Question

The distance between the points P(x, 2) and Q(3,-6) is 10units, find the value of x.

Answer 1

P(x, 2),Q(3,-6)
PQ=10units
By distance formula
x=9or-3

Answer 2

Given points : P( x , 2 ) ; Q( 3 , - 6 )

Distance between P and Q : 10 units

By Distance formula : -

$\boxed{\mathsf{\sqrt{(x_2 - x_1 )^2 + (y_2 - y_1)^2}= Distance\:between\:(x_1 , y_1) \:and\:(x_2 , y_2 )}}$

Assume P( x , 2 ) = ( x₁ , y₁ )

             Q( 3 , - 6 ) = ( x₂ , y₂ )

    Applying Distance formula

$\implies \sqrt{(3 -x )^2 + (-6-2)^2 } = 10\\\\\implies (3-x)^2 + ( - 8 )^2 = 10^2 \\\\\implies( 3 -x )^2 + 64 = 100\\\\\implies (3-x)^2 = 100 - 64\\\\\implies (3-x)^2=36\\\\\implies (3-x)^2=6^2\\\\\implies 3-x= 6 \: \mathsf{Or} \: 3-x=-6\\\\\implies 3 - 6 = x \: \mathsf{Or} \: 3+6= x$

⇒ - 3 = x  Or  9 = x

Therefore the value of x is - 3 or 9.