Draw the graphs of the following system of linear equations 3x + y = 3; 3x + y = 9.
Find the coordinates of the vertices of the trapezium formed by x-axis, y-axis and these lines. Also, find the
area of the trapezium so formed
Answers
EXPLANATION.
Graph of the equation.
⇒ 3x + y = 3. - - - - - (1).
⇒ 3x + y = 9. - - - - - (2).
As we know that,
From equation (1), we get.
⇒ 3x + y = 3. - - - - - (1).
Put the value of x = 0 in the equation, we get.
⇒ 3(0) + y = 3.
⇒ y = 3.
Their Co-ordinates = (0,3).
Put the value of y = 0 in the equation, we get.
⇒ 3x + (0) = 3.
⇒ 3x = 3.
⇒ x = 1.
Their Co-ordinates = (1,0).
From equation (2), we get.
3x + y = 9. - - - - - (2).
Put the value of x = 0 in the equation, we get.
⇒ 3(0) + y = 9.
⇒ y = 9.
Their Co-ordinates = (0,9).
Put the value of y = 0 in the equation, we get.
⇒ 3x + (0) = 9.
⇒ 3x = 9.
⇒ x = 3.
Their Co-ordinates = (3,0).
To find area of trapezium.
As we know that,
⇒ Area of triangle = 1/2 x Base x Height.
Area of trapezium(AXYC) = Area of triangle(ABC) - Area of triangle(XBY).
Area of triangle(ABC) = 1/2 x 3 x 9.
Area of triangle(ABC) = 27/2 sq. units.
Area of triangle(XBY) = 1/2 x 1 x 3 = 3/2 sq. units.
Area of trapezium(AXYC) = 27/2 - 3/2.
Area of trapezium(AXYC) = 24/2 = 12 sq. units.
Given :-
3x + y = 3; 3x + y = 9.
To Find :-
Coordinates
Solution :-
By putting x = 0 in 1
3(0) + y = 3
0 + y = 3
y = 3
Coordinate = (0,3)
By putting y = 0 in 1
3x + 0 = 3
3x = 3
x = 3/3
x = 1
Coordinate = (1,0)
Now
By putting x = 0 in 2
3(0) + y = 9
0 + y = 9
y = 9
Coordinate = (0,9)
By putting y = 0
3x + 0 = 9
3x = 9
x = 9/3
x = 3
Coordinate = (3,0)
Now
Area = 1/2 × 3 × 9 - 1/2 × 1 × 3
Area = 1/2 × 27 - 1/2 × 3
Area = 1/2(27 - 3)
Area = 1/2 × 24
Area = 12 cm²