find the area of the triangle whose vertices are (4,7) (1,3) (5,1)
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Let A(4,7), B(1,3), C(5,1)
Here , x1= 4, y1=7
x2=1, y2=3
x3=5, y3=1
Area of ∆ = 1/2 [(x1(y2-y3) +x2(y3-y1)+ x3(y1-y2)
Area of ∆ = 1/2[ 4(3-1)+1(1-7)+5(7-3)
= 1/2[ 4×2+1×-6+5×4]
= 1/2[8-6+20]
=1/2[ 2+20]
=1/2[22]
= 22/2
= 11
Hence, area of ∆ = 11 sq.units
==================================================================
Hope this will help you...
Here , x1= 4, y1=7
x2=1, y2=3
x3=5, y3=1
Area of ∆ = 1/2 [(x1(y2-y3) +x2(y3-y1)+ x3(y1-y2)
Area of ∆ = 1/2[ 4(3-1)+1(1-7)+5(7-3)
= 1/2[ 4×2+1×-6+5×4]
= 1/2[8-6+20]
=1/2[ 2+20]
=1/2[22]
= 22/2
= 11
Hence, area of ∆ = 11 sq.units
==================================================================
Hope this will help you...
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