Question

the coordinates of the vertices of a triangle are A(-4,6) B(4,-2) C(-6,-4). find the length AB, BC, and, AC

Answer 1

The length of the line AB, BC, and, AC are;

`\begin{gathered} AB=11.31\text{ units} \\ BC=10.20\text{ units} \\ AC=10.20\text{ units} \end{gathered}`

Step-by-step explanation:

We want to find the length of the line AB, BC, and, AC.

Applying the formula for calculating the distance between two points;

`d=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}`

We are given the coordinates of each point as;

`A\mleft(-4,6\mright),B\mleft(4,-2\mright),C\mleft(-6,-4\mright)`

So, the length AB is;

`\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}`

Length BC is;

`\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}`

Length AC is;

`\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}`

Therefore, the length of the line AB, BC, and, AC are;

`\begin{gathered} AB=11.31\text{ units} \\ BC=10.20\text{ units} \\ AC=10.20\text{ units} \end{gathered}`