Find the value of y for which the distance between the points $A( 3,\ -1)$ and $B( 11,\ y)$ is 10 units.

Given: Two points $\displaystyle A( 3,\ -1)$ and $\displaystyle B( 11,\ y)$ and distance between the two given points $=10$ units.

To do: To find out the value of y.

we know if there two points $(x_{1} ,\ y_{1} )$ and $(x_{2} ,\ y_{2} )$,

distance between the two points,$=\sqrt{( x_{2} -x_{1})^{2} +( y_{2} -y_{1})^{2}}$

Here on substituting the value of $A( 3,\ -1)$ and $B( 11,\ y$) and $AB=10\ units$

$\sqrt{( 3-11)^{2} +( -1-y)^{2})} =10$

$\Rightarrow \sqrt{( -8)^{2} +( -( 1+y)^{2} \ } =10$

$\Rightarrow \sqrt{( 64+( 1+y)^{2} \ } =10$

$\Rightarrow 64+( 1+y)^{2} =( 10)^{2}$

$\Rightarrow ( 1+y)^{2} =100-64=36$

$\Rightarrow 1+y=\pm \sqrt 36$

$\Rightarrow 1+y=\pm 6$

If $1+y=6$

$\Rightarrow y=6-1=5$

if $1+y=-6$

$\Rightarrow y=-6-1=-7$

$Therefore\ y=5,\ -7$