Question

solve graphically 4x - 3y + 4 = 0 , x + 2y + 7 = 0 find the area bounded of these lines 1. x- axis 2. y - axis​

Answer 1

EXPLANATION.

Graphically.

⇒ 4x - 3y + 4 = 0. - - - - - (1).

⇒ x + 2y + 7 = 0. - - - - - (2).

As we know that,

From equation (1), we get.

⇒ 4x - 3y + 4 = 0. - - - - - (1).

Put the value of x = 0 in the equation, we get.

⇒ 4(0) - 3y + 4 = 0.

⇒ - 3y + 4 = 0.

⇒ - 3y = - 4.

⇒ y = 4/3.

⇒ y = 1.33.

Their Co-ordinates = (0,1.33).

Put the value of y = 0 in the equation, we get.

⇒ 4x - 3(0) + 4 = 0.

⇒ 4x + 4 = 0.

⇒ x = - 1.

Their Co-ordinates = (-1,0).

From equation (2), we get.

⇒ x + 2y + 7 = 0. - - - - - (2).

Put the value of x = 0 in the equation, we get.

⇒ (0) + 2y + 7 = 0.

⇒ 2y + 7 = 0.

⇒ 2y = - 7.

⇒ y = - 7/2.

⇒ y = - 3.5.

Their Co-ordinates = (0,-3.5).

Put the value of y = 0 in the equation, we get.

⇒ x + 2(0) + 7 = 0.

⇒ x + 7 = 0.

⇒ x = - 7.

Their Co-ordinates = (-7,0).

As we know that,

Area of the triangle = 1/2 x Base x Height.

Base = - 1 - (-7) = - 1 + 7 = 6.

Height = -2.182.

Area of the triangle = 1/2 x 6 x (-2.182).

Area of the triangle = (-13.092)/2 = -6.546 sq. units.

Answer 2

Answer:

$\tt 4(0) - 3y + 4=0$

$\tt 0 - 3y+4=0$

$\tt 0-3y=-4$

$\sf -3y=-4$

$\tt y = \dfrac{-4}{-3}$

$\tt y = 1.33$

$\tt 4x-3(0)=-4$

$\tt 4x-0=-4$

$\tt 4x=-4$

$\tt x = \dfrac{-4}{4}$

$\tt x=-1$

$\sf x+2y = 0-7$

$\sf 0+2y=-7$

$\sf 2y=-7$

$\tt \dfrac{-7}{2}=y$

$\tt -3.5=y$

$\tt x+2(0)=-7$

$\tt x+0=-7$

$\tt x=-7$

$\bf Area = -6.546 sq. units.$