Find the area of the triangle whose vertices are (5, 2) (3, −5) and (−5, −1)
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Area can't be negative. So, Area of triangle = 32 sq. Unit
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Solution :
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
••••
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
••••
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