# 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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Updated on: 10-Oct-2022

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