Question

3. If the distance between two points (0, 2) and (x,2) is 4,then find the value of x?​

Answer 1

Answer:

$\sqrt{(0 - x) {}^{2} + (2 - 2) {}^{2} } = 4$

$\sqrt{ - x {}^{2} } = 4$

$x = 4$

$x = 4$

Answer 2

Given Point

⇒P(0 , 2) and Q(x , 2)

⇒Distance between PQ is 4

To Find Value of x

Formula

⇒D² = (x₂ -x₁)² + (y₂ - y₁)²

By Comparing with

⇒P(x₁ , y₁) and Q(x₂ , y₂)

We get

⇒x₁ = 0 , y₁ = 2 , x₂ = x and y₂ = 2

Now Put the value on formula

⇒PQ² = (x₂ -x₁)² + (y₂ - y₁)²

Where , PQ = 4

⇒(4)² = (x - 0)² + (2 - 2)

⇒(4)² = (x)² + 0

⇒x² = 4²

⇒x = 4

Answer

⇒x = 4