Given

Equation 1: 2x + 3y = 12 Equation 2: x – y = 1 Equation 3: x = 0

To calculate the Area at first we solve the Equation 1 & 2 Simultaneously by method of substitution.

We substitute the value of x from Equation 2 in Equation 1 to get the value of y

Equation 2: x – y = 1 ⇒ x = y + 1

Equation 1: 2x + 3y = 12

Substituting the value from equation 2 we get

2(y + 1) + 3y = 12 ⇒ 2y + 2 + 3y = 12 ⇒ 5y = 10 ⇒ y = 2

Putting the value in Equation 1 we get

x = 2 + 1 ⇒ x = 3

So both this lines passes through ( 3 , 2) Let this Coordinate name be P₁

Equation 3 is the equation for y axis

Equation 1 meets Y axis at (0 ,4) which is calculated by substituting x = 0 in Equation 1. Let this Coordinate name be P₂

Equation 2 meets Y axis at (0 , – 1) which is calculated by substituting x = 0 in Equation 2. Let this Coordinate name be P₃

So Area of the triangle = *| x₁ * (y₂ – y₃) + x₂ * (y₃ – y₁) + x₃ * (y₁ – y₂) |

Where x_{1 ,}y₁ are the coordinates of P₁

x_2, y₂ are the coordinates of P₂

x_{3 ,}y₃ are the coordinates of P₃

⇒ Area of the Given Triangle = *| 3* (4+ 1) + 0 * ( – 1 – 2) + 0* (2– 4) |

⇒ Area of the Given Triangle = *| 3* 5 |

⇒ Area = ⇒ Area = 7.5 Sq. Units

Area of the triangle is 7.5 sq. Units