[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]
[tex] \textbf{Let's use distance formula here } [/tex]~
[tex]\qquad \sf \dashrightarrow \: d = \sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} } [/tex]
[tex] \textsf{The given points are :} [/tex]
[tex] \qquad \qquad \tt A \: (x , 3) [/tex]
[tex] \textsf{and} [/tex]
[tex] \qquad \qquad \tt B \: ( - x , 2) [/tex]
━━━━━━━━━━━━━━━━━━━━━
[tex] \textsf{It's given that the distance between the points is} [/tex][tex] \textsf{ 5 units } [/tex]
[tex]\qquad \sf \dashrightarrow \: d = 5[/tex]
[tex] \textsf{that is : } [/tex]
[tex]\qquad \sf \dashrightarrow \: \sqrt{( - x - x) {}^{2} + (2 - 3) {}^{2} } = 5[/tex]
[tex]\qquad \sf \dashrightarrow \: ( - 2x) {}^{2} + ( - 1) {}^{2} = 25[/tex]
[tex]\qquad \sf \dashrightarrow \: 4 {x}^{2} + 1 = 25[/tex]
[tex]\qquad \sf \dashrightarrow \: 4 {x}^{2} = 25 - 1[/tex]
[tex]\qquad \sf \dashrightarrow \: {x}^{2} = 24 \div 4[/tex]
[tex]\qquad \sf \dashrightarrow \: x = \sqrt{6} [/tex]
[tex] \textbf{Let's verify the result} [/tex] ~
[tex] \textsf{plug the value of x as } [/tex][tex] \bf{\sqrt{6}} [/tex]
[tex]\qquad \sf \dashrightarrow \: \sqrt{( - x - x) {}^{2} + (2 - 3) {}^{2} } [/tex]
[tex]\qquad \sf \dashrightarrow \: \sqrt{( - \sqrt{6}- \sqrt{6}) {}^{2} + (2 - 3) {}^{2} } [/tex]
[tex]\qquad \sf \dashrightarrow \: \sqrt{( - 2\sqrt{6}) {}^{2} + ( - 1) {}^{2} } [/tex]
[tex]\qquad \sf \dashrightarrow \:\sqrt{( 4 ×6) + 1 }[/tex]
[tex]\qquad \sf \dashrightarrow \: \sqrt{24 + 1} [/tex]
[tex]\qquad \sf \dashrightarrow \sqrt{25}[/tex]
[tex]\qquad \sf \dashrightarrow \: 5 [/tex]
hence, what we got is correct ~
