The distance of the point P(-6, 8) from the o | vedclass

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(B) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$Here,the point

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$10$

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(B) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$Here,the point $P$ is $(-6, 8)$ and the origin $O$ is $(0, 0)$.So,$x_1 = -6, y_1 = 8$ and $x_2 = 0, y_2 = 0$.Substituting these values into the distance formula:$PO = \sqrt{(0 - (-6))^2 + (0 - 8)^2}$$PO = \sqrt{(6)^2 + (-8)^2}$$PO = \sqrt{36 + 64}$$PO = \sqrt{100}$$PO = 10$Thus,the distance of the point $P(-6, 8)$ from the origin is $10$ 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 $P(-6, 8)$ from the origin is

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