Find the area of △IJK with vertices I(9,5), J(–10,5), and K(9,1).
Question
Find the area of △IJK with vertices I(9,5), J(–10,5), and K(9,1).
Solution
To find the area of a triangle with vertices at given points, we can use the formula for the area of a triangle in the coordinate plane:
Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
where (x1, y1), (x2, y2), and (x3, y3) are the coordinates of the vertices of the triangle.
Here, the vertices are I(9,5), J(-10,5), and K(9,1). So, we have:
x1 = 9, y1 = 5 (coordinates of point I) x2 = -10, y2 = 5 (coordinates of point J) x3 = 9, y3 = 1 (coordinates of point K)
Substitute these values into the formula:
Area = 1/2 * |9(5 - 1) + -10(1 - 5) + 9(5 - 5)|
Simplify inside the absolute value:
Area = 1/2 * |94 + -10-4 + 9*0| Area = 1/2 * |36 + 40 + 0| Area = 1/2 * |76|
The absolute value of 76 is 76, so:
Area = 1/2 * 76 Area = 38 square units
So, the area of △IJK is 38 square units.
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