Question

triangle IJK with vertices I(-9-8),J(-5-6),and K(-7-3) is drawn on the coordinate grid below. what is the area in square units of triangle IJK

Answer 1

`\text{Area of }\Delta IJK\text{ is }\sqrt[]{65\text{ }}\text{ square units}`

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;

`\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}`

Thus, we have the area of the triangle as follows;

`\begin{gathered} \text{Area = }(1)/(2)* base* height \\ \\ \text{Area = }(1)/(2)*\sqrt[]{20}*\sqrt[]{13\text{ }}=\frac{\sqrt[]{260}}{2}\text{ = }\frac{2\sqrt[]{65}}{2}\text{ = }\sqrt[]{65\text{ }}\text{ square units} \end{gathered}`