Co-ordinate Geometry: Area Of Triangular Region

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

(x₁, y₁) is (2, –2)

⇒ x₁ = 2 and y₁ = -2

(x₂, y₂) is (4, 2)

⇒ x₂ = 4 and y₂ = 2

(x₃, y₃) is (–1, 3)

⇒ x₃ = -1 and y₃ = 3

Hence substituting values in formula for area we get

Area = × [2(2 – 3) + 4(3 – (-2)) + (-1)(-2 – 2)]

Area = × [-2 + 4(5) + 4]

Area = × [20 + 2]

Area = 11 unit²

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

(x₁, y₁) is (8, 9)

⇒ x₁ = 8 and y₁ = 9

(x₂, y₂) is (2, 6)

⇒ x₂ = 2 and y₂ = 6

(x₃, y₃) is (9, 2)

⇒ x₃ = 9 and y₃ = 2

Hence substituting values in formula for area we get

Area = × [8(6 – 2) + 2(2 – 9) + 9(9 – 6)]

Area = × [32 + (-14) + 27]

Area = × [32 + 13]

Area = × 45

Area = 22.5 unit²

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

(x₁, y₁) is (1, 2)

⇒ x₁ = 1 and y₁ = 2

(x₂, y₂) is (3, 0)

⇒ x₂ = 3 and y₂ = 0

(x₃, y₃) is origin which is (0, 0)

⇒ x₃ = 0 and y₃ = 0

Hence substituting values in formula for area we get

Area = × [1(0 – 0) + 3(0 – 2) + 0(2 – 0)]

Area = × [0 + (-6) + 0]

Area = -3

As area cannot be negative

Area = 3 unit²

Let the points be

A = (x₁, y₁) = (3, -2) and

B = (x₂, y₂) = (-5, 4) and

C = (x₃, y₃) = (-1, 1)

Now if the points A, B and C are collinear then the area formed by the triangle by joining these three points would be 0

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

So, if we prove that area of ΔABC = 0 then points A, B and C are collinear

⇒ Area = × [3(4 – 1) + (-5)(1 – (-2)) + (-1)(-2 – 4)]

⇒ Area = × [9 + (-5)(3) + 6]

⇒ Area = × [15 + (-15)]

⇒ Area = 0

Hence points (3, –2), (–5, 4) and (–1, 1) are collinear.

Method 2

Another method is that you can find distances between every pair of two points and if any distance is equal to sum of other two distances then points are collinear. Here AC + BC = AB.

Let the points be

A = (x₁, y₁) = (1, -1) and

B = (x₂, y₂) = (2, -1) and

C = (x₃, y₃) = (K, -1)

These points lie on a straight line which means points A, B and C are collinear

As these points are collinear the area of triangle formed by these points would be 0

⇒ Area = 0

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

⇒ × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0

⇒ × [1(-1 – (-1)) + 2(-1 – (-1)) + K(-1 – (-1))] = 0

⇒ × [1(-1 + 1) + 2(-1 + 1) + K(-1 + 1)] = 0

⇒ K × 0 = 0 … (a)

Here we can put any value for K from negative infinity to infinity and our equation (a) is satisfied

Hence K can take any value for the points (1, –1), (2, –1) and (K, –1) to lie on same straight line.

Method 2

If we plot the given points we can observe that K can take any value since the y-coordinate for all the points A, B and C is the same

Let the points be

A = (x₁, y₁) = (1, 2) and

B = (x₂, y₂) = (-2, -4)

Let O be the origin

O = (x₃, y₃) = (0, 0)

Now if the area of triangle formed by joining point A, O and B i.e. ΔAOB is zero then we can say that points A, O and B are collinear which means they lie on the straight line which would imply that line passing through A and B will pass through origin O

So we have to prove that area(ΔAOB) = 0

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Substituting values

⇒ Area = × [1(-4 – 0) + (-2)(0 – 2) + 0(2 – (-4))]

⇒ Area = × [-4 + 4 + 0]

⇒ Area = 0

Hence, line joining two points (1, 2) and (–2, –4) passes through origin.

Le the points be M = (2, 1) and N = (6, 5)

Midpoint of line segment joining two points is given by

Where (M_x, M_y) are x and y coordinates of point M and (N_x, N_y) are x and y coordinates of point N

⇒ Midpoint =

⇒ Midpoint = (4, 3)

Let this point be A (4, 3)

Let the points be B and C as follows

B = (–4, –5) and C = (9, 8)

Now, we have to prove that the midpoint i.e. A lies on line joining points B and C

If we prove that area of triangle formed by joining points A, B and C is 0 then A, B and C will be collinear and thus we can say that point A lies on line joining points B and C

Thus the 3 vertices of triangle here are

A = (x₁, y₁) = (4, 3)

B = (x₂, y₂) = (-4, -5)

C = (x₃, y₃) = (9, 8)

So we have to prove that area(ΔACB) = 0

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Substituting values

⇒ Area = × [4(-5 – 8) + (-4)(8 – 3) + 9(3 – (-5))]

⇒ Area = × [-52 + (-20) + 72]

⇒ Area = × [- 72 + 72]

⇒ Area = 0

Hence midpoint of (2, 1) and (6, 5) lie on the line joining two points (–4, –5) and (9, 8)

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

Plot the points we get the quadrilateral as shown

Divide the quadrilateral in two triangles by joining points A and C thus by observing figure we can conclude that

area(ABCD) = area(ΔABC) + area(ΔACD)

let us find area(ΔABC)

vertices are

A = (x₁, y₁) = (1, 1)

B = (x₂, y₂) = (3, 4)

C = (x₃, y₃) = (5, -2)

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Substituting values

⇒ area(ΔABC) = × [1(4 – (-2)) + 3(-2 – 1) + 5(1 – 4)]

⇒ area(ΔABC) = × [6 + (-9) + (-15)]

⇒ area(ΔABC) = × [6 - 24]

⇒ area(ΔABC) = -9

As area cannot be negative

⇒ area(ΔABC) = 9 unit²

Let us find area(ΔACD)

Vertices are

A = (x₁, y₁) = (1, 1)

C = (x₂, y₂) = (5, -2)

D = (x₃, y₃) = (4, -7)

⇒ area(ΔACD) = × [1(-2 – (-7)) + 5(-7 – 1) + 4(1 – (-2))]

⇒ area(ΔACD) = × [5 + (-40) + 12]

⇒ area(ΔACD) = × [-40 + 17]

⇒ area(ΔACD) = 11.5 unit²

Thus, area(ABCD) = area(ΔABC) + area(ΔACD)

⇒ area(ABCD) = 9 + 11.5

⇒ area(ABCD) = 20.5 unit²

Therefore, area of quadrilateral region is 20.5 unit²

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

Plot the points we get the quadrilateral as shown

Divide the quadrilateral in two triangles by joining points A and C thus by observing figure we can conclude that

area(ABCD) = area(ΔABC) + area(ΔACD)

let us find area(ΔABC)

vertices are

A = (x₁, y₁) = (1, 4)

B = (x₂, y₂) = (-2, 1)

C = (x₃, y₃) = (2, -3)

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Substituting values

⇒ area(ΔABC) = × [1(1 – (-3)) + (-2)(-3 – 4) + 2(4 – 1)]

⇒ area(ΔABC) = × [4 + 14 + 6]

⇒ area(ΔABC) = × 24

⇒ area(ΔABC) = 12

⇒ area(ΔABC) = 12 unit²

Let us find area(ΔACD)

Vertices are

A = (x₁, y₁) = (1, 4)

C = (x₂, y₂) = (2, -3)

D = (x₃, y₃) = (3, 3)

⇒ area(ΔACD) = × [1(-3 – 3) + 2(3 – 4) + 3(4 – (-3))]

⇒ area(ΔACD) = × [(-6) + (-2) + 21]

⇒ area(ΔACD) = × 13

⇒ area(ΔACD) = 6.5 unit²

Thus, area(ABCD) = area(ΔABC) + area(ΔACD)

⇒ area(ABCD) = 12 + 6.5

⇒ area(ABCD) = 18.5 unit²

Therefore, area of quadrilateral region is 18 unit²

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

A = (x₁, y₁) = (3, 4)

B = (x₂, y₂) = (-4, 3)

C = (x₃, y₃) = (8, -6)

Substituting values

⇒ Area = × [3(3 – (-6)) + (-4)(-6 – 4) + 8(4 – 3)]

⇒ Area = × [27 + 40 + 8]

⇒ Area = × 75

⇒ Area = 37.5 unit² … (i)

Consider AD is the perpendicular dropped on BC as shown

So, for ΔABC BC is the base then AD is the height

Area of triangle = × base × height … (ii)

Length of BC can be calculated by distance formula because coordinates of B and C are known

⇒ Distance BC =

=

=

= √225

= 15 units

Distance BC = 15 units

Using (i) and (ii)

⇒ 37.5 = × BC × AD

⇒ 37.5 = × 15 × AD

⇒ 75 = 15 × AD

⇒ AD = 3 units

area of triangle is 37.5 unit² and the perpendicular length drawn from the point A on BC i.e. AD = 3 units

The ΔABC with coordinates of A as (2, 5) and assuming B and C coordinates to be (x₂, y₂) and (x₃, y₃) respectively as shown

M is the midpoint of segment BC with coordinates as shown and G is the centroid

The vertices of ΔABC with their coordinates are

A = (x₁, y₁) = (2, 5) and

B = (x₂, y₂) and

C = (x₃, y₃)

G = (-2, 1)

Centroid of a triangle is given by

G =

⇒ (-2, 1) =

⇒ (-2, 1) =

Equate x-coordinate and y-coordinate

⇒ = -2 and = 1

⇒ 2 + x₂ + x₃ = -6 and 5 + y₂ + y₃ = 3

⇒ x₂ + x₃ = -8 and y₂ + y₃ = -2

Divide by 2

⇒ = -4 and = -1

Thus midpoint M of BC is (-4, -1)

Let the points of triangle be

A = (x₁, y₁) = (4, -3) and

B = (x₂, y₂) = (-5, 2) and

C = (x₃, y₃) = (x, y)

Centroid of triangle is origin hence G be the centroid G (0, 0)

Centroid of a triangle is given by

G =

Substituting values

⇒ (0, 0) =

Equate x-coordinate and y-coordinate

⇒ = 0 and = 0

⇒ x – 1 = 0 and y – 1 = 0

⇒ x = 1 and y = 1

Hence (x, y) is (1, 1)

Let us find area of ΔABC

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

A = (x₁, y₁) = (-1, 5)

B = (x₂, y₂) = (3, 1)

C = (x₃, y₃) = (5, 7)

Substituting values

⇒ Area(ΔABC) = × [(-1)(1 – 7) + 3(7 – 5) + 5(5 – 1)]

⇒ Area(ΔABC) = × [6 + 6 + 20]

⇒ Area(ΔABC) = × 32

⇒ Area(ΔABC) = 16 unit² … (i)

Let us now find the midpoints of BC, AC and AB i.e. points D, E and F respectively

Point F is the midpoint of AB

⇒ F =

⇒ F = (1, 3)

Point D is the midpoint of BC

⇒ D =

⇒ D = (4, 4)

Point E is the midpoint of AC

⇒ E =

⇒ E = (2, 6)

Let us find area of ΔDEF

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

D = (x₁, y₁) = (1, 3)

E = (x₂, y₂) = (4, 4)

F = (x₃, y₃) = (2, 6)

Substituting values

⇒ Area(ΔDEF) = × [1(4 – 6) + 4(6 – 3) + 2(3 – 4)]

⇒ Area(ΔDEF) = × [-2 + 12 + (-2)]

⇒ Area(ΔDEF) = × 8

⇒ Area(ΔDEF) = 4 unit² … (ii)

From (i) and (ii) we can conclude that area(ΔABC) = 4 × area(ΔDEF)

If we plot these points roughly on the coordinate axis we can see that it’s a right angled triangle with base = 6 units and height = 4 units

Thus, area = × base × height

⇒ area = × 6 × 4 = 12 unit²

Another method is to use the formula.

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Let the points of triangle be

(x₁, y₁) = (7, -5) and

(x₂, y₂) = (-2, 5) and

(x₃, y₃) = (4, 6)

Centroid of a triangle is given by

G =

Substituting values

⇒ G =

⇒ G = (3, 2)

Point B can either be origin or (3, 4) as shown

Consider B to be the origin

Base of ΔABC = BC = 3 units

Height of ΔABC = AB = 4 units

Thus, area = × base × height

⇒ area = × 3 × 4 = 6 unit²

Let A = (0, 0), B = (4, -3) and C = (x, y)

As the points are collinear area of triangle formed by these points is 0

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

(x₁, y₁) = (0, 0)

(x₂, y₂) = (4, -3)

(x₃, y₃) = (x, y)

Substituting values

⇒ 0 = × [0(-3 – y) + 4(y – 0) + x(0 – (-3))]

⇒ 0 = 0 + 4y + 3x

⇒ 4y + 3x = 0

Here there are infinite values for x and y which will satisfy the equation 4y + 3x = 0 which means there are infinite points

This can also be seen geometrically that is if line is passing through (0, 0) and (4, -3) there are infinite points on this line.

We can select the correct option by substituting the values given in option in equation 4y + 3x = 0 and if it satisfies then that is the correct option

x = 8, y = -6 satisfies the equation 4y + 3x = 0

x = –8, y = 6 satisfies the equation 4y + 3x = 0

Now, C = (8, -6) or C = (-8, 6), but for A, B and C to be collinear

AB + BC = AC
We know, By distance formula, Distance between two points X(x₁, y₁) and Y(x₂, y₂) is

Case I: C = (8, -6)

AB

units

BC

units

AC

units

Clearly, AB + BC = AC

Hence, (0, 0), (4, -3) and (8, -6) are collinear

Case II: C = (-8, 6)

AB

units

BC

units

AC

units

Clearly, AB + BC ≠ AC

Hence, (0, 0), (4, -3) and (8, -6) are not collinear.

[In this case, BA + AC = BC, ⇒ B, A and C are collinear]

Hence, Correct option is (a)

The ΔABC with coordinates of A as (7, -4) and assuming B and C coordinates to be (x₂, y₂) and (x₃, y₃) respectively as shown

M is the midpoint of segment BC with coordinates as shown and G is the centroid

The vertices of ΔABC with their coordinates are

A = (x₁, y₁) = (7, -4) and

B = (x₂, y₂) and

C = (x₃, y₃)

G = (1, 2)

Centroid of a triangle is given by

G =

⇒ (1, 2) =

⇒ (1, 2) =

Equate x-coordinate and y-coordinate

⇒ = 1 and = 2

⇒ 7 + x₂ + x₃ = 3 and – 4 + y₂ + y₃ = 6

⇒ x₂ + x₃ = -4 and y₂ + y₃ = 10

Divide by 2

⇒ = -2 and = 5

Thus midpoint M of BC is (-2, 5)

Let the vertices of ΔABC be

A = (x₁, y₁) and

B = (x₂, y₂) and

C = (x₃, y₃)

Centroid of a triangle is given by

G =

Let (0, 1), (1, 1) and (1, 0) be midpoints of AB, BC and AC respectively

Point (0, 1) is the midpoint of AB

⇒ (0, 1) =

Equating x and y coordinates

⇒ = 0 and = 1

⇒ x₁ + x₂ = 0 … (a) and y₁ + y₂ = 2 … (i)

Point (1, 1) is the midpoint of BC

⇒ (1, 1) =

Equating x and y coordinates

⇒ = 1 and = 1

⇒ x₂ + x₃ = 2 … (b) and y₂ + y₃ = 2 … (ii)

Point (1, 0) is the midpoint of AC

⇒ (1, 0) =

Equating x and y coordinates

⇒ = 1 and = 0

⇒ x₁ + x₃ = 2 … (c) and y₁ + y₃ = 0 … (iii)

Adding equations (a), (b) and (c) & adding equations (i), (ii) and (iii) we get

⇒ x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 4 and y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 4

⇒ 2 (x₁ + x₂ + x₃) = 4 and 2 (y₁ + y₂ + y₃) = 4

⇒ x₁ + x₂ + x₃ = 2 and y₁ + y₂ + y₃ = 2

Divide by 3

⇒ = and =

Hence centroid is G =

Let the vertices of triangle be

A = (x₁, y₁) = (15, 0) and

B = (x₂, y₂) = (0, 10) and

C = (x₃, y₃)

Centroid G = (6, 9)

Centroid of a triangle is given by

G =

⇒ (6, 9) =

Substituting values

⇒ (6, 9) =

Equating x and y coordinates

⇒ = 6 and = 9

⇒ 15 + x₃ = 18 and 10 + y₃ = 27

⇒ x₃ = 3 and y₃ = 17

Hence the coordinates of third vertex is (3, 17)

Let the collinear points be

A = (x₁, y₁) = (a, 0) and

B = (x₂, y₂) = (0, b) and

C = (x₃, y₃) = (1, 1)

The area of triangle formed by joining points A, B and C will be 0 because A, B and C are collinear

Area of triangle is given by

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

⇒ 0 = × [a(b – 1) + 0(1 – 0) + 1(0 – b)]

⇒ 0 = ab – a + (-b)

⇒ a + b = ab

Divide throughout by ab

⇒ + =

⇒ + = 1

Hence proved

Area of triangle is given by formula

Area = × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Where (x₁, y₁), (x₂, y₂) and (x₃, y₃) are vertices of triangle

Here,

(x₁, y₁) is (1, 4)

⇒ x₁ = 1 and y₁ = 4

(x₂, y₂) is (-1, 2)

⇒ x₂ = -1 and y₂ = 2

(x₃, y₃) is (-4, 1)

⇒ x₃ = -4 and y₃ = 1

Hence substituting values in formula for area we get

Area = × [1(2 – 1) + (-1)(1 – 4) + (-4)(4 – 2)]

Area = × [1 + 3 + (-8)]

Area = × (-4)

Area = -2

Area cannot be negative

Area = 2 unit²

Let the vertices of triangle be

A = (x₁, y₁) = (x – y, y – z) and

B = (x₂, y₂) = (–x, –y) and

C = (x₃, y₃) = (y, z)

Centroid of a triangle is given by

G =

Substituting values

⇒ G =

⇒ G =

⇒ G = (0, 0)

Centroid is (0, 0)