Question

If $x$ has a non zero value, then match the value of $x$ in each case.

• A. 4
• B. $2\sqrt{2}$
• C. 6

Answer 1

Since the abscissa of both P and Q are equal,
$x$ = distance between P and Q = difference between ordinates of P and Q.

$\therefore x=6-2=4$


Since $x$ is the distance between P and Q,
by distance formula, distance
=$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

=$\sqrt{(4-2{)}^2+(1-3{)}^2}$ = $\sqrt{4+4}$= $2\sqrt{2}$
$\therefore x=2\sqrt{2}$


By distance formula, $3\sqrt{2}=\sqrt{(3-x{)}^2+(1-4{)}^2}$
$\Rightarrow 18=9-6x+x^2+9$
$\Rightarrow x^2=6x$
$\Rightarrow x^2-6x=0$
$\Rightarrow x(x-6)=0$
$Eitherx=0orx=6$
Given $x$ has a non zero value
$\therefore x=6$