Find the coordinates of the vertices and the area of the triangle enclosed by the axis and the graphs of x + 3y 12 and x - 3y =0.
Answers
EXPLANATION.
Linear equation in two variable.
⇒ x + 3y = 12. - - - - - (1).
⇒ x - 3y = 0. - - - - - (2).
As we know that,
From equation (1),
⇒ x + 3y = 12. - - - - - (1).
Taking y - axis it means x = 0.
Put the values of x = 0 in the equation, we get.
⇒ (0) + 3y = 12.
⇒ 3y = 12.
⇒ y = 4.
Their Co-ordinates = (0,4).
Taking x - axis it means y = 0.
Put the values of y = 0 in the equation, we get.
⇒ x + 3(0) = 12.
⇒ x = 12.
Their Co-ordinates = (12,0).
From equation (2),
⇒ x - 3y = 0. - - - - - (2).
Taking y - axis it means x = 0,
Put the values of x = 0 in the equation, we get.
⇒ (0) - 3y = 0.
⇒ y = 0.
Their Co-ordinates = (0,0).
Taking x - axis it means y = 0.
Put the value of y = 0 in the equation, we get.
⇒ x - 3(0) = 0.
⇒ x = 0.
Their Co-ordinates = (0,0).
Co-ordinates of the vertices of the triangle = (0,0), (12,0), (6,2).
Area of triangle = 1/2 x base x height.
Base = 12 cm.
Height = 2 cm.
Using this formula in the equation, we get.
⇒ 1/2 x 12 x 2 = 12.
Area of the triangle = 12 sq. units.
