Question

5.If (-3 ,1) , (1 ,-3 ) and (2 ,3 ) then the area of triangle is *

a) 14 sq.units
b) 7 sq.units
c) 28 sq.units
d) 35 sq.units​

Answer 1

Coordinate Geometry

While solving this type of question it is needed to have the knowledge of one simple formula, which is as follows:

• The formula to calculate the area of triangle with three vertices.

We know the area of a triangle having vertices is evaluated as:

The area of a ∆ABC with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is given by,

$\boxed{\rm{Area = \dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|}}$

The coordinates of the vertices of the given triangle are:

• A(x₁ = -3, y₁ = 1)
• B(x₂ = 1, y₂ = -3)
• C(x₃ = 2, y₃ = 3)

Solution:

By substituting the given values in the formula, we get the following results:

$\implies Area = \dfrac{1}{2}\bigg|-3(-3-3)+1(3-1)+2(1-(-3))\bigg|$

$\implies Area = \dfrac{1}{2}\bigg|-3(-3-3)+1(3-1)+2(1+3)\bigg|$

$\implies Area = \dfrac{1}{2}\bigg|-3(-6)+1(2)+2(4)\bigg|$

$\implies Area = \dfrac{1}{2}\bigg|18+2+8\bigg|$

$\implies Area = \dfrac{1}{2}\bigg|28\bigg|$

$\implies Area = \dfrac{28}{2}$

$\implies Area = \dfrac{28}{2}$

$\implies Area = 14$

Hence, the area of triangle is [a] 14 sq. units.

$\rule{300}{2}$

Extra Information:

1. Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane, then the distance between A and B is given by,

$\boxed{AB = \sqrt{ {(x_{2} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2}}}$

2. The distance of the point P(x, y) from the origin O(0, 0) is given by,

$\boxed{OP = \sqrt{x^2 + y^2}}$

3. The area of a ∆ABC with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is given by,

$\boxed{{Area = \dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|}}$

4. Three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are collinear only when,

$\boxed{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0}$

5. Any point on the x-axis is of the form (x, 0).

6. Any point on the y-axis is of the form (0, y).

Answer 2

Question:-

5. If (– 3, 1), (1, – 3) and (2, 3) then the area of triangle is _______ .

Options:

(a) 14 sq.units.

(b) 7 sq.units.

(c) 28 sq.units.

(d) 35 sq.units.

Given:-

The coordinates of the vertices of the given triangle are:

• A(x₁ = – 3, y₁ = 1).

• B(x₂ = 1, y₂ = – 3).

• C(x₃ = 2, y₃ = 3).

To Find:-

• Area of a triangle.

Solution:-

We know that the area of the triangle whose vertices are  (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

$\sf \large \frac{1}{2} |x_1(y_2 - y_3) + (x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

The given vertices of the triangle are

The given vertices of the triangle are A(– 3, 1), B(1, – 3), and C(2, 3).

So, by using the above formula,

$\sf \large Area \: of \: triangle = \frac{1}{2} | - 3( - 3 - 3) + 1(3 - 1) + 2(1 - ( - 3))|$

$\sf \large Area \: of \: triangle = \frac{1}{2} |18 + 2 + 8|$

$\sf \large Area \: of \: triangle = \frac{1}{2} |28|$

$\sf \large Area \: of \: triangle = {{ \cancel{\frac{28}{2} }}}$

$\sf \large Area \: of \: triangle = 14.$

Answer:-

${ \boxed{ \sf \large \red{ \underline{Hence, the \: area \: of \: triangle \: is (a) 14 \: sq.units. }}}}$

Hope you have satisfied. ⚘