Question

If the distance between the points ( x, 2 ) and ( 3, - 6 ) is 10 units. Then find the positive value of x.​

Answer 1

Formula :

The distance between the two points (a₁,b₁) and (a₂,b₂) is

= $\sqrt{(a_{1}-a_{2})^{2}+(b_{1}-b_{2})^{2}}$ units

Solution :

The two given points are (x,2) and (3,-6)

So, the required distance between the two points be

= $\sqrt{(x-3)^{2}+(2+6)^{2}}$ units

= $\sqrt{x^{2}-6x+9+64}$ units

= $\sqrt{x^{2}-6x+73}$ units

By the given condition,

$\sqrt{x^{2}-6x+73} = 10$

→ x² - 6x + 73 = 10²

  ( squaring both sides )

→ x² - 6x + 73 = 100

→ x² - 6x + 73 - 100 = 0

→ x² - 6x - 27 = 0

→ x² - (9 - 3) x - 27 = 0

→ x² - 9x + 3x - 27 = 0

→ x (x - 9) + 3 (x - 9) = 0

→ (x - 9) (x + 3) = 0

So, either x - 9 = 0 or, x + 3 = 0

→ x = 9, - 3

Hence, the positive value of x is 9

Answer 2

Answer:

$\huge{\boxed{\boxed{x=9}}}$

Distance formula :

Distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula :

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

Comparing the above points :

$x_1=x\\y_1=2\\\\x_2=3\\y_2=-6$

Distance = 10 units .

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=10\\\\\implies \sqrt{(3-x)^2+(-6-2)^2}=10\\\\\implies \sqrt{9+x^2-6x+(-8)^2}=10$

$\implies \sqrt{x^2-6x+64+9}=10\\\\\implies \sqrt{x^2-6x+73}=10$

Squaring both sides we get :

$\implies x^2-6x+73=10^2\\\\\implies x^2-6x+73=100\\\\\implies x^2-6x+73-100=0\\\\\implies x^2-6x-27=0\\\\\implies x^2+3x-9x-27=0\\\\\implies x(x+3)-9(x+3)=0\\\\\implies (x+3)(x-9)=0$

Either

$x+3=0\\\\\implies x=-3$

Or

$x-9=0\\\\\implies x=9$

The positive value of x will be 9 .