Question

Find values of ${}^{\prime }{\alpha }^{\prime }$ if distance between $A(3,2)$ and $B(\alpha +3,4-\alpha )$ is $2$.

Answer 1

Distance between 2 points X(a,b) and Y(c,d) is given by:

$\sqrt{{\left(c-a\right)}^2+{\left(d-b\right)}^2}$

So, distance between 2 points A(3,2) and $B\left(\alpha +3,4-\alpha \right)$ is given by:

$\begin{array}{c}\sqrt{{\left(\left(\alpha +3\right)-3\right)}^2+{\left(\left(4-\alpha \right)-2\right)}^2}=\sqrt{{\alpha }^2+{\left(2-\alpha \right)}^2}\\ =\sqrt{{\alpha }^2+{\alpha }^2+4-4\alpha }\\ =\sqrt{2{\alpha }^2+4-4\alpha }\end{array}$

As distance is 2

So,

$\sqrt{2{\alpha }^2+4-4\alpha }=2$

Squaring both the sides,

$\begin{array}{c}2{\alpha }^2+4-4\alpha =4\\ 2{\alpha }^2-4\alpha =0\\ \alpha =0,2\end{array}$