Question

distance between (2,-2) and (-1,x) is 5 find the value of x.

Answer 1

given points A (2,-2) B (-1 , x)
distance = under root (x2 - x1)^2+(y2-y1)^2
5=(-2-(-1))^2+ (x-(-2))^2
5=1+x^2+2x+4
5-5=x^2+2x
x^2+2x=0

You can factor x2+2x=0 as:

x(x+2)=0

Now, we can solve each term for 0:

x=0 - no other work needed,

and

x+2=0

x+2−2=0−2

x+0=−2

x=−2

hope it helps u

Answer 2

Answer:

The value of x = 2 or -6

Step-by-step explanation:

Given,

The distance between two points (2,-2) and (-1,x) is 5

To find,

The value of 'x'

Recall the formula

Distance between two points $(x_1,y_1) \ and \ (x_2,y_2)$

=$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Solution:

Here the given points are (2,-2) and (-1,x)

Hence we have $x_1 = 2$ and $y_1 = -2$

$x_2 = -1$ and $y_2 = x$

Substituting the values of $x_1,y_1,x_2,y_2$ we get

Distance between (2,-2) and (-1,x)

$\sqrt{(-1-2)^2+(x-(-2))^2}$

$\sqrt{9+(x+2)^2}$

Given, distance between (2,-2) and (-1,x) = 5

Then we have

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

Squaring on both sides

9+(x+2)² = 25

9+x²+4x+4 = 25

x²+4x-12 = 0

x²+6x-2x-12 = 0

x(x+6)-2(x+6) = 0

(x-2)(x+6) = 0

x = 2 or -6

The value of x = 2 or -6

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