Question 27040: Hello my name is Charles and I have a question how do you fine the area of a triangle with vertices of (x1,y1),(x2,y2) and (x3,y3) I already know that the area = 1/2basexheight but it doesn't have equal sides Thank you for answering this question if you do
Found 2 solutions by longjonsilver, venugopalramana:
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! Use Heron's formula on
http://en.wikipedia.org/wiki/Triangle_(geometry)#Using_coordinates
you will need to find the lengths of the 3 sides, a,b and c using Coordinate geometry. Then put the values in Heron's formula to find the area.
jon.
Answer by venugopalramana(3286)  (Show Source):
You can put this solution on YOUR website! DRAW TRIANGLE ABC.DROP PERPENDICULARS FROM A,B,C TO X AXIS.LET THEM BE AL,BM,CN.NOW WE HAVE 3 TRAPEZIUMS,ALMB,ALNC,BMNC.IF WE TAKE A AS THE TOP VERTEX OF THE TRIANGLE ,B THE NEXT LOWER LEVEL TO IT AND C THE LOWEST...THEN...
AREA OF TRIANGLE ABC =AREA OF TRAPEZIUM ALMB+AREA OF TRAPEZIUM ALNC-AREA OF TRAPEZIUM BMNC
AREA OF TRAPEZIUM IS =(1/2)*(SUM OF PARALLEL SIDES)*HEIGHT...SO WE HET
TRIANGLE ABC AREA = (1/2){(AL+BM)LM+(AL+NC)LN-(BM+CN)MN}
=(1/2){(Y1+Y2)(OM-OL)+(Y1+Y3)(OL-OM)-(Y2+Y3)(ON-OM)}
=(1/2){(Y1+Y2)(X2-X1)+(Y1+Y3)(X1-X2)-(Y2+Y3)(X3-X2)}
=(1/2)|X1,X2,X3|=(1/2)*DETERMINANT OF(X1,X2,X3)
......|Y1,Y2,Y3|.....................(Y1,Y2,Y3)
......|1,1,1...|......................(1,1,1..)
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