Question

if distance between points (x,3) and (5,7) then find the value of x

Answer 1

$Let \: A(x_{1} , y_{1} ) = ( x , 3) \:and \\B(x_{2} , y_{2} ) = ( 5 , 7)$

$\underline { \blue { Distance \: Formula :}}$

$\boxed { \pink { AB = \sqrt{ ( x_{2} - x_{1})^{2} + ( y_{2} - y_{1})^{2}}}}$

$AB = \sqrt{ (5-x)^{2} + (7-3)^{2}} \\= \sqrt{ 5^{2} + x^{2} - 2 \times 5 \times x + 4^{2} } \\=\sqrt{ 25 + x^{2} - 10x + 16} \\= \sqrt{x^{2} - 10x + 41}$

Therefore.,

$\red { Distance \: between \:points\: (x,3)\: and\: (5,7) }\\\green { = \sqrt{x^{2} - 10x + 41}}$

•••♪

Answer 2

Step-by-step explanation:

\begin{gathered} AB = \sqrt{ (5-x)^{2} + (7-3)^{2}} \\= \sqrt{ 5^{2} + x^{2} - 2 \times 5 \times x + 4^{2} } \\=\sqrt{ 25 + x^{2} - 10x + 16} \\= \sqrt{x^{2} - 10x + 41} \end{gathered}

AB=

(5−x)

2

+(7−3)

2

=

5

2

+x

2

−2×5×x+4

2

=

25+x

2

−10x+16

=

x

2

−10x+41

Therefore.,

\begin{gathered} \red { Distance \: between \:points\: (x,3)\: and\: (5,7) }\\\green { = \sqrt{x^{2} - 10x + 41}} \end{gathered}

Distancebetweenpoints(x,3)and(5,7)

=

x

2

−10x+41

•••♪