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

Explanation

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

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