Find k if the area of the triangle with vertices at A(k, 2), B(-2, 1), C(1,-3) is23/2 sq. units
Question
Find k if the area of the triangle with vertices at A(k, 2), B(-2, 1), C(1,-3) is23/2 sq. units
Solution
The area of a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3) is given by the formula:
Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Given that the area of the triangle is 23/2, we can substitute the given points into the formula:
23/2 = 1/2 * |k(1 - (-3)) + (-2)(-3 - 2) + 1(2 - 1)|
Solving the equation gives:
23 = |k(4) + (-2)(-5) + 1|
23 = |4k + 10 + 1|
23 = |4k + 11|
We can split this into two cases, one where the absolute value is positive and one where it is negative:
Case 1: 4k + 11 = 23 4k = 23 - 11 4k = 12 k = 12 / 4 k = 3
Case 2: -(4k + 11) = 23 -4k = 23 + 11 -4k = 34 k = 34 / -4 k = -8.5
So, the possible values for k are 3 and -8.5.
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