It helps to Find the Area Of triangle when Vertices of triangle are given...this presentation is Made By Mohan Pudasaini ...To find Me follow the link in Facebook|: www.facebook.com/mohanpud .

1. Mr.MOHAN PUDASAINI

2. Mr.MOHAN PUDASAINI
zLif{laGb'x? Yfxf ePsf cj:yfdf lqe'sf]
Y
Ô]qkmn kQf nufpg] A(x1,y1
)
lrqdf ABC lqe'h 5 h;sf] C(x3,y3
A(x1,y1), B(x2,y2) / C(x3,y3) 5g .
lgb]{zfªs )
B(x2,y2
)
lqe'h ABC sf]
Ô]qkmn = ?
X’ X
Y’

3. Mr.MOHAN PUDASAINI
zLif{laGb'x? Yfxf ePsf cj:yfdf lqe'sf]
Y
Ô]qkmn kQf nufpg] A(x1,y1
)
C(x3,y3
X-cÔdf A,B,C af6 AD, BM / CN )
nDa lvrf}“ . B(x2,y2
)
ABC= ;nr (ABMD +ADCN -BMNC )
X’ X
M D N
Y’

4. Mr.MOHAN PUDASAINI
zLif{laGb'x? Yfxf ePsf cj:yfdf lqe'sf]
Ô]qkmn kQf nufpg] Y A(x1,y1
)
C(x3,y3
ABC= ;nr (ABMD +ADCN -BMNC ) B(x2,y2 )
)
=½(MB+DA)MD 1/2(DA+NC)DN ½(MB+NC)MN
+ –
=1/2[(y2+y1)(x1-x2)+(y1+y3)(x3-x1) (y2+y3)(x3-x2)]
;/n k5L -
=1/2[x1y2-x2y1)+(x2y3-x3y2)+(x3y1-x1y3)]
Ô]qkmn wgfTds x'g]
ePsf]n]
ABC=1/2 [x1y2-x2y1)+(x2y-x3y2)+(x3y1-x1y3)]
X
X’ M D N
Y’

5. Mr.MOHAN PUDASAINI
x1 x2 x3 x1
y1 y2 y3 y1
lqe'hsf] (x1y2 x2y1) +(x2y3 x3y2 )+(x3y1 x1y3 )
Ô]qkmn =1/2 - - -

6. Mr.MOHAN PUDASAINI
(2,-3), (3,2) / (-2,5) lzif{laGb'x? ePsf] lqe'hsf]
Ô]qkmn slt xG5<
2 3 -2 2
-3 2 5 -3
lqe'hsf] {4 -(-9)} + - (-4)}+{6 - 10} ju{ PsfO{
{15
Ô]qkmn =1/2=1/2 {4+9} + {15+4} + {-4} ju{ PsfO{
=1/2 13+19+4 ju{ PsfO{
=14 ju{ PsfO{