Question 11 Mark
If distance between points $P (x,-8,4)$ and $Q(-3,-5,4)$ is 5 unit, then find the value of $x$.
Answer
View full question & answer→Given points are $P \equiv(x,-8,4)$ and $Q \equiv(-3,-5,4)$
$\therefore \quad(PQ)^2=(x+3)^2+(-8+5)^2+(4-4)^2$
According to question, $PQ =5$ unit
$\begin{array}{l}\Rightarrow\quad(PQ)^2=25 \\\Rightarrow \quad(x+3)^2+(-3)^2+0=25 \\\Rightarrow \quad(x+3)^2=25-9=16 \\\Rightarrow \quad(x+3)= \pm 4 \\\Rightarrow \quad x=1,-7\end{array}$
$\therefore \quad(PQ)^2=(x+3)^2+(-8+5)^2+(4-4)^2$
According to question, $PQ =5$ unit
$\begin{array}{l}\Rightarrow\quad(PQ)^2=25 \\\Rightarrow \quad(x+3)^2+(-3)^2+0=25 \\\Rightarrow \quad(x+3)^2=25-9=16 \\\Rightarrow \quad(x+3)= \pm 4 \\\Rightarrow \quad x=1,-7\end{array}$