Question

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

Answer 1

$\boxed{ \red{\huge \mathfrak{solution}}}$

Given:

points are

•(2,-2)or(x1,y1)

and

•(-1,x)or(x2,y2)

•Distance=5

As we know that, D=

$\sqrt{(x2 - x1) {}^{2} + (y2 - y1 {})^{2} }$

Put the values :

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

Square both sides,we get

$= > 9 + {x}^{2} + 4 + 4x = 25 \\ = > {x}^{2} + 4x - 12 = 0 \\ = > {x}^{2} + 6x - 2x - 12 = 0 \\ = > x(x + 6) - 2(x +6) = 0 \\ = > (x - 2)(x + 6) = 0$

Now,

$x - 2 = 0 \\ = > x = 2$

and

$x + 6 = 0 \\ = > x = - 6$

which is not possible . So the value of x is

$\boxed{ \purple{2}}$

Answer 2

D=√(x2-x1)^2+(y2-y1)^2

5=√(-1-2)^2+(x+2)^2

square both sides ,

25=9+(x+2)^2

25=9+x^2+4+4x

x^2+4x-12=0

x^2+6x-2x-12=0

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

x=2,-6

then negative is not possible

x=2