Question

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The question is
By which formula we can calculate the area of triangle in this situation
situation = the vertices of triangle are x1,y1 ; x2,y2 ; x3y3
$...$
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Answer 1

Answer:Area of Triangle =$\frac{1}{2}(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))$

Now, we can easily derive this formula using a small diagram shown below.

how to find area of triangle using coordinates of three vertices of triangle

Suppose, we have a triangle ABC as shown in the diagram and we want to find its area.

Let the coordinates of vertices are (x1, y1), (x2, y2) and (x3, y3).

We draw perpendiculars AP, BQ and CR to x-axis.

Area of $\triangle ABC$ = Area of Trapezium ABQP + Area of Trapezium BCRQ - Area of Trapezium ACRP

Area of $\triangle ABC$ =$\frac{1}{2}(y1+y2)(x1-x2)+\frac{1}{2}(y1+y3)(x3-x1)$

$-\frac{1}{2}(y2+y3)(x3-x2)$ = $\frac{1}{2}(x1(y2-y3) + x2(y3-y1) + x3(y1 -y2))$

Example: Find area of triangle whose vertices are (1, 1), (2, 3) and (4, 5)

Solution:

We have (x1, y1) = (1, 1), (x2, y2) = (2, 3) and (x3, y3) = (4, 5)

Using formula:

Area of Triangle = $\frac{1}{2}(x1(y2-y3) + x2(y3-y1) + x3(y1 -y2))$

=$\frac{1}{2}(1(3-5) + 2(5-1) + 4(1-3))$ =$\frac{1}{2}(-2 + 8 -8)$ =$\frac{1}{2}(-2)$=-1

Because, Area cannot be negative. We only consider the numerical value of answer. Therefore, area of triangle = 1 sq units.

Step-by-step explanation:

Answer 2

Answer:

Answer:Area of Triangle =\frac{1}{2}(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))

2

1

(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))

Now, we can easily derive this formula using a small diagram shown below.

how to find area of triangle using coordinates of three vertices of triangle

Suppose, we have a triangle ABC as shown in the diagram and we want to find its area.

Let the coordinates of vertices are (x1, y1), (x2, y2) and (x3, y3).

We draw perpendiculars AP, BQ and CR to x-axis.

Area of \triangle ABC△ABC = Area of Trapezium ABQP + Area of Trapezium BCRQ - Area of Trapezium ACRP

Area of \triangle ABC△ABC =\frac{1}{2}(y1+y2)(x1-x2)+\frac{1}{2}(y1+y3)(x3-x1)

2

1

(y1+y2)(x1−x2)+

2

1

(y1+y3)(x3−x1)

-\frac{1}{2}(y2+y3)(x3-x2)−

2

1

(y2+y3)(x3−x2) = \frac{1}{2}(x1(y2-y3) + x2(y3-y1) + x3(y1 -y2))

2

1

(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))

Example: Find area of triangle whose vertices are (1, 1), (2, 3) and (4, 5)

Solution:

We have (x1, y1) = (1, 1), (x2, y2) = (2, 3) and (x3, y3) = (4, 5)

Using formula:

Area of Triangle = \frac{1}{2}(x1(y2-y3) + x2(y3-y1) + x3(y1 -y2))

2

1

(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))

=\frac{1}{2}(1(3-5) + 2(5-1) + 4(1-3))

2

1

(1(3−5)+2(5−1)+4(1−3)) =\frac{1}{2}(-2 + 8 -8)

2

1

(−2+8−8) =\frac{1}{2}(-2)

2

1

(−2) =-1

Because, Area cannot be negative. We only consider the numerical value of answer. Therefore, area of triangle = 1 sq units.