Question

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

Answer 1

Answer:

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 = 1 2 12 x | x1 x (y2 – y3) + x2 x (y3 – y1) + x3 x (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 = 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.

Answer 2

Answer:

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 = ½ x | x¹ x (y² - y³) + x² x (y³ - y¹) + x³ x (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 = ½ × | 3 × (4 + 1) + 0 x (-1 -2) + 0 x (2 - 4) |

⇒ Area of the Given Triangle = ½ × | 3 ½ 5|

⇒ Area = 15/2

⇒ Area = 7.5 Sq.

Units Area of the triangle is 7.5 sq. Units