Question

If the distance between the points P(5, 2) and Q(x,-2) is 5 units. Find the value of x.​

Answer 1

Answer:

let 5,2 be x1 and y1

let x,-2 be x2 and y2.

We know that distance formula:

$\sqrt{(x2 - x1)^{2} +{(y2 - y1)^{2} }} = 5$

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

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

${x}^{2} + 25 - 10x + 16 = 25$

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

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

$x(x - 2) - 8(x - 2)$

$(x - 8)(x - 2) = 0$

$thus \\ x - 2 = 0 \: and \: x - 8 = 0$

$x = 2 \: and \: \: x = 8$

Thus, distance between the points could be 2 units or 8 units.

HOPE THIS HELPS :D