# The area of the triangle formed by the lines 2x + 3y = 12, x – y – 1 = 0 and x = 0 (as shown in Fig. 3.23) , isA. 7 sq. unitsB. 7.5 sq. unitsC. 6.5 sq. unitsD. 6 sq. units

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 x1 ,y1 are the coordinates of P1

x2, y2 are the coordinates of P2

x3 ,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

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