Question

Prove that the area of a triangle with vertices (t, t-2),(t+2, t+2) and (t+3) is independent of t.​

Answer 1

error in ur question.......

Answer 2

$\large{ \underline{Solution\: :-}}$

There is a mistake in the question

It should be Prove that the area of a triangle with vertices (t, t-2),(t+2, t+2) and (t+3 , t) is independent of t.​

$\textbf{ Now as we know :- }$

Area of triangle = 1/2 × | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |

= 1/2 × | t((t + 2) - t) + (t + 2)(t - (t - 2)) + (t + 3)((t - 2) - (t + 2)) |

= 1/2 × | t(2) + (t +2)(2) + (t + 3)(-4)|

= 1/2 × | 2t + 2t + 4 - 4t - 12 |

= 1/2 × | 4t - 4t + 4 - 12 |

= 1/2 × | -8 |

= 1/2 × 8

= 4 unit square

Hence proved that it is independent of "t"