Answer:
√221 or 14.866
Step-by-step explanation:
The distance formula is
[tex]d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]
Using our coordinates, we have
[tex]d=\sqrt{(12-2)^2+(3-14)^2}\\\\=\sqrt{10^2+(-11)^2}\\\\=\sqrt{100+121}\\\\=\sqrt{221}\approx 14.866[/tex]
The distance between the points (3, 12) and (14, 2) is 14.86 units.
The distance between two points [tex](x_1, y_1)[/tex] and [tex](x_2,y_2)[/tex] is given by,
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
For given example,
We need to find the distance between points (3, 12) and (14, 2)
Consider given points as,
[tex](x_1,y_1)=(3,12)\\\\(x_2,y_2)=(14,2)[/tex]
Using the distance formula, the distance between two points would be,
[tex]\Rightarrow d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\\\Rightarrow d=\sqrt{(14-3)^2+(2-12)^2}\\\\\Rightarrow d=\sqrt{(11)^2+(-10)^2}\\\\\Rightarrow d=\sqrt{121+100}\\\\\Rightarrow d=\sqrt{221}\\\\\Rightarrow d=14.86~units[/tex]
Therefore, the distance between the points (3, 12) and (14, 2) is 14.86 units.
Learn more about the distance between points here:
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