The distance of the point P(2, 3) from the x- | vedclass

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(D) We know that for any point $P(x, y)$ in the Cartesian plane:$1$. The perpendicular distance of the point from the $y$-axis is given by the absolut

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$3$

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(D) We know that for any point $P(x, y)$ in the Cartesian plane:$1$. The perpendicular distance of the point from the $y$-axis is given by the absolute value of its $x$-coordinate (abscissa),i.e.,$|x|$.$2$. The perpendicular distance of the point from the $x$-axis is given by the absolute value of its $y$-coordinate (ordinate),i.e.,$|y|$.For the point $P(2, 3)$,the $x$-coordinate is $2$ and the $y$-coordinate is $3$.Therefore,the distance of the point $P(2, 3)$ from the $x$-axis is equal to the ordinate of the point,which is $3$ units.

$1$. $\square ABCD$ is a rhombus.	$a$. $\overline{AC}$ and $\overline{BD}$ bisect each other.
$2$. $\square ABCD$ is a parallelogram.	$b$. $\overline{AC}$ and $\overline{BD}$ bisect each other at right angles.
$3$. $\square ABCD$ is a rectangle.	$c$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other at right angles.
$4$. $\square ABCD$ is a square.	$d$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other.

The distance of the point $P(2, 3)$ from the $x$-axis is

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