Find the value of k, if the points A (8, 1) B(3, −4)


Find the value of k, if the points A (8, 1) B(3, −4) and C(2, k) are collinear.


The formula for the area 'A' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula,

$\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are A(8,1), B(3,4) and C(2,k). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\Delta=\frac{1}{2}|(8 \times-4+3 \times k+2 \times 1)-(3 \times 1+2 \times-4+8 \times k)|$

$0=\frac{1}{2}|(-32+3 k+2)-(3-8+8 k)|$

$0=\frac{1}{2}|-25-5 k|$


Hence the value of ' $k$ ' for which the given points are collinear is $k=-5$.

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