Question

Two points A, B have co-ordinates (2, 3) and (4, x) respectively. If $AB^2$ =13, the possible value of x is :
(A) −6
(B) 0
(C) 9
(D) 12

Answer 1

Answer:

(B) 0

Step-by-step explanation:

Find the length of AB:

$\text {Length = }\sqrt{(Y_2-Y_1)^2 + (X_2-X_1)^2}$

$\text {AB = }\sqrt{(x-3)^2 + (4-2)^2}$

$\text {AB = }\sqrt{(x-3)^2 + 4}$

Solve x:

Given that AB² = 13

$(\sqrt{(x-3)^2 + 4})^2 = 13$

$(x-3)^2 + 4 = 13$

$(x-3)^2 = 9$

$x-3 = \pm \sqrt{9}$

$x-3 = \pm 3$

$x = \pm 3 + 3$

$x = 6 \ or \ x = 0$

Answer: (B) 0

Answer 2

HELLO DEAR,

GIVEN:- AB² = 13,

the length of AB is

$\bold{\sf{\sqrt{(Y_2-Y_1)^2 + (X_2-X_1)^2}}}$

$\bold{\sf{\sqrt{(x-3)^2 + (4-2)^2}}}$

$\bold{\sf{\sqrt{(x-3)^2 + 4}}}$

As, AB² = 13

$\sf{(\sqrt{(x-3)^2 + 4})^2 = 13}$

$\sf{(x-3)^2 + 4 = 13}$

$\sf{(x-3)^2 = 9}$

$\sf{x-3 = \pm \sqrt{9}}$

$\sf{x-3 = \pm 3}$

$\sf{x = \pm 3 + 3}$

$\sf{x = 6 \ or \ x = 0}$

Hence, option (b) is correct

I HOPE IT'S HELP YOU DEAR,

THANKS