4)

= √10 units

Therefore, the distance of AB = √10 units.

Let the point be X(x,0) and the other two points are given as A(2, – 5) and B(– 2,9)

Given XA = XB

By distance formula, as shown below:

XA = √{(2 – x)² + (– 5 – 0)²}

= √{(2 – x)² + (– 5)²}

= √{4 + x² – 4x + 25}

⇒ XA = √{29 + x² – 4x}

XB = √{(– 2 – x)² + (9 – 0)²}

= √{(– 2 – x)² + (9)²}

= √{4 + x² + 4x + 81}

⇒ XB = √{85 + x² + 4x}

Now since

XA = XB

Squaring both sides, we get,

(29 + x² – 4x) = (85 + x² + 4x)

56 = – 8x

x = – 7

The point on × axis is (– 7, 0)

Let the point be X(x,0)

XA = 10

By distance formula, as shown below:

XA = √{(11 – x)² + (– 8 – 0)²}

10 = √{(11 – x)² + (– 8)²}

10 = √{121 + x² – 22x + 64}

10 = √{185 + x² – 22x}

Squaring both sides we get

100 = (185 + x² – 22x)

⇒ 85 + x² – 22x = 0

⇒ x² – 22x + 85 = 0

⇒ (x – 5)(x – 17)

⇒ × = 5 or × = 17

The points are (5, 0) and (17, 0)

Let the point be Y(0,y) and the other two points given as A(6,5) and B(– 4,3)

Given YA = YB

By distance formula, as shown below:

YA = √{(6 – 0)² + (5 – y)²}

= √{(6)² + (5 – y)²}

= √{36 + 25 + y² – 10y}

⇒ YA = √{61 + y² – 10y}

YB = √{(– 4 – 0)² + (3 – y)²}

= √{(– 4)² + (9 + y² – 6y)}

= √{16 + 9 + y² – 6y}

⇒ YB = √{25 + y² – 6y}

Now, YA = YB

Squaring both sides, we get,

(61 + y² – 10y) = (25 + y² – 6y)

36 = 4y

⇒ y = 9

The point is (0, 9)

The point P(x, y) is equidistant from the points A(5, 1) and B(– 1, 5), means PA = PB

By distance formula, as shown below:

PA = √{(5 – x)² + (1 – y)²}

= √{(25 + x² – 10x) + (1 + y² – 2y)}

⇒ PA = √{26 + x² – 10x + y² – 2y}

PB = √{(– 1 – x)² + (5 – y)²}

= √{(1 + x² + 2x + 25 + y² – 10y)}

⇒ PB = √{(26 + x² + 2x + y² – 10y)}

Now, PA = PB

Squaring both sides, we get

26 + x² – 10x + y² – 2y = 26 + x² + 2x + y² – 10y

⇒ 12x = 8y

⇒3x = 2y

Hence proved.

By distance formula, as shown below:

PA = √{(6 – x)² + (– 1 – y)²}

= √{(36 + x² –12x) + (1 + y² + 2y)}

⇒ PA = √{37 + x² – 12x + y² + 2y}

PB = √{(2 – x)² + (3 – y)²}

= √{(4 + x² – 4x + 9 + y² – 6y)}

⇒ PB = √{(13 + x² – 4x + y² – 6y)}

Given: PA = PB

Squaring both sides, we get

(37 + x² – 12x + y² + 2y) = (13 + x² – 4x + y² – 6y)

24 = 8x – 8y

Dividing by 8

x – y = 3

Hence proved.

Let the point be P(x,y), then since all three points are equidistant therefore

PA = PB = PC

By distance formula, as shown below:

We have, PA = √{(5 – x)² + (3 – y)²}

= √{25 + x² – 10x + 9 + y² – 6y}

⇒ PA = √{34 + x² – 10x + y² – 6y}

PB = √{(5 – x)² + (– 5 – y)²}

= √{25 + x² – 10x + 25 + y² + 10y}

⇒ PB = √{50 + x² – 10x + y² + 10y}

PC = √{(1 – x)² + (– 5 – y)²}

= √{1 + x² – 2x + 25 + y² + 10y}

⇒ PC = √{26 + x² – 2x + y² + 10y}

Squaring PA and PB we get

{34 + x² – 10x + y² – 6y} = {50 + x² – 10x + y² + 10y}

⇒ – 16 = 16y

⇒ y = – 1

Squaring PB and PC we get

{50 + x² – 2x + y² + 10y} = {26 + x² – 10x + y² + 10y}

24 = – 8x

x = – 3

P(– 3, – 1)

OA = √{(4 – 2)² + (3 – 3)²}

= √4

= 2

OB = √{(x – 2)² + 4 }

= √{x² + 4 – 4x + 4}

√{ 8 + x² – 4x}

OA² = OB²

4 = 8 + x² – 4x

⇒ x² – 4x + 4 = 0

⇒ x² – 2x – 2x + 4 = 0

⇒ x(x– 2) – 2(x – 2) = 0

⇒ (x – 2) (x – 2) = 0

x = 2

By distance formula

AC = √{(3 – (– 2))² + (– 1 – 3)²}

= √{(5)² + (– 4)²}

= √{25 + 16}

= √{41}

BC = √{(x –(– 2))² + (8 – 3)²}

= √{(x + 2)² + 5² }

= √{x² + 4 + 2x + 25}

= √{x² + 2x + 29}

AB = BC

√{x² + 2x + 29} = √{41}

× = 2 or × = – 6

Since, AB = BC

BC = √41 units

AP = BP

AP = √{(– 2 – 2)² + (k – 2)²}

= √{16 + k² – 4k + 4}

= √(k² – 2k + 20)

BP = √{(– 2k – 2)² + (– 3 – 2)²}

= √{4k² + 8k + 4 + 25}

= √(4k² + 8k + 29)

Squaring AP and BP and equating them we get

k² – 4k + 20 = 4k² + 8k + 29

3k² + 12k + 9 = 0

(k + 3)(k + 1) = 0

⇒ k = – 3

⇒ AP = √41units

Or k = – 1

⇒ AP = 5 units

Let point P(x,y) , A(a + b,a – b) , B(a – b,a + b)

Then AP = BP

AP = √{((a + b) – x)² + ((a – b) – y)²}

= √{(a + b)² + x² – 2(a + b)x + (a – b)² + y² – 2(a – b)y}

= √(a² + b² + 2ab + x² – 2(a + b)x + b² + a² – 2ab + y² – 2(a – b)y)

BP = √{((a – b) – x)² + ((a + b) – y)²}

= √{(a – b)² + x² – 2(a – b)x + (a + b)² + y² – 2(a + b)y}

= √(a² + b² – 2ab + x² – 2(a – b)x + b² + a² + 2ab + y² – 2(a + b)y)

Squaring and Equating both we get

a² + b² + 2ab + x² – 2(a + b)x + b² + a² – 2ab + y² – 2(a – b)y = a² + b² – 2ab + x² – 2(a – b)x + b² + a² + 2ab + y² – 2(a + b)y

– 2(a + b)x – 2(a – b)y = – 2(a – b)x – 2(a + b)y

ax + bx + ay – by = ax – bx + ay + by

Hence

bx = ay

Three or more points are collinear, if slope of any two pairs of points is same. With three points A, B and C if Slope of AB = slope of BC = slope of AC

then A, B and C are collinear points.

Slope of any two points is given by:

(y₂ – y₁)/(x₂ – x₁).

(i) Slope of AB = (2 – (– 1))/(5 – 1) = 3/4

Slope of BC = (5 – 2)/(9 – 5) = 3/4

Slope of AB = slope of BC

Hence collinear.

(ii) Slope of AB = (1 – 9)/(0 – 6) = 8/6 = 4/3

Slope of BC = (– 6 – 0)/(– 7 – 1) = 6/6 = 1

Slope of AC = (– 7 – 9)/(– 6 – 6) = – 16/ – 12 = 4/3

Slope of AB = slope of AC

Hence collinear.

(iii) Slope of AB = ((3 – (– 1))/((2 – (– 1)) = 4/3

Slope of BC = (11 – 2)/(8 – 3) = 9/5 = 1

Slope of AC = ((11 – (– 1))/((8 – (– 1)) = 12/9 = 4/3

Slope of AB = slope of AC

Hence collinear.

(iv) Slope of AB = (1 – 5)/((0 – (– 2)) = – 4/2 = – 2

Slope of BC = (– 3 – 1)/(2 – 0) = – 4/2 = – 2

Slope of AB = slope of AB

Hence collinear.

In an isosceles triangle any two sides are equal.

AB = √{(– 2 – 7)² + (5 – 10)²}

= √{(– 9)² + (– 5)²}

= √{81 + 25}

= √{106}

BC = √{(– 4 – 5)² + (3 – (– 2))²}

= √{(– 9)² + (5)²}

= √{81 + 25}

= √{106}

AB = BC

∴ It is an isosceles triangle.

In an isosceles triangle any two sides are equal.

AB = √{(6 – 3)² + (4 – 0)²}

= √{(3)² + (4)²}

= √{9 + 16}

= √{25} = 5 units

BC = √{(– 1 – 6)² + (3 – 4)²}

= √{(– 7)² + (– 1)²}

= √{49 + 1}

= √{50}

AC = √{(– 1 – 3)² + (3 – 0)²}

= √{(– 4)² + (3)²}

= √{16 + 9}

= √{25} = 5 units

AB = AC

∴ It is an isosceles triangle.

Given: A(5, 2), B(2, – 2) and C(– 2, t) are the vertices of a right triangle with ∠B = 90°
To find: The value of t.
Solution:


From the fig we have B = 90°,
so by Pythagoras theorem we have AC² = AB² + BC²

AC² = (– 2 – 5)² + (t – 2)²

= (– 7)² + t² + 4 – 2t
= 49 +t² + 4 - 2t

= 53 + t² – 2t

AB² = (2 – 5)² + (– 2 – 2)²
=(-3)² + (–4)²

= 9 + 16

= 25

BC² = (– 2 – 2)² + (t + 2)²
= (– 4)² + (t + 2)²

= 16 + t² + 4 + 2t

= 20 + t² + 2t

AB² + BC² = 25 + 20 + t² + 2t
= 45 + t² + 2t

AC² = 53 + t² – 2t

⇒ 53 + t² – 2t = 45 + t² + 2t

⇒ 8 = 4t
⇒ t = 2

For an equilateral triangle

AB = BC = AC

AB = √{(6 – 4)² + (2 – 2)²}

= √{(2)² + 0}

= √{4 + 0}

= √{4} = 2 units

BC = √{(2 + √3 – 2)² + (5 – 6)²}

= √{3 + (– 1)²}

= √{4} = 2 units

AC = √{(2 + √3 – 2)² + (5 – 4)²}

= √{3 + (– 1)²}

= √{4} = 2 units

Hence , AB = BC = AC

∴ ABC is an equilateral triangle.

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

Then, AB = √(3 + 3)²+( 3 + 3)²

=√(-6)²+(6)²

= √36+36

= √72

= 3√8

BC=√(-3√3+3)²+(3√3-3)²

= √(1-√3)²3²+(√3+1)²3²

= 3√[ 1+3-2√3+3+1+2√3]

= 3√8

CA = √(-3√3-3)²+(3√3-3)²

= √(-√3-1)²3²+(√3-1)²3²

= 3√[3+1+2√3+3+1-2√3]

=3√8

∵ AB = BC = CA

⇒ A, B, C are the vertices of an equilateral triangle.

AB = √{(0 – 6)² + (3 – (– 5))²}

= √{(– 6)² + (8)²}

= √{36 + 64}

= √{100} = 10 units

BC = √{(9 – 3)² + (8 – 0)²}

= √{(6)² + (8)²}

= √{36 + 64}

= √{100} = 10 units

AC = √{(9 – (– 5))² + (8 – 6)²}

= √{(14)² + (2)²}

= √{196 + 4}

= √{200}

For the right angled triangle

AC² = AB² + BC²

AC² = 200

AB² + AC² = 100 + 100 = 200

Since AB = BC

∴ ABC is an isosceles triangle.

Area = 1/2 (AB) (BC)

= 1/2 (10) (10)

= 1/2 (100)

= 50 sq units

OA = √{(√3)² + (3 – 0)²}

= √{(3) + (3)²}

= √{3 + 9}

= √{12}

AB = √{(– √3 – √3)² + (3 – 3)²}

= √{ – 2√3)²}

= √{12}

OB = √{(3 – 0)² + (– √3 – 0)²}

= √{9 + 3}

= √{12}

Since OA = AB = OB , ∴ equilateral triangle.

Area = 1/2 [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

= 1/2[ – 3√3 – 3√3 ]

= – 3√3 sq units

AB = √{(0 – 3)² + (5 – 2)²} = √{9 + 9} = √18 units

BC = √{(– 3 – 0)² + (2 – 5)²} = √{9 + 9} = √18 units

CD = √{(0 – (– 3))² + (– 1 – 2)²} = √{9 + 9} = √18 units

DA = √{(0 – 3)² + (– 1 – 2)²} = √{9 + 9} = √18 units

AC = √{(– 3 – 3)²} = √36 = 6 units

BD = √{(– 1 – 5)²} = √36 = 6 units

Since AB = BC = CD = DA and AC = BD

∴ ABCD is a square.

AB = √{(2 – 6)² + (1 – 2)²} = √{16 + 1} = √17 units

BC = √{(1 – 2)² + (5 – 1)²} = √{1 + 16} = √17 units

CD = √{(5 – 1)² + (6 – 5)²} = √{16 + 1} = √17 units

DA = √{(5 – 6)² + (6 – 2)²} = √{16 + 1} = √17 units

AC = √{(1 – 6)² + (5 – 2)²} = √{25 + 9} = √34 units

BD = √{(5 – 2)² + (6 – 1)²} = √{25 + 9} = √34units

Since AB = BC = CD = DA and AC = BD

∴ ABCD is a square.

AB = √{(3 – 0)² + (1 – (– 2))²} = √{9 + 9} = √18 units

BC = √{(0 – 3)² + (4 – 1)²} = √{9 + 9} = √18 units

CD = √{(– 3 – 0)² + (1 – 4)²} = √{9 + 9} = √18 units

DA = √{(– 3 – 0)² + (1 – (– 2))²} = √{9 + 9} = √18 units

AC = √{ (4 – (– 2))²} = √{36} = 6 units

BD = √{(– 3 – 3)² + (1 – 1)²} = √{36} = 6units

Since AB = BC = CD = DA and AC = BD

∴ ABCD is a square.

AC = √{(2 – (– 3))² + (– 32)²} = √{25 + 25} = √50 units

BD = √{(4 – (– 5))² + (4 – (– 5))²} = √{81 + 81} = √162 units

Area = 1/2× (product of diagonals)

= 1/2 × √50 × √162

= 45 sq units

AB = √{(4 – 3)² + (5 – 0)²} = √{1 + 25} = √26 units

BC = √{(– 1 – 4)² + (4 – 5)²} = √{25 + 1} = √26 units

CD = √{(– 2 – (– 1))² + (– 1 – 4)²} = √{1 + 25} = √26 units

DA = √{(– 2 – 3)² + (0 – 1)²} = √{25 + 1} = √26 units

AC = √{ (– 1 – 3)² + (4 – 0)²} = √{32}

BD = √{(– 2 – 4)² + (– 1 – 5)²} = √{36 + 36} = 6√2units

Since AB = BC = CD = DA

Hence, ABCD is a rhombus

Area = 1/2 × (product of diagonals)

= 1/2 × 4√2 × 6√2

= 24 sq units

AB = √{(8 – 6)² + (2 – 1)²} = √{4 + 1} = √5 units

BC = √{(9 – 8)² + (4 – 2)²} = √{1 + 4} = √5 units

CD = √{(7 – 9)² + (3 – 4)²} = √{4 + 1} = √5 units

DA = √{(7 – 6)² + (3 – 1)²} = √{1 + 4} = √5 units

AC = √{ (9 – 6)² + (4 – 1)²} = √(9 + 9) = 3√2 units

BD = √{(7 – 8)² + (3 – 2)²} = √{1 + 1} = √2 units

Since AB = BC = CD = DA

Hence, ABCD is a rhombus

Area = 1/2 × (product of diagonals)

= 1/2 × 3√2 × √2

= 3 sq units

AB = √{(5 – 2)² + (2 – 1)²} = √{9 + 1} = √10 units

BC = √{(6 – 5)² + (4 – 2)²} = √{1 + 4} = √5 units

CD = √{(3 – 6)² + (3 – 4)²} = √{9 + 1} = √10 units

DA = √{(3 – 2)² + (3 – 1)²} = √{1 + 4} = √5 units

Since AB = CD and BC = DA

∴ ABCD is Parallelogram

AC = √{(6 – 2)² + (4 – 1)²} = √{16 + 9} = 5 units

For a Rectangle

AC² = AB² + BC²

Here AC² = 25

But AB² + BC² = 15

∴ ABCD is not a rectangle

AB = √{(4 – 1)² + (3 – 2)²} = √{9 + 1} = √10 units

BC = √{(6 – 4)² + (6 – 3)²} = √{4 + 9} = √13 units

CD = √{(6 – 3)² + (5 – 6)²} = √{9 + 1} = √10 units

DA = √{(3 – 1)² + (5 – 2)²} = √{4 + 9} = √13 units

AB = CD and BC = DA

∴ ABCD is a parallelogram ∴

AC = √{(6 – 1)² + (6 – 2)²} = √{25 + 16} = √41 units

For a Rectangle

AC² = AB² + BC²

Here AC² = 41

But AB² + BC² = 23

∴ ABCD is not a rectangle

A(– 4, – 1), B(– 2, – 4), C(4, 0) and D(2, 3)

AB = √{(– 2 – (– 4))² + (– 4 – (– 1))²}

= √{4 + 9} = √13units

BC = √{(4 – (– 2))² + (0 – (– 4))²}

= √{36 + 16} = √52units

CD = √{(2 – 4)² + (3 – 0)²}

= √{4 + 9} = √13 units

DA = √{(2 – (– 4))² + (3 – (– 1))²}

= √{36 + 16} = √52units

AB = CD and BC = DA

AC = √{(4 – (– 4))² + (0 – (– 1))²}

= √{64 + 1} = √65 units

For a Rectangle

AC² = AB² + BC²

Here AC² = 65

But AB² + BC² = 13 + 52 = 65

∴ ABCD is a rectangle

AB = √{(14 – 2)² + (10 – (– 2))²}

= √{144 + 144} = √288

BC = √{(11 – 14)² + (10 – 13)²}

= √{9 + 9} = √18 units

CD = √{(– 1 – 11)² + (1 – 13)²}

= √{144 + 144}

= √288 units

DA = √{(– 1 – 2)² + (1 – (– 2))²}

= √{9 + 9} = √18units

AB = CD and BC = DA

AC = √{(11 – 2)² + (13 – (– 2))²}

= √{81 + 225}

= √306 units

For a Rectangle

AC² = AB² + BC²

Here AC² = 306

But AB² + BC² = 288 + 18 = 306

∴ ABCD is a rectangle

AB = √{(6 – 0)² + (2 – (– 4))²}

= √{36 + 36}

= √72units

BC = √{(3 – 6)² + (5 – 2)²}

= √{9 + 9}

= √18units

CD = √{(3 – (– 3))² + (– 1 – 5)²}

= √{36 + 36}

= √72 units

DA = √{(– 3 – 0)² + (– 1 – (– 4))²}

= √{9 + 9}

= √18units

AB = CD and BC = DA

AC = √{(3 – 0)² + (5 – (– 4))²}

= √{9 + 81}

= √90 units

For a Rectangle

AC² = AB² + BC²

Here AC² = 90

But AB² + BC² = 72 + 18 = 90

∴ ABCD is a rectangle

Let the point P(x,y) divides AB

Then

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (2 × 4 + 3 × (– 1))/2 + 3

= (8 – 3) /5

= 5/5 = 1

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (2 × (– 3) + 3 × 7)/ 5

= (– 6 + 21)/5

= 15 / 5 = 3

= (1, 3)

Let the point P(x,y) divides AB

Then

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (7 × 4 + 2 × (– 5))/7 + 2

= (28 – 10) /9

= 18/9 = 2

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (7 × (– 7) + 2 × 11)/ 9

= (– 49 + 22)/9

= – 27 / 9 = – 3

= (2, – 3)

Let the point P(x,y) divides AB

Then

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (3 × 2) + 4x (– 2))/ 3 + 4

= (6 – 8)/7

= – 2/7

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (3 × (– 4) + 4 × (– 2))/ 7

= (– 12 – 8)/ 7

= – 20 / 7

Let the point P(x,y) divides AB

Then

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (2 × (– 4) + 3 × 6)/2 + 3

= (– 8 + 18) /5

= 10/5 = 2

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (2 × (– 1) + 3 x( – 6))/ 5

= (– 2 – 18)/5

= – 20 / 5 = – 4

If the point A also lies on the line 3x + k(y + 1) = 0

Then

3 × 2 + k(– 4 + 1) = 0

6 – 3k = 0

6 = 3k

k = 2

P divides the segment AB in ratio 1:4

Q divides the segment AB in ratio 2:3

R divides the segment AB in ratio 3:2

For coordinates of P

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × 6 + 4 × 1)/1 + 4

= (6 + 4) /5

= 10/5 = 2

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1x 7 + 4 × 2)/5

= (7 + 8)/5

= 15 / 5 = 3

= (2, 3)

For coordinates of Q

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (2x 6 + 3x 1)/5

= (12 + 3) /5

= 15/5 = 3

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (2 × 7 + 3 × 2)/ 5

= (14 + 6)/5

= 20 / 5 = 4

= (3,4)

For coordinates of R

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (3 × 6 + 2 × 1)/5

= (18 + 2) /5

= 20/5 = 4

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (3 × 7 + 2 × 2)/ 5

= (21 + 4)/5

= 25 / 5 = 5

= (4,5)

Hence

P(2, 3), Q(3, 4), R(4, 5)

P divides the segment AB in ratio 1:3

Q divides the segment AB in ratio 2:2

R divides the segment AB in ratio 3:1

For coordinates of P

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × 5 + 3 × 1)/1 + 3

= (5 + 3) /4

= 8/4 = 2

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1× (– 2) + 3 × 6)/4

= (– 2 + 18)/5

= 16 / 4 = 4

= (2, 4)

For coordinates of Q

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (2x 5 + 2x 1)/4

= (10 + 2) /4

= 12/4 = 3

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (2 × (– 2) + 2 × 6)/ 4

= (– 4 + 12)/4

= 8 / 4 = 2

= (3,2)

For coordinates of R

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (3x 5 + 1x 1)/4

= (15 + 1) /4

= 16/4 = 4

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (3 × (– 2) + 1 × 6)/ 4

= (– 6 + 6)/4

= 0/ 4 = 0

= (4,0)

∴ the coordinates are P(2, 4), Q(3, 2), R (4, 0)

P divides the segment AB in ratio 1:2

Q divides the segment AB in ratio 2:1

For coordinates of P

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × 1 + 2 × 3)/1 + 2

= (1 + 6) /3

= 7/3 = p

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1x2 + 2 × (– 4))/3

= (2 – 8)/3

= – 6/ 3 = – 2

For coordinates of Q

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (2x 1 + 1x 3)/3

= (2 + 3) /3

= 5/3

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (2 × 2 + 1 × (– 4))/3

= (4 – 4)/3

= 0/ 3

= 0 = q

p = 7/3 , q = 0

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × (– 5) + 1x 3)/1 + 1

= (– 5 + 3) /2

= – 2/2 = – 1

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1x 4 + 1x 0)/2

= (4 + 0)/2

= 4 / 2 = 2

(– 1, 2)

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × 8 + 1x (– 11)/1 + 1

= (8 – 11) /2

= – 3/2

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1x (– 2) + 1x – 8)/2

= (– 2 – 8)/2

= – 10 / 2 = – 5

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × (– 2) + 1x 6)/1 + 1

= (– 2 + 6) /2

= 4/2 = 2

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1x 11 + 1x (– 5))/2

= (11 – 5)/2

= 6 / 2 = 3

p = 3

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × (– 2) + 1× 2a)/1 + 1

= (– 2 + 2a) /2

(– 2 + 2a)/2 = 1

– 2 + 2a = 2

2a = 4

a = 2

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1 × 3b + 1 ×4)/2

= (3b + 4)/2

(3b + 4)/2 = 2a + 1

(3b + 4)/2 = 5

(3b + 4) = 10

3b = 6

b = 2

a = 2, b = 2

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × 6 + 1x (– 2)/1 + 1

= (6 – 2) /2

= 4/2 = 2

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1x 3 + 1x 9)/2

= (3 + 9)/2

= 12 / 2 = 6

C(2,6)

Let the coordinates of A be × & y. So A(X,Y) and B(1,4)

2 = (m₁x₂ + m₂x₁)/ m₁ + m₂

2 = (1 × 1 + 1 × X)/1 + 1

2 = (1 + X) /2

1 + X = 4

× = 3

– 3 = (m₁y₂ + m₂y₁)/ m₁ + m₂

– 3 = (1× 4 + 1 × Y)/2

– 3 = (4 + Y)/2

(4 + Y) = – 6

Y = – 10

A(3, – 10)

2 = (m₁x₂ + m₂x₁)/ m₁ + m₂

2 = (m₁ × (– 6) + m₂ 8)/ m₁ + m₂

2 = (– 6m₁ + 8m₂ ) / m₁ + m₂

– 6m₁ + 8m₂ = 2(m₁ + m₂ )

– 8m₁ + 6m₂ = 0

5 = (m₁y₂ + m₂y₁)/ m₁ + m₂

5 = (m₁ × 9 + m₂ 2)/ m₁ + m₂

5 = (9m₁ + 2m₂ ) / m₁ + m₂

9m₁ + 2m₂ = 5(m₁ + m₂)

4m₁ + 3m₂ = 0

Solving for m₁ and m₂ we get

m₁ = 3

m₂ = 4

3:4

3/4 = (m₁x₂ + m₂x₁)/ m₁ + m₂

3/4 = (m₁ × 2 + m₂ (1/2))/ m₁ + m₂

3/4 = (2m₁ + m₂ /2) / m₁ + m₂

6m₁ + 6m₂ = 16m₁ + 4m₂

6m₁ – 2m₂ = 0

5/12 = (m₁y₂ + m₂y₁)/ m₁ + m₂

5/12 = (m₁ × (– 5) + m₂ (3/2))/ m₁ + m₂

5/12 = (– 5m₁ + 3m₂ /2) / m₁ + m₂

– 120m₁ + 36m₂ = 10(m₁ + m₂)

130m₁ – 26m₂ = 0

Solving for m₁ and m₂ we get

m₁ = 1

m₂ = 5

1:5

6 = (m₁y₂ + m₂y₁)/ m₁ + m₂

6 = (m₁ × 8 + m₂ 3)/ m₁ + m₂

6 = (8m₁ + 3m₂ ) / m₁ + m₂

8m₁ + 8m₂ = 6(m₁ + m₂)

2m₁ – 3m₂ = 0

m₁:m₂ = 3:2

Now,

m = (m₁x₂ + m₂x₁)/ m₁ + m₂

m = (m₁ × 2 + m₂ (– 4))/ m₁ + m₂

m = (2m₁ – 4m₂ ) / m₁ + m₂

2m₁ – 4m₂ = m(m₁ + m₂ )

Putting the values of m₁ & m₂

m = – 2/5

Hence, 3:2, m = – 2/5

– 3 = (m₁x₂ + m₂x₁)/ m₁ + m₂

– 3 = (m₁ × (– 2) + m₂ (– 5))/ m₁ + m₂

– 3 = (– 2m₁ – 5m₂ ) / m₁ + m₂

– 2m₁ – 5m₂ = – 3(m₁ + m₂ )

2m₁ + 5m₂ = 3(m₁ + m₂ )

m₁ – 2m₂ = 0

m₁:m₂ = 1:2

Now,

K = (m₁y₂ + m₂y₁)/ m₁ + m₂

K = (m₁ × 3 + m₂(– 4))/ m₁ + m₂

K = (3m₁ – 4m₂ ) / m₁ + m₂

3m₁ – 4m₂ = k(m₁ + m₂)

Putting the values of m₁ & m₂

k = 2/3

Hence,2:1, k = 2/3

The segment is divided by x – axis i.e the coordinates are (x,0)

x = (m₁x₂ + m₂x₁)/ m₁ + m₂

x = (m₁ × 5 + m₂ 2)/ m₁ + m₂

x = (5m₁ + 2m₂ ) / m₁ + m₂

5m₁ + 2m₂ = x(m₁ + m₂ )

(5 – x)m₁ + (2 – x)m₂ = 0

0 = (m₁y₂ + m₂y₁)/ m₁ + m₂

0 = (m₁ × 6 + m₂(– 3))/ m₁ + m₂

0 = (6m₁ – 3m₂ ) / m₁ + m₂

6m₁ – 3m₂ = 0

Solving for m₁ and m₂ we get

m₁ = 1

m₂ = 2

(1 : 2),

Putting the values of m₁ and m₂

x = 3

Hence coordinates are (3,0)

The segment is divided by y – axis i.e the coordinates are (0,y)

0 = (m₁x₂ + m₂x₁)/ m₁ + m₂

0 = (m₁ × 3 + m₂ (– 2))/ m₁ + m₂

0 = (3m₁ – 2m₂ ) / m₁ + m₂

3m₁ – 2m₂ = 0

m₁ = 2

m₂ = 3

(2:3)

y = (m₁y₂ + m₂y₁)/ m₁ + m₂

y = (m₁ × 7 + m₂(– 3))/ m₁ + m₂

y = (7m₁ – 3m₂ ) / m₁ + m₂

7m₁ – 3m₂ = y(m₁ + m₂)

Putting the values of m₁ and m₂

y = 1

The line segment joining any two points (x₁, y₁) and (x₂, y₂) y₂ is given as:

⇒

⇒ y + 1 = 10/5 (x-3)

⇒ y + 1 = 2(x-3)

⇒ y + 1 = 2x – 6
⇒ 2x – y = 7..eq(1) is the equation of line segment.

Now, we have to find the point of intersection of eq (1) & the given line: x – y- 2 = 0

2x – y = 7

& x – y – 2 = 0

2x – 7 = x – 2

⇒ x = 7- 2

⇒ x = 5

And, y = 3

Let us say this point divides the line segment in the ratio of k₁:k₂

Then,

⇒ 5k₁ + 5k₂ = 8k₁ + 3k₂

⇒ 5k₁ - 8k₁ + 5k₂ - 3k₂= 0

⇒ -3k₁ + 2k₂ = 0

⇒

For coordinates of median AD segment BC will be taken

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × 0 + 1x 2)/1 + 1

= (0 + 2) /2

= 2/2 = 1

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1x 3 + 1x 1)/2

= (3 + 1)/2

= 4 / 2 = 2

D(1,2)

By distance Formula

AD = √(1 – 0)² + (2 + 1)²

= √1 + 9

= √10

For coordinates of BE, segment AC will be taken

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × 0 + 1x 0)/1 + 1

= (0 + 0) /2

= 0/2 = 0

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1x 3 + 1x (– 1))/2

= (3 – 1)/2

= 2 / 2 = 1

∴ E(0,1)

By distance Formula

BE = √(0 – 2)² + (1 – 1)²

= √4 + 0

= √4 = 2

For coordinates of median CF segment AB will be taken

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × 2 + 1x 0)/1 + 1

= (2 + 0) /2

= 2/2 = 1

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1x(– 1) + 1x 1)/2

= (– 1 + 1)/2

= 0 / 2 = 0

F(1,0)

By distance Formula

CF = √(1 – 0)² + (0 – 3)²

= √1 + 9

= √10

AD = √10 units, BE = 2 units, CF = √10 units

First we need to calculate the coordinates of median

For coordinates of median AD segment BC will be taken

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (1 × 8 + 1x 5)/1 + 1

= (8 + 5) /2

= 13/2

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (1x 2 + 1x (– 2))/2

= (0)/2

= 0 / 2 = 0

D(13/2,0)

The centroid of the triangle divides the median in the ratio 2:1

By section formula,

X = (m₁x₂ + m₂x₁)/ m₁ + m₂

= (2 × 13/2 + 1x (– 1))/2 + 1

= (13 – 1) /3

= 12/3 = 4

Y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (2x 0 + 1x 0)/2 + 1

= 0/3

= 0

∴ G coordinate is (4, 0)

The figure is shonw as:

– 2 = (m₁x₂ + m₂x₁)/ m₁ + m₂

– 2 = (2 × x + 1x 1)/2 + 1

– 2 = (2x + 1) /3

– 6 = 2x + 1

– 7 = 2x

⇒ x = – 7/2

1 = (m₁y₂ + m₂y₁)/ m₁ + m₂

1 = (2x y + 1x (– 6))/3

1 = (2y – 6)/2

2 = 2y – 6

8 = 2y

⇒ y = 4

D(– 7/2,4)

Now for BC

– 7/2 = (m₁x₂ + m₂x₁)/ m₁ + m₂

– 7/2 = (1 × x + 1x (– 5))/1 + 1

– 7/2 = (x – 5) /2

– 7 = x – 5

– 7 + 5 = x

⇒ x = – 2

4 = (m₁y₂ + m₂y₁)/ m₁ + m₂

4 = (1 × y + 1x 2)/2

4 = (y + 2)/2

8 = y + 2

⇒ y = 6

Hence, C(– 2, 6)

Coordinate of D on median on BC

x = (m₁x₂ + m₂x₁)/ m₁ + m₂

x = (1 × 0 + 1x (– 3))/1 + 1

x = (0 – 3) /2

x = – 3/2

y = (m₁y₂ + m₂y₁)/ m₁ + m₂

y = (1 × (– 2) + 1x 1)/2

y = (– 2 + 1)/2

2y = – 1

y = – 1/2

D(– 3/2, – 1/2)

Now for AD we have D(– 3/2, – 1/2) and Centroid C(0,0)

0 = (m₁x₂ + m₂x₁)/ m₁ + m₂

0 = (2 × (– 3/2) + 1x x)/2 + 1

0 = (– 3 + x) /3

– 3 + x = 0

x = 3

0 = (m₁y₂ + m₂y₁)/ m₁ + m₂

0 = (2 × (– 1/2) + 1x y)/2 + 1

0 = (– 1 + y)/3

– 1 + y = 0

y = 1

Hence, A(3, 1)

We know that if diagonals of a quadrilateral bisect each other, then the quadrilateral is parallelogram

So,
If ABCD is a parallelogram,
the coordinates of the mid-point of the AC = Coordinates of the mid-point of the BD

Coordinates of mid-point of AC

Coordinates of mid-point of BD

= (2, 1)

Hence, ABCD is a parallelogram.

We know that the diagonals of a parallelogram bisect each other

So the coordinates of the mid – point of the PR = Coordinates of the mid – point of the QS

{(2 + a)/2,(15 – 11)/2} = {(5 + 1)/2,(b + 1)/2}

2 + a = 6

a = 4

15 – 11 = b + 1

4 = b + 1

b = 3

Hence, a = 4, b = 3

Coordinate of mid – point of AC = {(1 + 5)/2,(– 2 + 10)/2}

implies (3,4)

This is equal to the coordinates of mid – point of BD

3 = (3 + x)/2

6 = 3 + x

x = 3

4 = (6 + y)/2

8 = (6 + y)

y = 2

Hence, D(3, 2)

Let the coordinate of the point on y axis be (0,y)

0 = (m₁x₂ + m₂x₁)/ m₁ + m₂

0 = (m₁3 + m₂(– 4))/ m₁ + m₂

0 = (3m₁ – 4m₂)/ m₁ + m₂

(3m₁ – 4m₂) = 0

3m₁ = 4m₂

m₁: m₂ = 4:3

Given: The points P(1/2, y) lies on the line AB.

Then,

1/2 = (m₁x₂ + m₂x₁)/ m₁ + m₂

1/2 = (m₁(– 7) + m₂3)/ m₁ + m₂

1/2 = (– 7m₁ + 3m₂)/ m₁ + m₂

(m₁ + m₂) = – 14 m₁ + 6 m₂

15m₁ = 5m₂

m₁: m₂ = 3:5

y = (m₁y₂ + m₂y₁)/ m₁ + m₂

y = (3 × 9 + 5x (– 5))/3 + 5

y = (27 – 25)/8

y = 2/8

y = 1/4

Let the coordinate of the point on x axis be (x,0)

0 = (m₁y₂ + m₂y₁)/ m₁ + m₂

0 = (m₁7 + m2(– 3)/ m₁ + m₂

0 = (7m₁ – 3m₂)/ m₁ + m₂

7m₁ – 3m₂ = 0

7 m₁ = 3m₂

m₁ : m₂ = 3:7

x = (m₁x₂ + m₂x₁)/ m₁ + m₂

x = (3 x(– 2) + 7 × 3)/ 10

× = (– 6 + 21)/ 10

x = 15/10

x = 3/2

Hence the coordinate of the point be (3/2, 0)

Let QR be the base

Since origin is mid – point O(0,0) of QR

Then the coordinates of R(x,y) is given by

(– 4 + x)/2 = 0

x = 4

(0 + y)/2 = 0

y = 0

R(4,0)

Distance of QR = √(4 + 4)² + 0

QR = 8

∴ PR = 8

Let P(x,y)

8 = √(4 – x)² + (0 – y)²

64 = 16 + x² – 8x + y²

Since it will lie on x axis

∴ × = 0

64 = 16 + y²

48 = y²

y = 4√3 or – 4√3

Hence,

P(0, 4√3) or P(0, – 4√3) and R(4, 0)

Given: The base (BC) of the equilateral triangle ABC lies on y - axis, where, C has the coordinates: (0, - 3).The origin is the midpoint of the base.

Now, Δ ABC is an equilateral triangle

∴ AB = AC = BC …(1)

By symmetry the coordinate A lies on x axis.

Also D is another point such that ABCD is rhombus and every side of rhombus is equal to each other.

So For this condition to be possible D will also lie on x axis.

Now,

Let coordinates of A be (x,0),B be (0,y) and D be ( -x,0).

The figures are shown below:

BC = √(0 - 0)² + ( - 3 - y)²

⇒ BC = √0 + 9 + y² + 6y

⇒ BC = √9 + y² + 6y

Now, AC = √(0 - x)² + ( - 3 - 0)²

⇒ AC = √x² + ( - 3)²


⇒ AC = √(x² + 9)

And

AB = √(0 - x)² + (y - 0)²

⇒ AB = √x² + y²


From (1)

AB = AC

⇒ √x² + y² = √x² + 9

Taking square on both sides we get,

x² + y² = x² + 9

⇒ y² = 9

⇒ y = ± 3

Since B lies in positive y direction.

∴ The coordinates of B are (0,3)

Now from (1) AB = BC

⇒√ x² + y² = √9 + y² + 6y

Take square on both sides

⇒ x² + y² = 9 + y² + 6y

⇒ x² = 9 + 6y

Put the value of y to get,

⇒ x² = 9 + 6(3)

⇒ x² = 9 + 18

⇒ x² = 27

⇒ x = ± 3√3

Hence the coordinates of A can be ( 3√3,0) or ( - 3√3,0)

Also, ABCD is a rhombus.

⇒ AB = BC = DC = BD

So coordinates of D will be ( -3√3,0) or ( 3√3,0)

Hence coordinates are A( 3√3,0) , B(0,3) , D( - 3√3,0)

Or coordinates are A( -3√3,0) , B(0,3) , D( 3√3,0)

– 1 = (m₁x₂ + m₂x₁)/ m₁ + m₂

– 1 = (m₁6 + m₂(– 3))/ m₁ + m₂

– 1 = (6m₁ – 3m₂)/ m₁ + m₂

(6m₁ – 3m₂) = – m₁ – m₂

7m₁ = 2 m₂

m₁: m₂ = 2:7

y = (m₁y₂ + m₂y₁)/ m₁ + m₂

= (2x(– 8) + 7 × 10)/9

= (– 16 + 70)/9

= 54 / 9

y = 6

The figure is shown below:

P(x,y) = (– 1 – 1)/2 , (4 – 1)/2

= (– 1,3/2)

Q(x,y) = (5 – 1)/2 , (4 + 4)/2

= (2,4)

R(x,y) = (5 + 5)/2 , (– 1 + 4)/2

= (5,3/2)

S(x,y) = (5 – 1)/2 , (– 1 – 1)/2

= (2, – 1)

Coordinates of mid – point of PR = Coordinates of mid – point of QS

Coordinates of mid – point of PR = {(5 – 1)/2 ,(3/2 + 3/2)/2} = (2,3/2)

Coordinates of mid – point of QS = {(2 + 2)/2 , (– 1 + 4)/2 = (2,3/2)

Hence PQRS is a Rhombus.

For P(x,y)

X = (– 10 – 2)/2 = – 6

Y = (4 + 0)/2 = 2

Thus, P(– 6,2)

Now

– 6 = (m₁x₂ + m₂x₁)/ m₁ + m₂

– 6 = (m₁(– 4) + m₂(– 9))/ m₁ + m₂

– 6 = (– 4m₁ – 9m₂)/ m₁ + m₂

– 6(m₁ + m₂) = – 4 m₁ – 9 m₂

– 2m₁ = – 3m₂

m₁:m₂ = 3:2,

2 = (m₁y₂ + m₂y₁)/ m₁ + m₂

2 = (3 × y + 2x (– 4))/5

2 = (3y – 8)/5

10 = 3y – 8

3y = 18

y = 6

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(1(–2 + 3)–2(–4–2)–3(2–3))

= 1/2(1 + 12 + 3)

= 8 sq units

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(–5(–5–5)–4(5–7) + 4(7 + 5))

= 1/2(–50 + 8 + 48)

= 5 sq units

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(3(2 + 1)–4(–1–8) + 5(8–2))

= 1/2(9 + 36 + 30)

= 1/2(75)

= 37.5 sq units

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(10(5–3) + 2(3 + 6)–1(–6–5))

= 1/2(20 + 18 + 11)

= 1/2(49)

= 24.5 sq units

For triangle ABC

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(3(–5–0) + 9(0 + 1) + 14(–1 + 5))

= 1/2(–15 + 9 + 56)

= 1/2(50)

= 25

For triangle ACD

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(3(0–19) + 14(19 + 1) + 9(–1–0))

= 1/2(–57 + 280–9)

= 1/2(214)

= 107

Area of ABCD = Area of ABC + Area of ACD

= 25 + 107

= 132 sq units

For triangle PQR

Area of triangle

= l1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))l

= 1/2(–5(–6 + 3)–4(–3 + 3) + 2(–3 + 6))

= 1/2(15 + 0 + 6)

= 1/2(21)

For triangle PRS

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(–5(–3–2) + 2(2–(–3)) + 1(–3 + 3))

= 1/2(25 + 10 + 0)

= 1/2(35)

Area of ABCD = Area of ABC + Area of ACD

= 21/2 + 35/ 2

= 28 sq units

We divide quadrilateral in two triangles, such that
Area of ABCD = Area of ΔABC + Area of ΔACD

We know area of a triangle, if it’s coordinates are A(x­₁, y₁), B(x₂, y₂) and C(x₃, y₃) is

Area of ACD

Area of ABCD = Area of ABC + Area of ACD

= 28 sq units

For triangle ABC

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(–5(–5 + 6)–4(–6–7)–1(7 + 5))

= 1/2(–5 + 52–12)

= 1/2(35)

For triangle ACD

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(–5(–6–5)–1(5–7) + 4(7 + 6))

= 1/2(–55 + 2 + 52)

= 1/2(1)

Area of ABCD = Area of ABC + Area of ACD

= 18 sq units

By applying section formula we get the coordinates of mid points of AB,BC and AC.

Mid point of AB = P = {(2 + 4)/2,(1 + 3)/2}

P = (3,2)

Mid point of BC = Q = {(4 + 2)/2,(3 + 5)/2}

Q = (3,4)

Mid point of AC = R = {(2 + 2)/2,(1 + 5)/2}

R = (2,3)

For triangle PQR

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(3(4–3) + 3(3–2) + 2(2–4))

= 1/2(3 + 3–4)

= 1/2(2)

= 1 sq unit

D = {(3 + 5)/2,(3–1)/2} = (4,1)

For triangle ABD

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(7(3–1) + 5(1 + 3) + 4(–3–3))

= 1/2(14 + 20–24)

= 1/2(10)

= 5 sq unit

For triangle ACD

Area of triangle

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(7(–1–1) + 3(1 + 3) + 4(–3 + 1))

= 1/2(–14 + 12–8)

= 1/2(10)

= 5 sq unit

Hence area of triangle ABD and ACD is equal.

The diagram is given below:

Coordinates of B

2 = (1 + x)/2 [by section formula]

4 = 1 + x

X = 3

–1 = (–4 + y)/2

–2 = (–4 + y)

Y = 2

∴ the coordinates of B(3,2)

Coordinates of C [by section formula]

0 = (1 + x)/2

0 = (1 + x)

x = –1

–1 = (–4 + y)/2

–2 = (–4 + y)

Y = 2

∴ the coordinates of point C are (–1,2)

Now, Area of triangle ABC

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(1(2–2) + 3(2 + 4)–1(–4–2))

= 1/2(0 + 18 + 6)

= 1/2(24)

= 12 sq unit

Let (x, y) be the coordinates of D and ( x^’, y' ) be the coordinates of E. since the diagonals of a parallelogram bisect each other at the same point, therefore

(x + 8)/2 = (6 + 9)/2

X = 7

(y + 2)/2 = (1 + 4)/2

Y = 3

Thus, the coordinates of D are (7,3)

E is the midpoint of DC,

therefore

x^’ = (7 + 9)/2 = 8

y^’ = (3 + 4)/2 = 7/2

Thus, the coordinates of E are ( 8,7/ 2)

Let A(x₁,y₁) = A(6,1), E(x₂,y₂) = (8,7/2) and D(x₃,y₃) = D(7,3)

Now Area

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(6(7/2–3) + 8(3–1) + 7(1–7/2))

= 1/2(3/2)

= 3/4 sq unit

Hence, the area of the triangle ΔADE is 3/4 sq. units.

Area = 15

⇒ Δ = 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

15 = 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

15 = 1/2(1(p–7) + 4(7 + 3)–9(–3–p))

15 = 1/2(10p + 16)

|10p + 16| = 30

10p + 16 = 30 or –30

Hence, p = −9 or p = −3.

Δ = 6

⇒ Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

6 = 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

6 = 1/2(k + 1(–3 + k) + 4(–k–1) + 7(1 + 3))

6 = 1/2(k²–2k–3–4k–4 + 28)

k² – 6k + 9 = 0

k = 3

Given the area of triangle, Δ = 53

⇒ Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

53 = 1/2{–2(–4–10) + k(10–5) + 2k + 1(5 + 4)}

53 = 1/2{28 + 5k + 9(2k + 1)}

106 = (28 + 5k + 18k + 9)

37 + 3k = 106

23k = 69

k = 3

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Given, the area of the triangle, Δ = 0

⇒ Δ = 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

⇒ Δ = 1/2{2 (8– 4) + (–3) (4 + 2) –1 (2– 8)}

⇒ Δ = 1/2 {8–18 + 10}

⇒ Δ = 0

Hence the points A(2, – 2), B(–3, 8) and C(–1, 4) are collinear.

To show that the points are collinear, we show that the area of triangle is equilateral = 0

⇒ Δ = 1/2 {(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))}

⇒ Δ = 1/2{–5(5– 7) + 5 (7–1) + 10 (1–5) }

⇒ Δ = 1/2{10 + 30–40}

⇒ Δ = 0

Hence collinear.

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

⇒ Δ = 1/2{5(–1– 4) + 1 (4– 1) + 11 (1 + 1)}

⇒ 1/2{–25 + 3 + 22}

= 0

Hence collinear

A(8, 1), B(3, –4) and C(2, –5)

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

⇒ 1/2{8(–4 + 5) + 3 (–5–1) + 2 (1 + 4)}

⇒ 1/2{8–18 + 10}

= 0

Hence collinear.

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2 {x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

⇒ Δ = 1/2{x(–4 + 5)–3 (–5– 2) + 7 (2 + 4)} = 0

⇒ Δ = 1/2{x + 21 + 42} = 0

x = –63

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

⇒ Δ = 1/2{–3(6–9) + 7 (9 – 12) + x(12– 6)} = 0

⇒ (–3)(–3) + 7(–3) + 6x = 0

⇒ 9–21 + 6x = 0

6x = 12

x = 2

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

⇒ Δ = 1/2{1(y–16) + 3 (16–4)–3 (4–y)} = 0

⇒ y–16 + 36–12 + 3y = 0

⇒ 8 + 4y = 0

⇒ 4y = –8

y = –2

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

⇒ Δ = 1/2{–3(y + 5) + 2 (–5–9) + 4 (9–y)} = 0

⇒ –3y–15–28 + 36–4y = 0

⇒ 7y = 36–43

y = –1

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

⇒ Δ = 1/2{8(–2k + 5) + 3 (–5–1) + k (1 + 2k)} = 0

⇒ –16k + 40–18 + k + 2k² = 0

⇒ 2k² + 15k + 22 = 0

⇒ 2k²–11k–14k + 22 = 0

⇒ K(2k–11)–2(2k–11) = 0

k = 2 or k =

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

⇒ Δ = 1/2{2(y–5) + x (5– 1) + 7 (1– y)}

⇒ 2y–10 + 4x–7–7y = 0

⇒ 4x –5y – 3 = 0

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

⇒ Δ = 1/2{x (7–5) + (–5) (–5–y) –4 (y– 7)}

⇒ 7x–5x–25 + 5y–4y + 28 = 0

⇒ 2x + y + 3 = 0

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)} = 0

⇒ Δ = 1/2{a(b– 1) + 0 (1– 0) + 1 (0– b)} = 0

⇒ (ab–a–b) = 0

Dividing the equation by ab.

1–1/b–1/a

1–(1/a + 1/b)

1–1 = 0

Hence collinear.

To show that the points are collinear, we show that the area of triangle is equilateral = 0

Δ = 0

Δ = 1/2{x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)}

⇒ Δ = 1/2{–3 (b + 5) + a (–5–9) + 4 (9–b)} = 0

⇒ –3b–150–14a + 36–4b = 0

2a + b = 3

Now solving a + b =1 and 2a + b = 3 we get a = 2 and b = −1.

Hence a = 2, b = –1

The distance of any point which lies on the circumference of the circle from the centre of the circle is called radius.

∴ OA = OB = Radius of given Circle

taking square on both sides, we get-

OA² = OB²

⇒ (-1-2)² + [y-(-3y)]² = (5-2)² + [7-(-3y)]²

[using distance formula, the distance between points (x₁,y₁) and (x₂,y₂) is equal to units.]

⇒ 9 + 16y² = 9 + (7 + 3y)²

⇒ 16y² = 49 + 42y + 9y²

⇒ 7y² - 42y - 49 = 0

⇒ 7(y²-6y-7) = 0

⇒ y²-7y + y-7 = 0

⇒ y(y-7) + 1(y-7) = 0

⇒ (y + 1)(y-7) = 0

∴ y = 7 or y = -1

Thus, possible values of y are 7 or -1.

According to question-

AB = AC

taking square on both sides, we get-

AB² = AC²

⇒ (0-3)² + (2-p)² = (0-p)² + (2-5)²

[using distance formula, the distance between points (x₁,y₁) and (x₂,y₂) is equal to units.]

⇒ 9 + 4 + p² - 4p = p² + 9

⇒ 4p-4 = 0

⇒ 4p = 4

∴ p = 1

Thus, the value of p is 1.

fig.1

Clearly from fig.1, One of the diagonals of the rectangle ABCD is BD.

Length of diagonal BD is given by-

√(16 + 9)

= √25

= 5 units

According to question-

AP = BP

taking square on both sides, we get-

AP² = BP²

⇒ (k-4)² + (2-k)² = (-1)² + (2-5)²

[using distance formula, the distance between points (x₁,y₁) and (x₂,y₂) is equal to units.]

⇒ k² - 8k + 16 + 4 + k² - 4k = 1 + 9

⇒ 2k² - 12k + 20 = 10

⇒ 2k² - 12k + 10 = 0

⇒ 2(k²-6k + 5) = 0

⇒ (k²-5k-k + 5) = 0

⇒ k(k-5)-1(k-5) = 0

⇒ (k-1)(k-5) = 0

∴ k = 1 or k = 5

Thus, the value of k is 1 or 5.

Let the point P(x, 2) divides the join of A(12, 5) and B(4, -3) in the ratio of m:n.

fig.2

Recall that if (x,y) ≡ (a,b) then x = a and y = b

∴ assume that

(x,y) ≡ (x,2)

(x₁,y₁) ≡ (12,5)

and, (x₂,y₂) ≡ (4,-3)

Now, Using Section Formula-

⇒ 2m + 2n = -3m + 5n

⇒ 5m = 3n

∴ m:n = 3:5

Thus, the required ratio is 3:5.

fig.3

Length of diagonal AC is given by-

√(9 + 49)

= √58 units

Length of diagonal BD is given by-

= √(9 + 49)

= √58 units

Clearly, the length of the diagonals of the rectangle ABCD are equal.

Mid-point of Diagonal AC is given by

Similarly, Mid-point of Diagonal BD is given by

Clearly, the coordinates of mid-point of both the diagonals coincide i.e. diagonals of the rectangle bisect each other.

A median of a triangle is a line segment joining a vertex to the midpoint of the opposing side, bisecting it.

fig.4

Mid-point of side BC opposite to vertex A i.e. coordinates of point D is given by-

= (4,1)

Mid-point of side AC opposite to vertex B i.e. coordinates of point E is given by-

= (5,-2)

Length of Median AD is given by-

= √(9 + 16)

= √25

= 5 units

Length of Median BE is given by-

= √(0 + 5² )

= √25

= 5 units

Thus, Length of Medians AD and BE are same which is equal to 5 units.

Given that point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3.

∴ m:n = 2:3

Recall that if (x,y) ≡ (a,b) then x = a and y = b

Let (x,y) ≡ (k,4)

(x₁,y₁) ≡ (2,6)

and, (x₂,y₂) ≡ (5,1)

Now, Using Section Formula-

On dividing numerator and denominator of R.H.S by n, we get-

∴ k = (16/5)

Thus the value of k is (16/5).

Let the point on the x-axis which is equidistant from points A(-1,0) and B(5,0) i.e. the point which divides the line segment AB in the ratio 1:1 be C(x,0).

∴ m:n = 1:1

Recall that if (x,y) ≡ (a,b) then x = a and y = b

Let (x,y) ≡ (x,0)

(x₁,y₁) ≡ (-1,0)

and (x₂,y₂) ≡ (5,0)

Using Section Formula,

⇒ x = (4/2) = 2

Thus, the point on the x-axis which is equidistant from points A(-1,0) and B(5,0) is P(2,0).

The distance between the points and is given by- [using distance formula, the distance between points (x₁,y₁) and (x₂,y₂) is equal to units.]

= 2 units

Since the point (3, a) lies on the line represented by 2x - 3y = 5

Thus, the point (3,a) will satisfy the above linear equation

∴ 2×(3) - 3×(a) = 5

⇒ 3a = 6-5

⇒ 3a = 1

∴ a = (1/3)

Thus, the value of a is (1/3).

The distance of any point which lies on the circumference of the circle from the centre of the circle is called radius.

∴ OA = OB = Radius of given Circle

taking square on both sides, we get-

OA² = OB²

⇒ (2-4)² + (3-3)² = (2-x)² + (3-5)²

[using distance formula, the distance between points (x₁,y₁) and (x₂,y₂) is equal to units.]

⇒ (-2)² + 0 = x²-4x + 4 + (-2)²

⇒ x²-4x + 4 = 0

⇒ (x-2)² = 0

∴ x = 2

Thus, the value of x is 2.

According to question-

AP = BP

taking square on both sides, we get-

AP² = BP²

⇒ (7-x)² + (1-y)² = (3-x)² + (5-y)²

[using distance formula, the distance between points (x₁,y₁) and (x₂,y₂) is equal to units.]

⇒ x² - 14x + 49 + y² - 2y + 1 = x² - 6x + 9 + y² - 10y + 25

⇒ -8x + 8y + 16 = 0

⇒ -8(x-y-2) = 0

⇒ x-y-2 = 0

∴ x-y = 2

This is the required relation between x and y.

Every triangle has exactly three medians, one from each vertex, and they all intersect each other at a common point which is called centroid.

fig.5

In the fig.5, Let AD, BE and CF be the medians of ΔABC and point G be the centroid.

We know that-

Centroid of a Δ divides the medians of the Δ in the ratio 2:1.

Mid-point of side BC i.e. coordinates of point D is given by

Let the coordinates of the centroid G be (x,y).

Since centroid G divides the median AD in the ratio 2:1 i.e.

AG:GD = 2:1

∴ using section-formula, the coordinates of centroid is given by-

Now, according to question-

Centroid of ΔABC having vertices A(a, b), B(b, c) and C(c, a) is the origin.

Thus, the value of a + b + c is 0.

The centroid of a Δ whose vertices are (x_1,y₁), (x_2,y₂) and (x_3,y₃) is given by-

∴ centroid of the given ΔABC ≡ [ (2-4 + 5)/3 , (2-4-8)/3 ]

≡ (1,-10/3)

Thus, the centroid of the given triangle ABC is (1,-10/3).

Let the ratio in which the point C(4, 5) divide the join of A(2, 3) and B(7, 8) be m:n.

Recall that if (x,y) ≡ (a,b) then x = a and y = b

Let (x,y) ≡ (4,5)

(x₁,y₁) ≡ (2,3)

and, (x₂,y₂) ≡ (7,8)

Now, Using Section Formula-

⇒ 4m + 4n = 7m + 2n

⇒ 3m = 2n

∴ m:n = 2:3

Thus, the required ratio is 2:3.

If the three points are collinear then the area of the triangle formed by them will be zero.

Area of a Δ ABC whose vertices are A(x₁,y₁), B(x₂,y₂) and C(x₃,y₃) is given by-

∴ Area of given Δ ABC = 0

⇒ ^{√(2(k-(-3)) + 4(-3-3) + 6(3-k) ) = 0}

squaring both sides, we get-

2(k + 3) + 4(-6) + 6(3-k) = 0

⇒ 2k + 6-24 + 18-6k = 0

⇒ -4k + 24-24 = 0

∴ k = 0

Thus, the value of k is zero.

The distance between any two points P₁(x₁, y₁) and P₂(x₂, y₂) is given by the following formula:

From the question we have,

⇒ P₁(x₁, y₁) = (0, 0)…………co-ordinates of origin

⇒ P₂(x₂, y₂) = (-6, 8)…………co-ordinates of point

⇒

⇒ d = √(36 + 64 )

⇒ d = √100

⇒ d = 10 units

Therefore the distance between the point and origin is 10 units.

The distance of any point from x-axis can be determined the modulus or absolute value of the y-coordinate of that point and in similar manner the distance of any point from y-axis can be determined the modulus or absolute value of the x-coordinate of that point

The modulus of y-coordinate is taken because distance cannot be negative.

In this case the y-coordinate is 4 and hence the distance of point from x-axis is 4 units.

⇒ For the point to be equidistant, the point has to be the midpoint of the line joining the points A and B.

⇒ If P(x, y) is the midpoint of the line joining AB then By Midpoint Formula we have,

⇒ and y

Finding x co-ordinate of midpoint:

⇒

⇒

⇒ x = 2

Finding y- co-ordinate of midpoint:

⇒

⇒ y = 0

Therefore the point which is equidistant from A and B is P(2,0).

⇒ If P(x, y) is the midpoint of the line joining AB then By Midpoint Formula we have,

⇒ and y

Finding the value of y:

⇒

⇒

⇒ 12 = 5 + y

⇒ y = 12 – 5

⇒ y = 7

Therefore the value of y is 7

⇒ If P(x, y) is the dividing point of the line joining AB then By Section Formula we have,

⇒ and y

where m and n is the ratio in which the point C divides the line AB

Finding the value of k:

⇒ m = 2 and n = 3

⇒

⇒

The value of k is 16/5.

The perimeter is the addition of lengths of all sides.

Let the points be A = (0, 4), B = (0, 0) and C = (3, 0).

The distance between any two points P₁(x₁, y₁) and P₂(x₂, y₂) is given by the following formula:

⇒

⇒ Distance AB =

=

= 4

⇒ Distance BC =

=

= 3

⇒ Distance AC =

= √(9 + 16)

= √25

= 5

∴ Perimeter = 3 + 4 + 5

= 12

Therefore the perimeter of triangle is 12.

Since the given quadrilateral is a parallelogram, the length of parallel sides is equal.

So by distance formula,

⇒ Distance AB =

= √(4 + 1)

= √5

⇒ Distance CD =

=

⇒ Distance CD = Distance AB

⇒ =

Squaring both sides

⇒ 5 = (x-2)² + 1

⇒ 4 = (x-2)²

Taking square root of both sides

⇒ 2 = x-2

⇒ x = 4

or

⇒ -2 = x-2

⇒ x = 0

Therefore the value of x can be 0 or 4.

Three points A, B, C are said to be collinear if,

Area of triangle formed by three points is zero

The formula of Area of Triangle of three points is given as follows:

⇒ A = 0

= 0

⇒ 1/2 × {[(x + 3) × 1] – [6× -10]} = 0

⇒ x + 3 + 60 = 0

⇒ x = -63

Therefore the value of x is -63.

⇒ Formula of Area of Triangle of three points is given as follows:

⇒ Area, A

=

= 1/2 × {[-3× -4] -0}

= 1/2 × 12

= 6

Therefore the area of a triangle in square units is 6.

⇒ Formula of Area of Triangle of three points is given as follows:

⇒ Area, A

=

= (ab)/2

Therefore the area of the triangle is ab/2.

If P(x, y) is the midpoint of the line joining AB then By Midpoint Formula we have,

⇒ and y

⇒

⇒ a = -8

Therefore the value of a is -8.

Distance BD is the length of one of its diagonal.

⇒ So by distance formula,

⇒ Distance BD =

=

=

= 5

Therefore the length of diagonal is 5 units.

If P(x, y) is the dividing point of the line joining AB then By Section Formula we have,

⇒ and y

where m and n is the ratio in which the point C divides the line AB

Finding the x-coordinate of P:

⇒

⇒

⇒ x = 3

Finding the y-coordinate of P:

⇒

⇒

⇒ y = 5

Therefore the coordinates of P is (3,5).

Since the center divides the diameter into two equal halves.

⇒ Therefore by Midpoint Formula we have,

⇒ and y

Finding the coordinates of another end of diameter:

Finding x-coordinate:

⇒

⇒ -4 = 2 + x₂

⇒ x₂ = -4-2

⇒ x₂ = -6

Finding y-coordinate:

⇒

⇒ 10 = 3 + y₂

⇒ y₂ = 10-3

⇒ y₂ = 7

Therefore the coordinates of another end of diameter are (-6, 7).

From the given diagram, we come to know

⇒ AP = PQ = QB

⇒ Therefore the point P divides the line internally in the ratio 1:2 and Q divides the line in the ratio 2: 1

⇒ Then by section formula the y-coordinate of point Q which divide the line AB is given as

⇒

⇒ y = -12/3

⇒ y = -4

Therefore the value of y is -4.

⇒ Therefore by Midpoint Formula we have,

⇒ and y

Finding the coordinates of the end of A:

⇒ Finding x-coordinate:

⇒

⇒ x₂ = 2

Finding y-coordinate:

⇒

⇒ 8 = 3 + y₂

⇒ y₂ = 8-3

⇒ y₂ = 5

Therefore the coordinates of the end of A are (2, 5).

If P(x, y) is the dividing point of the line joining AB then By Section Formula we have,

⇒ and

⇒ where m and n is the ratio in which the point C divides the line AB

Finding the x-coordinate of P:

⇒

⇒

⇒ x = 16/3

Finding the y-coordinate of P:

⇒

⇒

⇒ y = -11/3

Therefore the coordinates of P is (16/3, -11/3).

Since in fourth quadrant x-coordinate is positive and y-coordinate is negative.

Therefore the point P lies in the fourth quadrant.

⇒

⇒ Distance AB =

=

= √(25 + 144 )

= √(169 )

= 13

⇒ Distance 2AB = 2×13

= 26 units.

Therefore the distance 2AB is 26 units.

⇒ Point on x-axis means its y-coordinate is zero.

⇒ Let the point be P(x, 0)

Using the distance formula,

⇒

⇒ Distance AP = Distance BP

⇒ (x-7)² + (0-6)² = (x + 3)² + (0-4)²
⇒ x² + 49-14x + 36 = x² + 9 + 6x + 16

⇒ 49-9 + 36-16 = 6x + 14x

⇒ 40 + 20 = 20x

⇒ x = 60/20

x = 3

Therefore the coordinate of P is (3,0).

The distance of any point from x-axis can be determined the modulus or absolute value of the y-coordinate of that point and in a similar manner, the distance of any point from y-axis can be determined the modulus or absolute value of the x-coordinate of that point

The modulus of y-coordinate is taken because distance cannot be negative.

In this case, the y-coordinate is 4 and hence the distance of the point from x-axis is 4 units.

⇒ Let the ratio be k:1.

⇒ Then by section formula the coordinates of point which divide the line AB is given as

⇒ Since the point lies on x-axis its y-coordinate is zero.

⇒ = 0

⇒ 6k = 3

⇒ k = 1/2

Therefore the ratio in which x-axis divide the line AB is 1:2.

⇒ Let the ratio be k:1.

⇒ Then by section formula the coordinates of point which divide the line AB is given as

⇒ Since the point lies on y-axis its x-coordinate is zero.

⇒ = 0

⇒ 8k = 4

⇒ k = 1/2

Therefore the ratio in which x-axis divide the line AB is 1:2.

∴ by Midpoint Formula we have,

⇒ and y

Finding value of b:

⇒

⇒ 2 = 2b + 4

⇒ 2 – 4 = 2b

⇒ b = -2/2

⇒ b = -1

Therefore the value of b is -1.

⇒ Let 2x + y = 4 ………….. (1)

Finding the equation of line formed by AB:

Finding slope:

⇒

⇒

⇒ m = 9

The equation of line AB:

⇒ y – y₁ = m×(x-x₁)

⇒ y-(-2) = 9×(x-2)

⇒ y + 2 = 9x – 18

⇒ 9x – y = 20……………… (2)

When we solve the two equations simultaneously, we get point of intersection of two lines.

⇒ Adding (1) and (2)

⇒ 11x = 24

⇒ x = 24/11

⇒ Substituting the value of x in (1)

⇒ 2×24/11 + y = 4

⇒ y = 4 – 48/11

⇒ y = -4/11

let us assume the line divides the segment AB in the ratio k:1

Then by section formula, the coordinates of point which divide the line AB is given as

Since we know x-coordinate of the point

⇒

⇒ 33k + 22 = 24k + 24

⇒ 9k = 2

⇒ k = 2:9

Therefore the line 2x + y -4 = 0 divides the line segment AB into the ratio 2:9.

Since the AD is median, it divides the line BC into two equal halves. So D acts as the midpoint of line BC.

If D(x, y) is the midpoint of the line joining BC then By Midpoint Formula we have,

⇒ and y

Finding x co-ordinate of midpoint:

⇒

⇒

⇒ x = 7/2

Finding y- co-ordinate of midpoint:

⇒

⇒ y = 9/2

Therefore the point which is equidistant from A and B is P(7/2,9/2).

Let P(x, y) be the centroid of the triangle

⇒ Finding the x-coordinate of P:

⇒

⇒

⇒ x = 4

Finding the y-coordinate of P:

⇒

⇒ y = 0

Therefore the coordinates of P are (4, 0).

Finding the x-coordinate of C:

⇒

⇒ x = -4

Finding the y-coordinate of P:

⇒

⇒ -9 = 6 + y

⇒ y = -15

Therefore the coordinates of P are (-4, -15).

The distance between any two points P₁(x₁, y₁) and P₂(x₂, y₂) is given by the following formula:

⇒

⇒ Distance AB =

=

= 8

⇒ Distance BC =

=

= 5

⇒ Distance AC =

= √(9 + 16)

= √25

= 5

Since the length of two sides is equal, given triangle is an isosceles triangle.