Question

If the distance between the point
A(x, 3) | B(-x, 2) is 5. Find the value of x

Answer 1

`\qquad\qquad\huge\underline{{\sf Answer}}`

`\textbf{Let's use distance formula here }`~

`\qquad \sf  \dashrightarrow \: d = \sqrt{(x2 - x1) {}^(2) + (y2 - y1) {}^(2) }`

`\textsf{The given points are :}`

`\qquad \qquad \tt A \: (x , 3)`

`\textsf{and}`

`\qquad \qquad \tt B \: ( - x , 2)`

━━━━━━━━━━━━━━━━━━━━━

`\textsf{It's given that the distance between the points is}`
`\textsf{ 5 units }`

`\qquad \sf  \dashrightarrow \: d = 5`

`\textsf{that is : }`

`\qquad \sf  \dashrightarrow \: \sqrt{( - x - x) {}^(2) + (2 - 3) {}^(2) } = 5`

`\qquad \sf  \dashrightarrow \: ( - 2x) {}^(2) + ( - 1) {}^(2) = 25`

`\qquad \sf  \dashrightarrow \: 4 {x}^(2) + 1 = 25`

`\qquad \sf  \dashrightarrow \: 4 {x}^(2) = 25 - 1`

`\qquad \sf  \dashrightarrow \: {x}^(2) = 24 / 4`

`\qquad \sf  \dashrightarrow \: x = √(6)`

`\textbf{Let's verify the result}` ~

`\textsf{plug the value of x as }`
`\bf{√(6)}`

`\qquad \sf  \dashrightarrow \: \sqrt{( - x - x) {}^(2) + (2 - 3) {}^(2) }`

`\qquad \sf  \dashrightarrow \: \sqrt{( - √(6)- √(6)) {}^(2) + (2 - 3) {}^(2) }`

`\qquad \sf  \dashrightarrow \: \sqrt{( - 2√(6)) {}^(2) + ( - 1) {}^(2) }`

`\qquad \sf  \dashrightarrow \:√(( 4 ×6) + 1 )`

`\qquad \sf  \dashrightarrow \: √(24 + 1)`

`\qquad \sf  \dashrightarrow √(25)`

`\qquad \sf  \dashrightarrow \: 5`

hence, what we got is correct ~