Question

Question 38
Find the Area of a Triangle whose vertices
are (-3, 1), (1, -3) and (2,3)
7 Sq.Units
9 Sq.Units
11 Sq.Units
14 Sq.Units​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given:-

Given points are (-3, 1), (1, -3) and (2,3)

To find:-

Find the Area of a Triangle whose vertices

are (-3, 1), (1, -3) and (2,3) ?

Solution:-

Given points are (-3, 1), (1, -3) and (2,3)

Let (x1, y1)=(-3,1)=>x1=-3 and y1 = 1

Let (x2, y2)=(1,-3)=>x2=1 and y2 = -3

Let (x3, y3)=(2,3)=>x3=2 and y3=3

We know that

The Area of a triangle formed by whose vertices are( x1 ,y1 )and (x2, y2 )and (x3, y3) is

∆=(1/2) | x1(y2-y3) +x2(y3-y1) + x3(y1-y2) | sq.units

On Substituting the values in the above formula

=>∆=(1/2) | (-3)(-3-3)+(1)(3-1)+(2)(1-(-3)) | sq.units

=>∆=(1/2) | (-3)(-6)+(1)(2)+(2)(4) | sq.units

=>∆=(1/2) | 18+2+8 | sq.units

=>∆=(1/2) | 28 | sq.units

=>∆=28/2 sq.units

=>∆=14 sq.units

Answer:-

The area of the given triangle formed by the given vertices is 14 sq.units

Used formulae:-

1)

The Area of a triangle formed by whose vertices are( x1 ,y1 )and (x2, y2 )and (x3, y3) is

∆=(1/2) | x1(y2-y3) +x2(y3-y1) + x3(y1-y2) | sq.units