Question

Find the values of x if the distance b/w the points A(0,0)B(x,-4) =5 units

Answer 1

$\huge\bf\mathscr\pink{Your\: Answer}$

x = ±3

step-by-step explanation:

We know that the distance between two points (x1,y1) and (x2,y2) is given by,

Distance = √ {${( x2-x1)}^{2}$ + ${( y2-y1)}^{2}$}

Given in the question,

Pt. A = (0,0)

Pt. B = (x,-4)

Here,

x1 = 0

y1 = 0

x2 = x

y2 = -4

Also,

distance = 5 units

Now,

Putting these values in the distance formula,

we get,

√{${( x-0)}^{2}$ + ${( -4-0)}^{2}$ } = 5

squaring both sides,

we get,

=> ${(x-0)}^{2}$ + ${(-4-0)}^{2}$ = 25

=> ${(x)}^{2}$ + ${(-4)}^{2}$ = 25

=> ${(x)}^{2}$ + 16 = 25

=> ${(x)}^{2}$ = 25-16 = 9

=> x = ±3

Hence, the values of x = ±3

Answer 2

The value of x is ± 3.

Step-by-step explanation:

Here, $A(x_{1}=0,y_{1}=0)$ and $B(x_{2}=x,y_{2}=-4)$

The distance between the points A(0, 0) and B(x,- 4) = 5 units

To find, the value of x =?

We know that,

Distance formula =$\sqrt{(x_{2} -x_{1}) ^{2}+(y_{2} ^{2}-y_{1}) ^{2}}$

$\sqrt{(x -0) ^{2}+(-4-0) ^{2}}=5$

Squaring both sides, we get

$x -0) ^{2}+(-4-0) ^{2}=25$

⇒$x^{2}+16=25$

⇒ $x^{2}=25-16=9$

⇒ x = ± 3

Hence, the value of x is ± 3.