Question

Find the area of triangle whose vertices are (0 1) (2,1) (0,3) ​

Answer 1

GIVEN:

Vertices of a triangle = (0,1) (2,1) and (0, 3)

TO FIND:

Area of the triangle.

SOLUTION:

Let the points be A(0,1), B(2,1) and C(0,3)

We know that,

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

From the above vertices,

x1 = 0 ; y1 = 1

x2 = 2 ; y2 = 1

x3 = 0 ; y3 = 3

Substituting the values,

Ar( ∆ ABC ) = 1/2 | 0(1 - 3) + 2(3 - 1) + 0(1 - 1) |

= 1/2 | 0 (-2) + 2(2) + 0(0) |

= 1/2 | 0 + 4 |

= 1/2 | 4 |

= 2

Therefore, the Ar ( ∆ ABC ) is 2 square units.

Answer 2

Step-by-step explanation:

GIVEN:

Vertices of a triangle = (0,1) (2,1) and (0, 3)

TO FIND:

Area of the triangle.

SOLUTION:

Let the points be A(0,1), B(2,1) and C(0,3)

We know that,

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

From the above vertices,

x1 = 0 ; y1 = 1

x2 = 2 ; y2 = 1

x3 = 0 ; y3 = 3

Substituting the values,

Ar( ∆ ABC ) = 1/2 | 0(1 - 3) + 2(3 - 1) + 0(1 - 1) |

= 1/2 | 0 (-2) + 2(2) + 0(0) |

= 1/2 | 0 + 4 |

= 1/2 | 4 |

= 2

Therefore, the Ar ( ∆ ABC ) is 2 square units.

hope this helps you!!