Question

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

A. 1/2 sq. unit

B. 1 sq. unit

C. 2 sq. unit

D. None of these

Answer 1

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 = \(\frac{1}{2}\) 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 = \(\frac{1}{2}\) x| 3 x (4– 3) + 4 x (3 – 4) + 3 x (4– 4) |

Area = \(\frac{1}{2}\) x | 3 x (1) + 4 x ( – 1) + 3 x (0) |

Area = \(\frac{1}{2}\) x | 3– 4 |

⇒ Area = \(\frac{1}{2}\) sq. units

The Area of the triangle is \(\frac{1}{2}\) sq. units

Answer 2

Correct option is (A) \(\frac 12\) square units

Given x = 3, y = 4, x = y

We have plotting points as (3, 4), (3, 3), (4, 4) when x = y

Therefore, area of ΔABC = \(\frac 12\)(base × height)

= \(\frac 12\)(AB × AC)

= \(\frac 12\)(1 × 1)

= \(\frac 12\)

∴ Area of ΔABC = \(\frac 12\) square units.