If the distance between the points $ (x,-1) $ and $ (3,2) $ is 5 , then the value of $x$ is

Given:

The distance between the points \( (x,-1) \) and \( (3,2) \) is 5.

To do: 

We have to find the values of $x$.

Solution:

Let $x_1=x,\ y_1=-1,\ x_2=3,\ y_2=2$

Using the distance formula,

The distance between the points $(x_1, y_1)$ and $(x_2, y_2)=\sqrt{( x_2-x_1)^2+( y_2-y_1)^2}$

$\Rightarrow 5=\sqrt{( 3-x)^2+( 2-( -1))^2}$

$\Rightarrow 5=\sqrt{(3)^2+(x)^2-2(3)(x)+(3)^2}$

$\Rightarrow 25=18+x^2-6x$

$\Rightarrow x^2-6x+18-25=0$

$\Rightarrow x^2-6x-7=0$

$\Rightarrow x^2+x-7x-7=0$

$\Rightarrow x(x+1)-7(x+1)=0$

$\Rightarrow (x+1)(x-7)=0$

$x=-1$ or $x=7$

The values of $x$ are $-1$ and $7$.

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Updated on: 10-Oct-2022

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