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If the points $A( k + 1,\ 28),\ B( 3k,\ 2k + 3)$ and $C( 5k - 1,\ 5k)$ are collinear, then find the value of $k$.
Given: Points $A( k + 1,\ 28),\ B(3k, 2k + 3)$ and C( 5k - 1,\ 5k) are collinear.
To do: To find the value of k.
Solution:
Given $A( k+1,\ 2k) ,\ B( 3k,\ 2k+3)$ and $C( 5k-1,\ 5k)$ are collinear.
If three points are collinear then the area of the triangle formed by the given points will be zero,
And we know that area of a triangle with vertices $( x_{1} ,\ y_{1}) ,\ ( x_{2} ,\ y) \ and\ ( x_{3} ,\ y_{3} )$
$\frac{1}{2}[ x_{1}( y_{2} -y_{3}) +x_{2}( y_{3} -y_{1}) +x_{3}( y_{1} -y_{2})]$
by using the above formula, we have
$\frac{1}{2}[( k+1)( 2k+3-5k) +3k( 5k-2k) +( 5k-1)( 2k-2k-3)] =0$
$\Rightarrow ( k+1)( 3-3k) +9k^{2} -15k+3=0$
$\Rightarrow 3k-3k^{2} +3-3k+9k^{2} -15k+3=0$
$\Rightarrow 6k^{2} -15k+6=0$
$\Rightarrow 6\left( k^{2} -\frac{5}{2} k+1\right) =0$
$\Rightarrow k^{2} -2k-\frac{k}{2} +1=0$
$\Rightarrow k( k-2) -\frac{1}{2}( k-2) =0$
$\Rightarrow ( k-2)\left( k-\frac{1}{2}\right) =0$
If $k-2=0$
$\Rightarrow k=2$
If $k-\frac{1}{2} =0$
$\Rightarrow k=\frac{1}{2}$
Thus $k=2$, $\frac{1}{2}$.
Updated on: 10-Oct-2022
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