The area of the triangle formed by the lines x = 3, y = 4 and x = y is

Given :

Equation 1: x = 3

Equation 2: y = 4

Equation 3: x = y

Equation 1 is a line parallel to y axis

Equation 2 is a line parallel to x axis

So Equation 1 & 2 are mutually perpendicular to each other.

Hence the triangle formed is a right angled triangle.

First we solve the three lines simultaneously by method of substitution and get the three points of intersection or three coordinates of the triangle.

Solving Equation 1 & 2 we get the coordinate ( 3, 4 ). Let this Coordinate name be P₁

Solving Equation 2 & 3 we get the coordinate ( 4 ,4 ). Let this Coordinate name be P₂

Solving Equation 3 & 1 we get the coordinate ( 3 ,3 ). Let this Coordinate name be P₃

We now use the formula for Area of a triangle through 3 given points

Area = *| 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– 3) + 4 * (3 – 4) + 3* (4– 4) |

Area = *| 3* (1) + 4 * ( – 1) + 3* (0) |

Area = *| 3– 4 |⇒ Area = sq. units

The Area of the triangle is sq. units