MCQ
If the points $A (2,3), B (5, K)$ and $C (6,7)$ are collinear then .
  • A
    $K =4$
  • B
    $K =6$
  • C
    $K =\frac{-3}{2}$
  • $K =\frac{11}{4}$

Answer

Correct option: D.
$K =\frac{11}{4}$
Here, $\left( x _1=2, y _1=3\right),\left( x _2=5, y _2= K \right)$ and $\left( x _3=6, y _3=7\right.$ )
Since the given points are collinear, we must have:
$x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)=0$
$\Rightarrow 2(K-7)+5(7-3)+6(3-k)=0$
$\Rightarrow 2 K-14+20+18-6 K=0$
$\Rightarrow 4 K=24$
$\Rightarrow K=\frac{24}{4}=6$

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