Question

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

Answer 1

Step-by-step explanation:

Given:-

• Two points A( x , -1) and B(5 , 3)

• The distance between A( x , -1) and B(5 , 3) is 5 units.

To Find:-

• The value of x

Concept used:-

For given two points P(x₁ , x₂) and Q(y₁ , y₂)

The distance between P and is given by:-

$\: \: \: \: \: \sf PQ = \sqrt{ (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} }$

Solution:-

Given, Two points A(x , -1) and B(5 , 3)

Comparing A(x , -1) and B(5 , 3) with P(x₁ , x₂) and Q(y₁ , y₂)

Here:-

• x₁ = x $\:\:\:\:\:\:\:$ • x₂ = -1

• y₁ = 5 $\:\:\:\:\:\:\:$ • y₂ = 3

Now, the distance between AB = 5 units

$\sf AB = \sqrt{ (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} }$

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

$\sf \implies \sqrt{ {x}^{2} + 25 - 10x + ( -4)^{2} } = 5$

$\sf \implies \sqrt{ {x}^{2} + 25 - 10x + 16 } = 5$

$\sf \implies \sqrt{ {x}^{2} - 10x +41 } = 5$

Squaring on both sides:-

$\sf \implies (\sqrt{ {x}^{2} - 10x +41 } )^{2} = {5}^{2}$

$\sf \implies {x}^{2} - 10x +41 = 25$

$\sf \implies {x}^{2} - 10x +41 -25 = 0$

$\sf \implies {x}^{2} - 10x + 16 = 0$

By splitting the middle term:-

$\sf \implies {x}^{2} - 2x - 8x + 16 = 0$

$\sf \implies x(x - 2) - 8(x - 2)= 0$

$\sf \implies (x - 8) (x - 2)= 0$

$\sf \implies x - 8 = 0 \: \: \: (or) \: \: \: x - 2= 0$

$\sf \implies x = 8 \: \: \: (or) \: \: \: x = 2$

$\underline{\boxed{\sf \therefore The\: possible\: values\: of\: x\: are \: 8 \: and \: 2}}$

Answer 2

Step-by-step explanation:

Given :

• Distance between A(x,-1) and B(5,3) is 5 units

To find :

• The values of x

Solution :

Since distance between A(x,-1) and B(5,3) is 5 units,

we have

AB = 5

$: \implies \sf \: \: \sqrt{(x_1-x_2)^2+(y_1-y_1)^2}=5 \\ \\ \\ :\implies \sf \: \: \: \sqrt{(x-5)^2+(-1-3)^2}=5 \\ \\ \\ </p><p></p><p>: \implies \sf \: \:\sqrt{(x-5)^2+(-4)^2}=5 \\ \\ \\ </p><p></p><p>: \implies \sf \: \:\sqrt{(x-5)^2+16}=5 \: \\ \\ \\ \:$

$\underline \boldsymbol{Squaring \: on \: bothsides, }$

$: \implies \sf \: \: \: (x-5)^2+16=25 \\ \\ \\ </p><p></p><p>: \implies \sf \: \: \: \: (x-5)^2=25-16 \\ \\ \\ </p><p></p><p>: \implies \sf \: \: \: \: \: (x-5)^2=9 \\ \\ </p><p>$

Taking square root on bothsides, we get

$: \implies \sf \: \: \: \: x=5\pm{3} \\ \\ </p><p>: \implies \sf \: \: \: \: \: x = 5 - 3 \\ \\ : \implies \sf \: \: \: \:x = 2 \\ \\ </p><p> : \implies \sf \: \: \: \: \;x=5+3\;\text{(or)}\; \\ \\ </p><p></p><p> : \implies \sf \: \: \: \;x=8\;\text{(or)}\; \\ \\ </p><p>$

• The values of x are 8 and 2