If the distance between the points (x, -1) and (3, 2) is 5 units, then what will be the value of x.

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Bihar STET TGT (Maths) Official Paper-I (Held On: 04 Sept, 2023 Shift 2)
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  1. 1
  2. -1
  3. 7
  4. Both (2) and (3)

Answer (Detailed Solution Below)

Option 4 : Both (2) and (3)
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Bihar STET Paper 1 Mathematics Full Test 1
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Detailed Solution

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Concept -

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a 2D plane is given by the distance formula:

\( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Explanation -

Here, the points are (x, -1) and (3, 2), and the distance is given as 5 units. Using the distance formula:

\(5 = \sqrt{(3 - x)^2 + (2 - (-1))^2} \\ 5 = \sqrt{(3 - x)^2 + 3^2} \\ 5 = \sqrt{(3 - x)^2 + 9} \)

let's square both sides to eliminate the square root:

\( 5^2 = (3 - x)^2 + 9 \\ 25 = (3 - x)^2 + 9 \\ (3 - x)^2 = 25 - 9 \\ (3 - x)^2 = 16 \)

Now, take the square root of both sides:

\( 3 - x = \pm \sqrt{16} \\ 3 - x = \pm 4 \)

Solve for x in both cases:

When 3 - x = 4:
x = 3 - 4 
 x = -1 

When 3 - x = -4:
 x = 3 + 4 
 x = 7 

So, the values for x are x = -1 or x = 7.

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