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Sagot :
Answer:
The length of the line AB, BC, and, AC are;
[tex]\begin{gathered} AB=11.31\text{ units} \\ BC=10.20\text{ units} \\ AC=10.20\text{ units} \end{gathered}[/tex]Explanation:
We want to find the length of the line AB, BC, and, AC.
Applying the formula for calculating the distance between two points;
[tex]d=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]We are given the coordinates of each point as;
[tex]A\mleft(-4,6\mright),B\mleft(4,-2\mright),C\mleft(-6,-4\mright)[/tex]So, the length AB is;
[tex]\begin{gathered} AB=\sqrt[]{(-2_{}-6_{})^2+(4_{}-(-4))^2} \\ AB=\sqrt[]{(-8)^2+(8)^2} \\ AB=\sqrt[]{64^{}+64} \\ AB=\sqrt[]{128} \\ AB=11.31\text{ units} \end{gathered}[/tex]Length BC is;
[tex]\begin{gathered} BC=\sqrt[]{(-4_{}-(-2)_{})^2+(-6-4)^2} \\ BC=\sqrt[]{(-4_{}+2_{})^2+(-10)^2} \\ BC=\sqrt[]{(-2_{})^2+(-10)^2} \\ BC=\sqrt[]{4+100} \\ BC=\sqrt[]{104} \\ BC=10.20\text{ units} \end{gathered}[/tex]Length AC is;
[tex]\begin{gathered} AC=\sqrt[]{(-4_{}-6_{})^2+(-6-(-4))^2} \\ AC=\sqrt[]{(-10))^2+(-6+4)^2} \\ AC=\sqrt[]{(-10)^2+(-2)^2} \\ AC=\sqrt[]{100+4} \\ AC=\sqrt[]{104} \\ AC=10.20\text{ units} \end{gathered}[/tex]Therefore, the length of the line AB, BC, and, AC are;
[tex]\begin{gathered} AB=11.31\text{ units} \\ BC=10.20\text{ units} \\ AC=10.20\text{ units} \end{gathered}[/tex]
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