Question

if the distance between two points (x, -7) and (3, -3) is 5 units, then the values of x are a) 0 or 6 b) 2 or 3 c) -6 or 0 d) 5 or 1 (MCQ)​

Answer 1

Answer:

√(3-x)² + (-3+7)² = 5

sobs

( √(3-x)² + (4)² )² = 5²

(3-x)² + 16 = 25

9 +x²-6x = 25-16

x²-6x+9 = 9

x²-6x+9-9= 0

x²-6x= 0

x(x-6)=0

x=0 or x-6= 0

x= 6

so value of x is 0 or -6

OPTION A IS CORRECT

Answer 2

The value of x = 0 , 6

Given :

The distance between the two points (x, -7) and (3,-3) is 5 unit

To find :

The value of x

a) 0 or 6

b) 2 or 3

c) - 6 or 0

d) 5 or 1

Solution :

Step 1 of 2 :

Form the equation

Here it is given that distance between the two points (x, -7) and (3,-3)is 5 unit

By the given condition

$\displaystyle \sf{ \sqrt{ {(x - 3)}^{2} + {( - 7 + 3)}^{2} } = 5 }$

Step 2 of 2 :

Find the value of x

$\displaystyle \sf{ \sqrt{ {(x - 3)}^{2} + {( - 7 + 3)}^{2} } = 5 }$

$\displaystyle \sf{ \implies {(x - 3)}^{2} + {( - 7 + 3)}^{2} } = {5}^{2}$

$\displaystyle \sf{ \implies {(x - 3)}^{2} + {( - 4)}^{2} } = {5}^{2}$

$\displaystyle \sf{ \implies {(x - 3)}^{2} + 16 = 25}$

$\displaystyle \sf{ \implies {(x - 3)}^{2} = 25 - 16}$

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

$\displaystyle \sf{ \implies {(x - 3)}^{} = \pm \: 3}$

x - 3 = + 3 gives x = 6

x - 3 = - 3 gives x = 0

So the required value of x are 0 and 6

Hence the correct option is a) 0 or 6

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