Question

Calculate the perimeter and area of the triangle formed by the coordinates K (-4,-1), L (-2,2), and M (3,-1). Round your answer to two decimal places.

Answer 1

PERIMETER:

You can calculate the perimeter of a triangle knowing the coordinates, by calculating the distance between every point, as follows:

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

Where x1=-4, y1=-1, x2=-2 and y2=2, replace these values:

`\begin{gathered} \bar{KL}=\sqrt[]{((-2)-(-4))^2+(2-(-1))^2} \\ \bar{KL}=\sqrt[]{(2)^2+(3)^2} \\ \bar{KL}=\sqrt[]{4+9}=\sqrt[]{13}=3.61 \end{gathered}`

Now, you have to do the same for the other segments, let's continue with LM:

`\begin{gathered} \bar{LM}=\sqrt[]{(x_3-x_2)^2+(y_3-y_2)^2} \\ x_3=3\text{ and }y_3=-1\text{ by replacing:} \\ \bar{LM}=\sqrt[]{(3_{}-(-2))^2+((-1)-2)^2} \\ \bar{LM}=\sqrt[]{(5)^2+(-3)^2} \\ \bar{LM}=\sqrt[]{25+9}=\sqrt[]{34}=5.83 \end{gathered}`

Same for segment KM:

`\begin{gathered} \bar{KM}=\sqrt[]{(x_3-x_1)^2+(y_3-y_1)^2} \\ \bar{KM}=\sqrt[]{(3-(-4))^2+((-1)-(-1))^2} \\ \bar{KM}=\sqrt[]{(7)^2+(0)^2} \\ \bar{KM}=\sqrt[]{49+0^{}}=\sqrt[]{49}=7 \end{gathered}`

The perimeter can be calculated as KL+LM+KM:

`\begin{gathered} P=\bar{KL}+\bar{LM}+\bar{KM}=3.61+5.83+7 \\ P=16.44 \end{gathered}`

AREA:

The area can be calculated by using the next formula:

`A=(1)/(2)\mleft\lbrace\lbrack(x_1\cdot y_2)+(x_2\cdot y_3)+(x_3\cdot y_1)\mright]-\lbrack(x_1\cdot y_3)+(x_3\cdot y_2)+(x_2\cdot y_1)\rbrack\}`

Then, you have to replace the values to find the area:

`\begin{gathered} A=(1)/(2)\lbrace\lbrack((-4)_{}\cdot2)+((-2)\cdot(-1))+(3\cdot(-1))\rbrack-\lbrack((-4)\cdot(-1))+(3\cdot2)+((-2)\cdot(-1))\rbrack\} \\ A=(1)/(2)\lbrace\lbrack(-8)+(2)+(-3)\rbrack-\lbrack(4)+(6)+(2)\rbrack\} \\ A=(1)/(2)\lbrace\lbrack-9\rbrack-\lbrack12\rbrack\} \\ A=(1)/(2)\mleft\lbrace\lvert-21\rvert\mright\rbrace \\ A=(21)/(2)=10.50 \end{gathered}`