Question

The area of the triangle formed by the lines 2x + 3y = 12, x – y -1 = 0 and x = 0 ​

Answer 1

Answer:

7.5 sq. Units

Step-by-step explanation:

Given Equation 1: 2 x + 3 y = 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: 2 x +3 y = 12 Substituting the value from equation 2 we get 2(y + 1) + 3 y = 12 ⇒ 2 y + 2 + 3 y = 12 ⇒  5 y = 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 Pb 1 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 P 2 Equation 2 meets Y axis at (0 , – 1) which is calculated by substituting x = 0 in Equation 2. Let this Coordinate name be P 3 So Area of the triangle =  1 2 12 x | x 1 x (y 2 –  y 3) + x 2 x (y 3 – y 1) + x 3 x (y 1 – y 2) | Where x 1 ,y 1 are the coordinates of P 1 x 2, y 2 are the coordinates of P 2 x 3 ,y 3 are the coordinates of P 3 ⇒ Area of the Given Triangle =  1 2 12 x | 3 x (4 +1) + 0 x ( – 1 – 2) + 0 x (2– 4) | ⇒ Area of the Given Triangle =  1 2 12 x | 3  1 2 12 5 | ⇒ Area =  15 2 152 ⇒ Area = 7.5 Sq. Units Area of the triangle is 7.5 sq. Units

Answer 2

Answer:

7.5 sq.units

Step-by-step explanation:

Given Equation 1: 2 x + 3 y = 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: 2 x +3 y = 12 Substituting the value from equation 2 we get 2(y + 1) + 3 y = 12 ⇒ 2 y + 2 + 3 y = 12 ⇒  5 y = 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 Pb 1 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 P 2 Equation 2 meets Y axis at (0 , – 1) which is calculated by substituting x = 0 in Equation 2. Let this Coordinate name be P 3 So Area of the triangle =  1 2 12 x | x 1 x (y 2 –  y 3) + x 2 x (y 3 – y 1) + x 3 x (y 1 – y 2) | Where x 1 ,y 1 are the coordinates of P 1 x 2, y 2 are the coordinates of P 2 x 3 ,y 3 are the coordinates of P 3 ⇒ Area of the Given Triangle =  1 2 12 x | 3 x (4 +1) + 0 x ( – 1 – 2) + 0 x (2– 4) | ⇒ Area of the Given Triangle =  1 2 12 x | 3  1 2 12 5 | ⇒ Area =  15 2 152 ⇒ Area = 7.5 Sq. Units Area of the triangle is 7.5 sq. Units