Respuesta :

VictorZ
Hi there!

Use th' following formula
To find the distance between each pair of pts. :-

[tex]\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]

According to th' question :-

♦ (-2, -2) (-2, 10) :--

[tex]\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]

[tex]\sqrt{(-2 + 2)^2 + (10 + 2)^2}[/tex]

[tex]\sqrt{0^2 + 12^2}[/tex]

[tex]\sqrt{12^2}[/tex] = [tex]\sqrt{144}[/tex] = 12

Therefore,
distance between the following pts. is 12 units.

♦ (-5.3, -2.2) (-5.3, -8.7) :--

[tex]\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]

[tex]\sqrt{(-5.3 + 5.3)^2 + (-8.7 + 2.2)^2}[/tex]

[tex]\sqrt{0^2 + (-6.5)^2}[/tex]

[tex]\sqrt{-6.5^2}[/tex] = [tex]\sqrt{42.25}[/tex] = 6.5

Therefore,
distance between th' following pts. is 6.5 units.

~ Hope it helps!
iCarl
[tex]\sf Hello![/tex]

[tex]\sf To\: calculate\: the\: distance,[/tex]
[tex]\sf We\: have [/tex] :-

[tex]\sf \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}[/tex]

Then,

(i) [tex]\sf (-2, -2) and (-2, 10)[/tex] :

⇒ [tex]\sf \sqrt{(-2 + 2)^{2} + (10 + 2)^{2}}[/tex]

⇒ [tex]\sf \sqrt{0^{2} + 12^{2}}[/tex]

⇒ [tex]\sqrt{144}[/tex] = [tex]\sf 12[/tex]

[tex]\sf Therefore,[/tex]
[tex]\sf The\: distance\: is\: 12 \:units.[/tex]

(ii) [tex]\sf (-5.3, -2.2) and (-5.3, -8.7)[/tex]

⇒ [tex]\sf \sqrt{(-5.3 + 5.3)^{2} + (-8.7 + 2.2)^{2}}[/tex]

⇒ [tex]\sf \sqrt{0^{2} + (-6.5)^{2}}[/tex]

⇒ [tex]\sqrt{ 42.25}[/tex] = [tex]\sf 6.5[/tex]

[tex]\sf Therefore,[/tex]
[tex]\sf The\: distance\: is\: 6.5 \:units.[/tex]

~ [tex]\sf iCarl[/tex]

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