Question

Find all possible values of $x$ for which the distance between the points $A\left( x,-1 \right)$ and $B\left( 5,3 \right)$ is $5$ units.

Answer 1

Hint: Here we have been given two points and we have to find the value of the unknown variable in one of them by taking the distance between the two points as $5$ units. We know the distance between two points is calculated by the distance formula. So we will put the points and the distance given in the distance formula and simplify it to get our answer.

Complete step by step answer:
The two points are given as follows,
$A\left( x,-1 \right)$
$B\left( 5,3 \right)$
We have to find the possible value of $x$ if distance between the above points is $5$ units.
$d=5$….$\left( 1 \right)$
The distance formula for two general points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{1}},{{y}_{1}} \right)$ is given as follows,
$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$…..$\left( 2 \right)$
Comparing the general points by points $A$ and $B$ we get,
$\left( {{x}_{1}},{{y}_{1}} \right)=\left( x,-1 \right)$
$\left( {{x}_{2}},{{y}_{2}} \right)=\left( 5,3 \right)$
Substituting the above points and equation (1) value in equation (2) we get,
$5=\sqrt{{{\left( 5-x \right)}^{2}}+{{\left( 3-\left( -1 \right) \right)}^{2}}}$
$\Rightarrow 5=\sqrt{{{\left( 5-x \right)}^{2}}+{{\left( 4 \right)}^{2}}}$
Using the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ on first term above we get,
$\Rightarrow 5=\sqrt{\left( 25-2\times 5\times x+{{x}^{2}} \right)+16}$
$\Rightarrow 5=\sqrt{25-10x+{{x}^{2}}+16}$
So we get,
$\Rightarrow 5=\sqrt{41-10x+{{x}^{2}}}$
Squaring both sides we get,
$\Rightarrow {{5}^{2}}={{\left( \sqrt{41-10x+{{x}^{2}}} \right)}^{2}}$
$\Rightarrow 25=41-10x+{{x}^{2}}$
Taking all terms one side we get,
$\Rightarrow {{x}^{2}}-10x+41-25=0$
$\Rightarrow {{x}^{2}}-10x+16=0$
Using quadratic formula which state that for any equation in form $a{{x}^{2}}+bx+c=0$ we get the value of $x$ by using formula,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Comparing the general from by equation (3) we get,
$a=1,b=-10,c=16$
Putting the above value in the formula we get,
$x=\dfrac{-\left( -10 \right)\pm \sqrt{{{\left( -10 \right)}^{2}}-4\times 1\times 16}}{2\times 1}$
$x=\dfrac{10\pm \sqrt{100-64}}{2}$
Simplifying further we get,
$x=\dfrac{10\pm \sqrt{36}}{2}$
$\Rightarrow x=\dfrac{10\pm 6}{2}$
So we get the two values as,
$x=\dfrac{10+6}{2},x=\dfrac{10-6}{2}$
On simplifying we get,
$x=\dfrac{16}{2},x=\dfrac{4}{2}$
$\Rightarrow x=8,x=2$
So we obtain two values of $x$ which is $8$ and $2$
Hence there are $2$ possible values of $x$ for which the distance between the points $A\left( x,-1 \right)$ and $B\left( 5,3 \right)$ is $5$ units.

Note:
The distance between two points is the length of the line joining them. As the distance was given to us the only way to get the value of the unknown variable is to put all the values we have in the distance formula and then solve it. We can first square both sides and then solve the bracket and still get the same answer.