Question

12.
If the distance between the points A(2,-2) and B (-1, x) is equal to 5, then the value of x is
(A) 2
(B)-2
(C) 1
(D) -1​

Answer 1

Step-by-step explanation:

the correct option is a this is your answer

Answer 2

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• The points are A(2,-2) and B(-1,x).
• The distance between the two points are 5 units.

$\LARGE{\bf{\underline{\underline{TO:FIND:-}}}}$

• The value of x.

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

We can use the distance formula for this question:

$\boxed{\sf{\blue{d = \sqrt {\left( {x_1 - x_2 } \right)^2 + \left( {y_1 - y_2 } \right)^2 }}}}$

In points A(2,-2) and B(-1,x),

• x₁=2
• y₁= -2
• x₂= -1
• y₂=x.

We know that d (distance) is 5. Thus, after substitution:

$\sf 5=\sqrt {(-1 - 2))^2 + (- 2-x)^2}$

$\sf \to 5=\sqrt{9+(-2-x)^2}$

Square on both sides.

$\sf \to 5^2=(\not \sqrt{9+(-2-x)^2})^{\not{2}}$

$\sf \to 25=9+(-2-x)^2$

$\to \sf 25-9=(-2-x)^2$

Transpose square (²) to LHS.

$\sf \to \sqrt{16}=(-2-x)$

Here, √16 can be +4 or -4.

$\sf \to \pm4=-2-x$

Thus, two cases arise.

CASE 1:

$\sf \to +4=-2-x$

$\boxed{\sf{\blue{x=-6}}}$

CASE 2:

$\sf \to -4=-2-x$

$\boxed{\sf{\blue{x=2}}}$

∴x is either 2 or -6.

OPTION A IS CORRECT.