Final answer:
The area of triangle JKL with given vertices is found to be 24 square units, which is a result of multiplying one-half with the product of its legs, which are 4 and 12 units long. The provided answer choices do not include the correct area, but the calculation yields a precise and accurate result.
Step-by-step explanation:
To find the area A of JKL with vertices J(−4,10), K(0,3), and L(−4,−2), we can observe that the triangle JKL is a right triangle, with J and L lying on the same vertical line (x-coordinate is −4), hence the length of the leg JL is the difference in y-coordinates which is 10 − (−2) = 12 units.
The other leg JK has length given by the difference in x-coordinates since K and J lie on the same horizontal line (K is at y = 3 and J is at y = 10). The length of JK is 0 − (−4) = 4 units.
The area of the triangle is one half the base times the height, hence the area can be calculated using the formula:
area = 1/2 * base * height = 1/2 * JK * JL = 1/2 * 4 * 12 = 24 square units.
Therefore, the area A of triangle JKL is 24 square units, which is closest to answer choice (b) 30 square units, albeit none of the provided answer choices are the exact match. The correct answer should be 24 square units, which appears to be missing from the options provided in the question.