# AREA OF TRIANGLE WHEN VERTICES ARE GIVEN

## About "Area of triangle when vertices are given"

Area of triangle when vertices are given :

Here, we are going to see, how to find area of a triangle when coordinates of the three vertices are given.

Let us consider the triangle given below.

In the above triangle, A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) are the vertices.

To find area of the triangle ABC, now we have take the vertices A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) of the triangle ABC in order (counter clockwise direction) and write them column-wise as shown below.

And the diagonal products x₁y₂, x₂y₃ and x₃y₁ as shown in the dark arrows.

Also add the diagonal products xy, xy and xy as shown in the dotted arrows.

Now, subtract the latter product from the former product to get area of the triangle ABC.

Hence, area of the triangle ABC.

## Collinearity of three points

Three or more points in a plane are said to be collinear, if they lie on the same straight line.

In other words, three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are collinear, if any one of these points lies on the straight line joining the other two points.

Suppose that the three points A(x₁, y₁), B(x₂, y₂) and  C(x₃, y₃) are collinear, then they can not form a triangle . Hence, the area of triangle ABC  =  0

1/2 x  { (x₁y₂ + xy + xy) - (xy + xy + xy) }  =  0

(or)

x₁y₂ + x₂y₃ + x₃y   =   xy + x₃y₂ + xy₃

One can prove that the converse is also true.

Hence, the area of triangle ABC is zero if and only if the points A, B and C are collinear.

Problem 1 :

Find the area of the triangle whose vertices are (1, 2), (-3, 4) and (-5, -6)

Solution :

Plot the given points in a rough diagram as given below and take them in order (counter clock wise)

Let the vertices be A(1, 2), B(-3, 4) and C(-5, -6)

Then, we have

(x₁, y₁)  =  (1, 2)

(x₂, y₂)  =  (-3, 4)

(x₃, y₃)  =  (-5, -6)

Area of triangle ABC is

=  (1/2) x  { [x₁y₂ + xy + xy₁] - [xy + xy + xy₃] }

=  (1/2) x  { [1.4 + (-3).(-6) + (-5).2] - [(-3).2 + (-5).4 + 1.(-6)] }

=  (1/2) x  { [4 + 18 - 10] - [-6 - 20 -6] }

=  (1/2) x  { [12] - [-32] }

=  (1/2) x  { 12 + 32 }

=  (1/2) x  {  44 }

=  22 square units.

Hence, are of triangle ABC is 22 square units.

Let us look at the next problem on "Area of triangle when vertices are given"

Problem 2 :

If the area of the triangle ABC is 68 square units and the vertices are A(6, 7), B(-4, 1) and C(a, -9) taken in order, then find the value of "a".

Solution :

Let (x₁, y₁)  =  (6, 7),  (x₂, y₂)  =  (-4, 1) and (x₃, y₃)  =  (a, -9)

Given :  Area of triangle ABC   =  68 square units

(1/2) x  { [x₁y₂ + xy + xy₁] - [xy + xy + xy₃] }  =  68

Multiply by 2 on both sides,

{ [x₁y₂ + x₂y₃ + x₃y₁] - [x₂y₁ + x₃y₂ + x₁y₃] }  =  136

{ [6 + 36 + 7a] - [-28 + a - 54] }  =  136

[42 + 7a] - [a - 82]  =  136

42 + 7a -a +82  =  136

6a + 124  =  136

6a  =  12

a  =  2

Hence, the value of "a" is 2.

Let us look at the next problem on "Area of triangle when vertices are given"

Problem 3 :

Using the concept of area of triangle, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

Solution :

Using the concept of area of triangle, if the three points A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃)  are collinear, then

x₁y₂ + x₂y₃ + x₃y   =   xy + x₃y₂ + xy₃

Comparing the given points to A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃), we get

(x₁, y₁)  =  (5, -2)

(x₂, y₂)  =  (4, -1)

(x₃, y₃)  =  (1, 2)

x₁y₂ + x₂y₃ + x₃y₁  =  5x(-1) + 4x2 + 1x(-2)

x₁y₂ + x₂y₃ + x₃y₁  =  -5 + 8 -2

x₁y₂ + x₂y₃ + x₃y₁  =  1 --------(1)

xy + x₃y₂ + xy₃  =  4x(-2) + 1x(-1) + 5x(2)

xy + x₃y₂ + xy₃  =  -8 - 1 + 10

xy + x₃y₂ + xy₃  =  1 --------(2)

From (1) and (2), we get

x₁y₂ + x₂y₃ + x₃y   =   xy + x₃y₂ + xy₃

Hence, the three points A, B and C are collinear.

Let us look at the next problem on "Area of triangle when vertices are given"

Problem 4 :

If P(x, y) is any point on the line segment joining the points (a, 0), and (0, b), then prove that x/a + y/b  =  1. where  a ≠ b.

Solution :

Clearly, the points (x, y), (a, 0), and (0, b) are collinear.

Then, area of the triangle  =  0

Since, area of the triangle is zero, we have

x₁y₂ + x₂y₃ + x₃y   =   xy + x₃y₂ + xy₃ ------(1)

(x₁, y₁)  =  (x, y)

(x₂, y₂)  =  (a, 0)

(x₃, y₃)  =  (0, b)

Plugging the above points in (1),

x.0 + a.b + 0.y   =   a.y + 0.0 + x.b

0 + a.b + 0   =   a.y + 0 + x.b

a.b   =   a.y + x.b

Divide by ab on bothe sides,

1  =  y/b + x/a

or

x/a + y/b  =  1

Let us look at the next problem on "Area of triangle when vertices are given"

Problem 5 :

If the points (k, -1), (2, 1) and (4, 5) are collinear, then find the value of "k".

Solution :

Since the given points are collinear, area of triangle  =  0.

Then, we have

x₁y₂ + x₂y₃ + x₃y   =   xy + x₃y₂ + xy₃ ------(1)

(x₁, y₁)  =  (k, -1)

(x₂, y₂)  =  (2, 1)

(x₃, y₃)  =  (4, 5)

Plugging the above points in (1),

k.1 + 2.5 + 4.(-1)   =   2.(-1) + 4.1 + k.5

k + 10 - 4   =   -2 + 4 + 5k

k + 6   =   2 + 5k

4  =  4k

1  =  k

Hence, the value of "k" is 1.

After having gone through the stuff given above, we hope that the students would have understood "Area of triangle when vertices are given".

Apart from the stuff given above, if you want to know more about "Area of triangle when vertices are given", please click here

Apart from the stuff, "Area of triangle when vertices are given", if you need any other stuff in math, please use our google custom search here.

You can also visit our following web pages on different stuff in math.

WORD PROBLEMS

Word problems on simple equations

Word problems on linear equations

Algebra word problems

Word problems on trains

Area and perimeter word problems

Word problems on direct variation and inverse variation

Word problems on unit price

Word problems on unit rate

Word problems on comparing rates

Converting customary units word problems

Converting metric units word problems

Word problems on simple interest

Word problems on compound interest

Word problems on types of angles

Complementary and supplementary angles word problems

Double facts word problems

Trigonometry word problems

Percentage word problems

Profit and loss word problems

Markup and markdown word problems

Decimal word problems

Word problems on fractions

Word problems on mixed fractrions

One step equation word problems

Linear inequalities word problems

Ratio and proportion word problems

Time and work word problems

Word problems on sets and venn diagrams

Word problems on ages

Pythagorean theorem word problems

Percent of a number word problems

Word problems on constant speed

Word problems on average speed

Word problems on sum of the angles of a triangle is 180 degree

OTHER TOPICS

Profit and loss shortcuts

Percentage shortcuts

Times table shortcuts

Time, speed and distance shortcuts

Ratio and proportion shortcuts

Domain and range of rational functions

Domain and range of rational functions with holes

Graphing rational functions

Graphing rational functions with holes

Converting repeating decimals in to fractions

Decimal representation of rational numbers

Finding square root using long division

L.C.M method to solve time and work problems

Translating the word problems in to algebraic expressions

Remainder when 2 power 256 is divided by 17

Remainder when 17 power 23 is divided by 16

Sum of all three digit numbers divisible by 6

Sum of all three digit numbers divisible by 7

Sum of all three digit numbers divisible by 8

Sum of all three digit numbers formed using 1, 3, 4

Sum of all three four digit numbers formed with non zero digits

Sum of all three four digit numbers formed using 0, 1, 2, 3

Sum of all three four digit numbers formed using 1, 2, 5, 6