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Ali points inside the triangle with vertices at $(1,3),(5,0)$ and $(-1,2)$ must necessarily satisfy.
Options:
Solution:
1307 Upvotes
Verified Answer
The correct answer is:
$3 x+2 y>0$
Let $A \equiv(1,3), B \equiv(5,0), C \equiv(-1,2)$
For point $A(1,3), 3 x+2 y=3+6=9>0$
For point $B(5,0), 3 x+2 y=15+0=15>0$
For point $C(-1 ; 2), 3 x+2 y=-3+4=1>0$
$\therefore$ Option (a) is rejected.
Now, for option (c) $2 x-3 y-12=2-9-12 < 0$
$\therefore$ Option $(\mathrm{c})$ is rejected.
Also, $2 x+y-13=4+3-13 < 0$
$\therefore$ Option (d) is rejected.
So, option (b) is correct.
For point $A(1,3), 3 x+2 y=3+6=9>0$
For point $B(5,0), 3 x+2 y=15+0=15>0$
For point $C(-1 ; 2), 3 x+2 y=-3+4=1>0$
$\therefore$ Option (a) is rejected.
Now, for option (c) $2 x-3 y-12=2-9-12 < 0$
$\therefore$ Option $(\mathrm{c})$ is rejected.
Also, $2 x+y-13=4+3-13 < 0$
$\therefore$ Option (d) is rejected.
So, option (b) is correct.
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