Question

If A(6, −1), B(1, 3) and C(k, 8) are three points such that AB = BC, find the value of k.

Answer 1

The given points are A(6, −1), B(1, 3) and C(k, 8).
Also, it is given that AB = BC.
Therefore, (AB)² = (BC)²
$$\begin{gathered} \Rightarrow {\left(1-6\right)}^2+{\left\{3-\left(-1\right)\right\}}^2={\left(k-1\right)}^2+{\left(8-3\right)}^2 \\ \Rightarrow {\left(1-6\right)}^2+{\left(3+1\right)}^2={\left(k-1\right)}^2+{\left(8-3\right)}^2 \\ \Rightarrow {\left(-5\right)}^2+{\left(4\right)}^2={\left(k-1\right)}^2+{\left(5\right)}^2 \\ \Rightarrow 25+16={\left(k-1\right)}^2+25 \\ \Rightarrow 25+16-25={\left(k-1\right)}^2 \\ \Rightarrow {\left(k-1\right)}^2=16 \\ \Rightarrow \left(k-1\right)=\pm \sqrt{16} \\ \Rightarrow k-1=\pm 4 \\ \Rightarrow k-1=4\mathrm{or}k-1=-4 \\ \Rightarrow k=5\mathrm{or}-3 \end{gathered}$$
Therefore, the value of k will be either 5 or −3.