MCQ
If $x$ is a positive integer such that the distance between points $P(x, 2)$ and $Q(3, -6)$ is $10$ units, then $x =$
- A$3$
- B$-3$
- ✓$9$
- D$-9$
✓
Answer
Correct option: C.
$9$
Distance between $P(x, 2)$ and $Q(3, -6) = 10$ units
$\Rightarrow\ \sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}=10$
$\Rightarrow\ \sqrt{(3-\text{x})^2+(-6-2)^2}=10$
$\Rightarrow\ \sqrt{(3-\text{x})^2+(-8)^2}=10$
$\Rightarrow\ \sqrt{(3-\text{x})^2+64}=10$
Squaring both sides,
$ (3-x)^2+64=100 $
$ \Rightarrow 9+x^2-6 x+64-100=0 $
$ \Rightarrow x^2-6 x-27=0 $
$ \Rightarrow x^2-9 x+3 x-27=0$
$\begin{Bmatrix}\because\ 27=-9\times3\\\ \ -6=-9+3\end{Bmatrix}$
$\Rightarrow x(x - 9) + 3(x - 9) = 0$
$\Rightarrow (x - 9)(x - 3) = 0$
Either $x - 9 = 0$, then $x = 9$ or $x + 3 = 0,$ then $x = -3$
$x$ is positive integer.
Hence $x = 9.$
$\Rightarrow\ \sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}=10$
$\Rightarrow\ \sqrt{(3-\text{x})^2+(-6-2)^2}=10$
$\Rightarrow\ \sqrt{(3-\text{x})^2+(-8)^2}=10$
$\Rightarrow\ \sqrt{(3-\text{x})^2+64}=10$
Squaring both sides,
$ (3-x)^2+64=100 $
$ \Rightarrow 9+x^2-6 x+64-100=0 $
$ \Rightarrow x^2-6 x-27=0 $
$ \Rightarrow x^2-9 x+3 x-27=0$
$\begin{Bmatrix}\because\ 27=-9\times3\\\ \ -6=-9+3\end{Bmatrix}$
$\Rightarrow x(x - 9) + 3(x - 9) = 0$
$\Rightarrow (x - 9)(x - 3) = 0$
Either $x - 9 = 0$, then $x = 9$ or $x + 3 = 0,$ then $x = -3$
$x$ is positive integer.
Hence $x = 9.$
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