Question

If the distance between the two points (x, -7) and (3,-3)is 5 units, then the value of x are

Answer 1

Answer:

X=0 or 6

Step-by-step explanation:

apply the distance formula

distance =√(x2-x1)^2+(y2-y1)^2

substitute the values

5=√(x-3)^2+(-7+3)^2

now squaring both sides

25=x^2+9-6x+16

25=x^2-6x+25

0=x^2-6x

0=x(x-6)

either X=0 or 6

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

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

Hence the required value of x are 0 and 6

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