Question

If the distance between A(9,x) and B(15,11) is 10 units then find x. ​

Answer 1

GIVEN :–

• The distance between A(9,x) and B(15,11) is 10 units.

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

• If two points are $\:\: { \bold{A(x_{1} , y_{1})}} \:\:$ and $\:\: { \bold{B(x_{2} , y_{2})}} \:\:$ then Distance –

$\\ \dashrightarrow \: \large{ \boxed{ \bold{AB = \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} }}}}$

• Here –

$\\ \:\: { \huge{.}} \: \: { \bold{x_{1} = 9}} \:\:$

$\\ \:\: { \huge{.}} \: \: { \bold{x_{2} = 15}} \:\:$

$\\ \:\: { \huge{.}} \: \: { \bold{y_{1} = x}} \:\:$

$\\ \:\: { \huge{.}} \: \: { \bold{y_{2} = 11}} \:\:$

• Now put the values –

$\\ \implies \:{ \bold{10 = \sqrt{ {(15 - 9)}^{2} + {(11 - y)}^{2} }}}$

$\\ \implies \:{ \bold{10 = \sqrt{ {(6)}^{2} + {(11 - x)}^{2} }}}$

$\\ \implies \:{ \bold{10 = \sqrt{36+ {(11 - x)}^{2} }}}$

• Square on both sides –

$\\ \implies \:{ \bold{ {(10)}^{2} = {36+ {(11 - x)}^{2} }}}$

$\\ \implies \:{ \bold{{36+ {(11 - x)}^{2} = 100}}}$

$\\ \implies \:{ \bold{{{(11 - x)}^{2} = 100 - 36}}}$

$\\ \implies \:{ \bold{{{(11 - x)}^{2} =64}}}$

$\\ \implies \:{ \bold{{11 - x= \sqrt{64}}}}$

$\\ \implies \:{ \bold{{11 - x= \pm \: 8}}}$

▪︎ Take (+) sign –

$\\ \implies \:{ \bold{{11 - x= 8}}}$

$\\ \implies \:{ \bold{{x= 11 - 8}}}$

$\\ \implies \large{ \boxed{ \bold{x= 3}}}$

▪︎ Take (-) sign –

$\\ \implies \:{ \bold{{11 - x= - 8}}}$

$\\ \implies \:{ \bold{{x= 11 + 8}}}$

$\\ \implies \large{ \boxed{ \bold{x= 19}}}$

Answer 2

Answer:

your answer is in above pic