Asked by vishal
Find the area of the triangle whose vertices are (2,3),(3,4),(-2,-4).
Answers
Answered by
Reiny
easiest way:
list the points in a column, repeat the one you started with
2 3
-2 -4
3 4
2 3
area = (1/2)[ sum of downproducts - sum of up-products]
= (1/2)[-8 -8 + 9 - (-6 -12 +8) ]
= (1/2)( -7 + 10]
= 3/2
or
label the points A(2,3) , B(3,4) and C(-2,-4)
BC = √(25 + 64) = √89
equation of BC:
slope BC = -8/-5 = 8/5
using the point (3,4)
y-4 = (8/5)(x-3)
5y - 20 = 8x - 24
8x - 5y - 4 = 0
consider BC as the base and we need the height from A to BC
distance = |8(2) -5(3)-4\/√(64+25) = 3/√89
area = (1/2)(√89)(3/√89)
= 3/2
list the points in a column, repeat the one you started with
2 3
-2 -4
3 4
2 3
area = (1/2)[ sum of downproducts - sum of up-products]
= (1/2)[-8 -8 + 9 - (-6 -12 +8) ]
= (1/2)( -7 + 10]
= 3/2
or
label the points A(2,3) , B(3,4) and C(-2,-4)
BC = √(25 + 64) = √89
equation of BC:
slope BC = -8/-5 = 8/5
using the point (3,4)
y-4 = (8/5)(x-3)
5y - 20 = 8x - 24
8x - 5y - 4 = 0
consider BC as the base and we need the height from A to BC
distance = |8(2) -5(3)-4\/√(64+25) = 3/√89
area = (1/2)(√89)(3/√89)
= 3/2
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