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5 years ago
#21
(Original post by BWV1007)
Honestly... I think you can just use the formula

Area of triangle =
| x1 x2 x3 x1 |
| y1 y2 y3 y1 |

= 1/2[(x1 x y2) + (x2 x y3) + (x3 x y1) - (y1 x x2) - (y2 x x3) - (y3 x x1)]
ARE YOU SERIOUS

The base = 7
The height = 3

What on earth are you suggesting the above for?
0
5 years ago
#22
(Original post by TenOfThem)
ARE YOU SERIOUS

The base = 7
The height = 3

What on earth are you suggesting the above for?
Observe.

= 1/2 |(0 x -7) + (0 x -6) + (3 x 0) - (0 x 0) - (-7 x 3) - (-6 x 0)|
= 1/2 |-(-7x3)|
= 1/2 (21)
= 10.5 units^2
0
5 years ago
#23
(Original post by BWV1007)
Observe.

= 1/2 |(0 x -7) + (0 x -6) + (3 x 0) - (0 x 0) - (-7 x 3) - (-6 x 0)|
= 1/2 |-(-7x3)|
= 1/2 (21)
= 10.5 units^2
I know it works

I am asking why you are suggesting something beyond the requirement of the OP

They are studying C1 at A Level and the question they posed is solvable directly from looking at the co-ordinates

So - why overcomplicate
1
5 years ago
#24
Bloody hell, I don't think this needs to be over complicated. Draw a diagram and then just work out the area of the triangle! No pythagoras is needed!
1
5 years ago
#25
(Original post by TenOfThem)
incorrectly, I assume

there is no need for Pythagoras in this question

I didn't use pythagoras
0
5 years ago
#26
(Original post by techno-thriller)
Find the lengths between the points

I.E distance= (x1-x2)^2+(y1-y2)^2 do for all sides the put each number on their respective side and use 1\2height x base
(Original post by techno-thriller)

I didn't use pythagoras

ok
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