Find the values of x if | x-1 2 0 2 x
-4 0 0 0 x-3 |=0
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Answer:
Given (a,b),(a
1
,b
1
) and (a−a
1
,b−b
1
)
We know the points are collinear if area (ΔABC)=0
Area of triangle =
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
Then, Area =
2
1
{a(b
1
−(b−b
1
)+a
1
{(b−b
1
)−b}+(a−a
1
)(b−b
1
)}=0
2
1
{ab
1
−(ab−ab
1
)+a
1
b−a
1
b
1
−a
1
b}+(ab−ab
1
−a
1
b+a
1
b
1
)=0
{ab+a
1
b
1
}=ab−ab
1
−a
1
b+a
1
b
1
−ab
1
−a
1
b=0
−ab
1
=a
1
b
Therefore, We can write as
a
1
a
=
b
1
b
Hence, proved.
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