Find the value of t, if the distance between the points P(–3, –14) and Q(t,...

Find the value of t, if the distance between the points P(–3, –14) and Q(t, –5) is 9 units.

The correct answer is: C

Let the given points be:

P(-3, -14) = (x\(_1\), y\(_1\))

Q(t, -5) = (x\(_2\), y\(_2\))

PQ = 9 units (given)

Using the distance formula,

d = √ [ \((x_2 - x_1)^2 + (y_2 - y_1)^2\)]

PQ = √ [ \((t - (-3))^2 + (-5 + 14)^2\)]

\(\implies\) √ [ \((t + 3)^2 + 81\)] = 9

Squaring on both sides,

⇒ (t + 3)\(^2\) + 81 = 81

⇒ (t + 3)\(^2\) = 0

⇒ t + 3 = 0

∴ t = -3

There is an explanation video available .