If the points A(2, 3), B(5, k) and C(6, 7) are collinear, then

Question:

If the points A(2, 3), B(5, k) and C(6, 7) are collinear, then

(a) $k=4$

(b) $k=6$

(c) $k=\frac{-3}{2}$

(d) $k=\frac{11}{4}$

 

Solution:

(b) k = 6
The given points are A(2, 3), B(5, k) and C(6, 7).

Here, $\left(x_{1}=2, y_{1}=3\right),\left(x_{2}=5, y_{2}=k\right)$ and $\left(x_{3}=6, y_{3}=7\right)$

Points A,B and C are collinear. Then,

$x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$

$\Rightarrow 2(k-7)+5(7-3)+6(3-k)=0$

$\Rightarrow 2 k-14+20+18-6 k=0$

$\Rightarrow-4 k=-24$

$\Rightarrow k=6$

 

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