Question

If A(3, 8), B(4, -2) and C(5, -1) are the vertices of Traingle ABC. Then, its area is ??​

Answer 1

Given

⇒A(3,8)=x1,y1

⇒B(4,-2)=x2,y2

⇒C(5,-1)=x3,y3

∴ By formula of area of triangle

$\frac{1}{2} x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)$

$\frac{1}{2} \times 3( - 2 - (- 1)) + 4( - 1 - 8) + 5(8 - (- 2 ))$

$\frac{1}{2} \times 3( - 2 + 1) + 4( - 9) + 5(10)$

$\frac{1}{2} \times - 3 - 36 + 50$

$\frac{1}{2} \times (- 39 + 50)$

$\frac{1}{2} \times 11$

$\frac{11}{2} = 5.5$

Answer 2

Answer: 5.5 sq units.

Explanation:

Coordinates of the vertices:

• A(3, 8)
• B(4, - 2)
• C(5, -1)

We know,

∆ = ½ | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |

=> ∆ = ½ | 3[- 2 - (- 1)] + 4[- 1 - 8] + 5[8 - (- 2)] |

=> ∆ = ½ | 3(- 1) - 36 + 50 |

=> ∆ = ½ | - 3 - 36 + 50 |

=> ∆ = ½ | 11 | = 5.5 sq units.