If $(k,2),(2,3)$ and $(3,4)$ are collinear, then the value of  $\text{'}k\text{'}$ is

Given $A(k,2),B(2,3),C(3,4)$ are collinear

$AB=\sqrt{{(k-2)}^2+{(3-2)}^2}=\sqrt{k^2+4-4k+1}=\sqrt{k^2-4k+5}$

$BC=\sqrt{{(3-2)}^2+{(4-3)}^2}=\sqrt{2}$

$AC=\sqrt{{(k-3)}^2+{(2-4)}^2}=\sqrt{k^2+9-6k+4}=\sqrt{k^2-6k+13}$

$AB+BC=AC$

$\sqrt{k^2-4k+5}+\sqrt{2}=\sqrt{k^2-6k+13}$

$k^2-4k+5+2+2\sqrt{(k^2-4k+5)2}=k^2-6k+13$

$2\sqrt{(k^2-4k+5)2}=-2k+6$

$\sqrt{(k^2-4k+5)2}=-k+3$

$2k^2-8k+10=k^2-6k+9$

$k^2-2k+1=0$

${(k-1)}^2=0$

$k=1$

(or) 2^nd method

Slope of AB= slope of BC

$slope=\frac{y_2-y_1}{x_2-x_1}$

$\frac{3-2}{2-k}=\frac{4-3}{3-2}=1$

$\frac{3-2}{2-k}=1$

$2-k=1$

$k=1$

Slope $AB=$Slope of BC

$\frac{3-2}{2-k}=\frac{4-3}{3-2}$

$\frac{1}{2-k}=\frac{1}{1}$

$2-k=1$

$k=1$