Question

Find the area of the triangle whose vertices are
(-5, -1), (3, -5), (5, 2)

Answer 1

Answer:

Let A( x1 , y1 ) = ( 5 , 2 ) ,

B( x2 , y2 ) = ( 3 , -5 ) ,

C( x3 , y3 ) = ( -5 , -1 ) are vertices of

∆ABC ,

Area of ∆ABC

= 1/2|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|

=1/2|5[(-5)-(-1)]+3[-1-2]+(-5)[2-(-5)] |

= 1/2| 5(-5+1)+3(-3)+(-5)(2+5) |

= 1/2| -20 - 9 - 35 |

= 1/2| -64 |

= 64/2

= 32

Therefore ,

Area of ∆ABC = 32 Square units

Answer 2

Step-by-step explanation:

area of triangle = 1/2 [ x1 ( y2-y3) + x2 ( y3 - y1 ) + x3 ( y1-y2 )]

= 1/2 [ (-5) ( -5-2) + 3 ( 2 -(-1) ) + 5 ( -1-(-5) )]

= 1/2 [ (-5) (-7) + 3(2) + 5 ( 5 ) ]

= 1/2 [ 35 + 6 + 25 ]

= 1/2 ( 66 )

= 33 sq.units