Answer:
k = 6 and k = -6.
Explanation:
To find the value(s) of k for which the area of the triangle with vertices (-3,0), (3,0), and (0,k) is 9 square units, we can use the formula for the area of a triangle:
Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
In this case, (-3,0), (3,0), and (0,k) correspond to (x1, y1), (x2, y2), and (x3, y3) respectively. So, we have:
x1 = -3
y1 = 0
x2 = 3
y2 = 0
x3 = 0
y3 = k
Now, we can plug these values into the area formula and set it equal to 9 square units:
9 = 1/2 * |-3(0 - k) + 3(0 - 0) + 0(0 - 0)|
9 = 1/2 * |-3k|
Now, we can simplify further:
9 = 1.5|k|
To solve for k, we need to isolate k:
|k| = 9 / 1.5
|k| = 6
Now, we consider both the positive and negative values for k, as the absolute value is involved:
k = 6 or k = -6
So, the values of k for which the area of the triangle is 9 square units are k = 6 and k = -6.