Area of the triangle with vertices P(0, 6), Q(0,2) and R(2, 0) is…………
Answers
Answer:
1/2(2(4))
=1/2*8
4 sq units
Step-by-step explanation:
Given :-
P(0, 6),
Q(0,2),
and R(2, 0)
To find:-
Find the area of the triangle with vertices P(0, 6), Q(0,2) and R(2, 0) ?
Solution:-
Given vertices of a triangle PQR are:
P(0, 6), Q(0,2) and R(2, 0)
Let (x1, y1) = (0,6) => x1 = 0 and y1 = 6
Let (x2, y2) = (0,2) => x2 = 0 and y2 = 2
Let (x3, y3) = (2,0) => x3 = 2 and y3 = 0
We know that
Area of a triangle formed by the points (x1, y1),
(x2, y2) and (x3, y3) is denoted by ∆ and it is
∆ = (1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units
Where the symbol | is modulus
On Substituting these values in the above formula then
=> ∆ = (1/2) | 0(2-0)+0(0-6)+2(6-2) | sq.units
=> ∆ = (1/2) | 0(2)+0(-6)+2(4) |
=> ∆ = (1/2) | 0+0+8 |
=> ∆ = (1/2) | 8 |
=> ∆ = (1/2)×8
=> ∆ = 8/2
=> ∆ = 4 sq.umits
Answer:-
Area of the given traingle PQR is 4 sq.units
Used formulae:-
Area of a triangle formed by the points (x1, y1),
(x2, y2) and (x3, y3) is denoted by ∆ and it is
∆ = (1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units
Where the symbol | is modulus