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If the points $A( -2,\ 1) ,\ B( a,\ b)$ and $C( 4,\ -1)$ are collinear and $a-b=1$, find the values of $a$ and $b$.
Given: points $A( -2.\ 1) ,\ B( a,\ b)$ and $C( 4,\ -1)$ are collinear and $a-b\ =1$
To do: To find the value of a and b.
Solution:
The given points are $A ( -2,\ 1),\ B ( a,\ b)$ and $C( 4,\ -1)$. Since the given points are collinear, the area of the triangle ABC is 0.
And we know that area of a triangle with vertices $( x_{1} ,\ y_{1}) ,\ ( x_{2} ,\ y_{2})$ and $( x_{3} ,\ y_{3} )$
$\frac{1}{2}[ x_{1}( y_{2} -y_{3}) +x_{2}( y_{3} -y_{1}) +x_{3}( y_{1} -y_{2})]$
Here on subtituting the values in the formula,
$\frac{1}{2}[ -2( b+1) +a( -1-1) +4( 1-b)] =0$
$\Rightarrow -2b-2-2a+4-4b=0$
$\Rightarrow -2a-6b+2=0$
$\Rightarrow -2( a+2b) =-2$
$\Rightarrow a+2b=1\ ...............( 1)$
And given $a-b=1..........( 2)$
On subtracting $( 2)$ from $( 1)$ ,
$a+2b-a+b=1-1$
$\Rightarrow 3b=0$
$\Rightarrow b=0$,
On subtituting $b=0$ in $( 1)$
$a-0=1$
$\Rightarrow a=1$
Thus we have $a=1,\ b=0$
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