Question

Find 'x' if the distance between (5,-1,7) and(x,5,1) is 9 units

Answer 1

Answer:

x = 8

x =  2

Step-by-step explanation:

given  points are :-

A (5,-1,7) and B (x,5,1)

also given that,

distance between these points is 9 units

now,

we know that,

distance between two points

A (x1,y1,z1) and  B (x2,y2,z2)

 is given by,

$D = \sqrt{(x2-x1)^{2}+(y2-y1)^{2}+(z2-z1)^{2}}$

So, putting th respective values,

we get,

$D = \sqrt{(x-5)^{2} +(5+1)^{2}+(1-7)^{2}}$

but, D = 9 units

so, putting the value and squaring both sides,

we get,

$81 = {x}^{2} + 25 - 10x + 72$

=> ${x}^{2} - 10x + 16 = 0$

=> ${x}^{2} - 8x - 2x + 16 = 0$

=> $x(x-8) -2(x-8) = 0$

=> $(x-8)(x-2) = 0$

=> $x = 8$ 'or' $x = 2$

Answer 2

Given:

The distance between (5, -1, 7) and(x, 5, 1) is 9 units.

To find:

The value of x.

Solution:

To answer this question, first of all, we should know that the distance between two points A and B having coordinates A(x, y, z) and B(l, m, n) is given by:

$= \sqrt{ {(l - x)}^{2} + {(m - y)}^{2} + {(n - z)}^{2} }$

Now, as given,

We have,

Coordinates of two points are (5, -1, 7) and(x, 5, 1).

The distance between two points = 9 units.

So,

$\sqrt{ {(x - 5)}^{2} + {(5 - ( - 1))}^{2} + {(1 - 7)}^{2} } = 9$

On solving the above we get

$\sqrt{ {(x - 5)}^{2} + {(5 + 1)}^{2} + {( - 6)}^{2} } = 9$

$\sqrt{ {(x - 5)}^{2} + 36 + 36 } = 9$

Now, after squaring on both sides, we get

${(x - 5)}^{2} + 72 = 81$

${(x - 5)}^{2} = 9$

Taking square root on both sides, we get

$x - 5 = 3 \: and \: x - 5 = - 3$

$x = 8 \: and \: x \: = 2$

Hence, the value of x can be 2 or 8.