Question

If the distance between the points P(x,2) and Q(3,-4) is 10 units, find the value of x​

Answer 1

Answer:

x = 11, -5

Step-by-step explanation:

P(x, 2) (x1, y1)

Q(3, -4) (x2, y2)

Distance formula:

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

$\sqrt{(3-x)^2+(-4-2)^2}$ = 10    

$\sqrt{(3^2 + x^2 - 2(3)(x)) - 36}$ = 10

$\sqrt{(9 + x^2 - 6x) +36} = 10\\ \sqrt{(9 + x^2 - 6x +36} = 10\\\\ \sqrt{(x^2 - 6x +36+9)} = 10\\\\ \sqrt{(x^2 - 6x +45)} = 10$

Now, we need to take a square on both sides.  

x^2 - 6x +45 = 10^2

x^2 - 6x +45-100 = 0

x^2 - 6x -55 = 0

(x-11)(x+5) = 0

Next, we have to solve for x.

x = 11, -5

Thus, the distance between the points could be 11 units or -5 units.