Find the value of k for which the points A

Solution:

Given: The points are $A(-5,1), B(1,2)$ and $C(k, 0)$

To find: value of k

$A B=\sqrt{(1+5)^{2}+(2-1)^{2}}=\sqrt{36+1}$

$=\sqrt{37}$ units

$B C=\sqrt{(k-1)^{2}+4}$

$A C=\sqrt{(k+5)^{2}+1}$

Since the points are collinear, $A B+B C=A C$

$\Rightarrow \sqrt{37}+\sqrt{(\mathrm{k}-1)^{2}+4}=\sqrt{(\mathrm{k}+5)^{2}+1}$

Squaring both sides and rearranging,

$\Rightarrow 37+(\mathrm{k}-1)^{2}+4-(\mathrm{k}+5)^{2}-1=-2 \sqrt{37} \sqrt{(\mathrm{k}-1)^{2}+4}$

On simplifying,

$\Rightarrow 40-2 \mathrm{k}+1-10 \mathrm{k}-25=-2 \sqrt{37} \sqrt{(\mathrm{k}-1)^{2}+4}$

$\Rightarrow 16-12 \mathrm{k}=-2 \sqrt{37} \sqrt{(\mathrm{k}-1)^{2}+4}$

$\Rightarrow 8-6 \mathrm{k}=-\sqrt{37} \sqrt{(\mathrm{k}-1)^{2}+4}$

Squaring both sides,

$\Rightarrow 64-96 k+36 k^{2}=37 \times\left(k^{2}-2 k+5\right)$

$\Rightarrow 64-96 k+36 k^{2}=37 k^{2}-74 k+185$

Rearranging,

$\Rightarrow 37 k^{2}-74 k+185=36 k^{2}-96 k+64$

$\Rightarrow \mathrm{k}^{2}+22 \mathrm{k}+121=0$

$\Rightarrow(\mathrm{k}+11)^{2}=0$

$\Rightarrow \mathrm{k}=-11$

Therefore, the value of k for which the points A, B and C are collinear is –11.