[tex]\text{Area of }\Delta IJK\text{ is }\sqrt[]{65\text{ }}\text{ square units}[/tex]
Here, we want to calculate the area of the given triangle
Mathematically, the product of the height and base of a triangle divided by 2 gives its area
Now, as we can see the shape, while IJ represents the base, KJ represents the height
We need to calculate the distance between these points before we can get the area of the triangle
To get the distances between the points, we have to use the distance formula
We can proceed with this as follows;
[tex]\begin{gathered} D\text{ = }\sqrt[]{(_{}x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{For IJ} \\ (x_1,y_1)\text{ = (-9,-8)} \\ (x_2,y_2)\text{ = (-5,-6)} \\ |IJ|\text{ = }\sqrt[]{(-5+9)^2+(-6+8)^2} \\ |IJ|\text{ = }\sqrt[]{16\text{ + 4}} \\ |IJ|\text{ = }\sqrt[]{20} \\ \\ \text{For KJ} \\ (x_1,y_1)\text{ = (-7,-3)} \\ (x_2,y_2)\text{ = (-5,-6)} \\ |KJ|\text{ = }\sqrt[]{(-5+7)^2+(-6+3)^2} \\ |KJ|\text{ = }\sqrt[]{4+9} \\ |KJ|\text{ = }\sqrt[]{13} \end{gathered}[/tex]
Thus, we have the area of the triangle as follows;
[tex]\begin{gathered} \text{Area = }\frac{1}{2}\times base\times height \\ \\ \text{Area = }\frac{1}{2}\times\sqrt[]{20}\times\sqrt[]{13\text{ }}=\frac{\sqrt[]{260}}{2}\text{ = }\frac{2\sqrt[]{65}}{2}\text{ = }\sqrt[]{65\text{ }}\text{ square units} \end{gathered}[/tex]
