Find the value of t, if the distance between the points P(–3, –14) and Q(t, –5) is 9 units.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
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
There is an explanation video available below.
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 below.