(0,1/2) (1/2,1/2) (1/2,0) are the mid ponts of The sides of a triangle. Its area in square units is :
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Answers
Step-by-step explanation:
Given :-
(0,1/2) (1/2,1/2) (1/2,0) are the mid ponts of the sides of a triangle.
To find :-
Find Its area in square units ?
Solution :-
Given that
(0,1/2) (1/2,1/2) (1/2,0) are the mid ponts of the sides of a triangle.
Let Consider ABC triangle
Let A = (x1, y1)
Let B = (x2, y2)
Let C = (x3, y3)
Let the mid point of AB = D(0,1/2)
We know that
The mid point of a line segment joining the points (x1, y1) and (x2, y2)
= ( ( x1+x2)/2 , (y1+y2)/2 )
Now
=> ( ( x1+x2)/2 , (y1+y2)/2 ) = (0,1/2)
On comparing both sides then
=> (x1+x2)/2 = 0 and (y1+y2)/2 = 1/2
=> x1+x2 = 0 and y1+y2 = 1
=> x1 = -x2 --------(1)
and y1 +y2 = 1 ------(2)
and
Let the mid point of BC = E(1/2,1/2)
The mid point of a line segment joining the points (x2 , y2 and (x3, y3)
= ( ( x2+x3)/2 , (y2+y3)/2 )
Now
( ( x2+x3)/2 , (y2+y3)/2 ) = (1/2,(1/2)
On comparing both sides then
=> (x2+x3)/2 = 1/2 and (y2+y3)/2 = 1/2
=> x2+x3 = 1 ---------(3)
and y2+y3 = 1--------(4)
and
Let the mid point of AC = F(1/2,0)
The mid point of a line segment joining the points (x1 , y1 and (x3, y3)
= ( ( x1+x3)/2 , (y1+y3)/2 )
Now
( ( x1+x3)/2 , (y1+y3)/2 ) = (1/2,0)
On comparing both sides then
=> (x1+x3)/2 = 1/2 and y1+y3/2 = 0
=> x1+x3 = 1 and y1+y3 = 0
=> x1+x3 = 1 -----------(5)
y1+y3 = 0
=> y1 = -y3 -------------(6)
From (1) and (5)
-x2+x3 = 1 ----------(7)
From (3) and (7)
x2+x3 = 1
-x2+x3 = 1
(+)
_________
0+2x3 = 2
________
=> 2x3 = 2
=> x3 = 2/2
=> x3 = 1
and From (7)
=> -x2 +1 = 1
=> -x2 = 1-1
=> x2 = 0
From (1)
x1 = -(0)
=> x1 = 0
From (2)&(6)
=> -y3 +y2 = 1 ----------(8)
From (4)&(8)
y2+y3 = 1
-y3 +y2 = 1
(+)
_________
2y2+0 = 2
________
=> 2y2 = 2
=> y2 = 2/2
=> y2 = 1
From (4)
=> 1+y3 = 1
=> y3 = 1-1
=> y3 = 0
From (6)
y1 = -0
=> y1 = 0
Therefore,
A (x1,y1) = (0,0)
B (x2, y2) = (0,1)
C (x2, y3) = (1,0)
Now
We know that
The area formed by the points (x1, y1) , (x2, y2) and (x2,y3) is defined by
∆ = (1/2)|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| sq.units
On substituting these values in the above formula
=> ∆ = (1/2) | 0(1-0)+0(0-0)+1(0-1) | sq.units
=> ∆ = (1/2) | 0(1)+0(0)+1(-1) |
=> ∆ = (1/2) | 0+0-1 |
=> ∆ = (1/2) | 0-1 |
=> ∆ = (1/2) | -1 |
=> ∆ = (1/2)×1
=> ∆ = 1/2
=> ∆ = 1/2 sq.units
Answer:-
The area of the given triangle is 1/2 sq.units
Used formulae:-
Mid Point formula :-
→ The mid point of a line segment joining the points (x1, y1) and (x2, y2)
= ( ( x1+x2)/2 , (y1+y2)/2 )
Area of a triangle formula :-
→ The area formed by the points (x1, y1) , (x2, y2) and (x2,y3) is defined by
∆ = (1/2)|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| sq.units