Question

find the distance between the point (-4 , 3) &(2-3) please tell the answer friends ​

Answer 1

Let's Understand the question:

Find the distance between the point (-4,3) & (2,-3). So,we should find the distance of the two points.

To Find:

$\: \: \: \: \: \: \: \: \: \:\to \sf{Distance \: between \: the \: points = \: ? }$

Solution:

We know that,

$\: \: \: \: \: \: \: \: \: \bullet \: \sf{AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Let us consider that,

$\sf{x_1\:=\:-4}$

$\sf{x_2\:=\:2}$

$\sf{y_1\:=\:3}$

$\sf{y_2\:=\:-3}$

Applying the values,

$\: \: \: \: \: \: \: \: \: \: \: \longmapsto \sf{ AB =\sqrt{\big(2-(-4)\big)^{2} + \big( - 3 - 3 \big)^{2}}}$

$\: \: \: \: \: \: \: \: \: \longmapsto\sf{AB=\sqrt{(2 + 4) ^{2} + (-3-3)^{2}}}$

$\: \: \: \: \: \: \: \: \: \longmapsto \sf{AB= \sqrt{(6)^{2} + ( - 6)^{2}}}$

$\: \: \: \: \: \: \: \: \: \: \longmapsto \sf{AB=\sqrt{36 + 36}}$

$\: \: \: \: \: \: \: \: \longmapsto \sf{AB= \sqrt{72}} \: units$

Thus,The distance between the two points is $\sqrt{72}$units.

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Answer 2

$\huge \sf \underline{Answer}:$
$\large \implies \sf\sqrt{72}Units$
$\\ \\ \huge \sf \underline{Solution}:$

$\large \sf \underline{Given}:$
Point (-4 , 3) &(2-3)

$\\ \large \sf \underline{To \: Find}:$
Distance between the points.

We know that,
$\bf {Distance \: between \: two \: points \: = }$
$\large \boxed{\tt XY = \sqrt{(x2-x1)^2 +(y2-y1)^2 }}$
so,
$\sf Distance = \sqrt{(2-(-4))^2 +(-3-3)^2 }$
$\\ \implies \sqrt{(6)^2 +(-6)^2 }$
$\\ \implies \sqrt{36 +36 }$
$\\ \sf \implies \sqrt{72 } Unit$

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Therefore, Distance between the point (-4 , 3) &(2-3) is $\sf \sqrt{72 } Unit$
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