The area of the triangle with vertices \(A(k,\,2)\), \(B(-1,\,-3)\) and \(C(3,\,1)\) is 10 square units. Which of the following could be the value of \(k\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(9\)
Write the area with the formula \(\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|\) in terms of \(k\) and set it equal to 10. Because of the modulus, the expression inside can be \(+20\) or \(-20\).
\(k(-3-1)+(-1)(1-2)+3(2-(-3))=-4k+1+15=16-4k\)
\(\frac{1}{2}\left|16-4k\right|=10\), so \(\left|16-4k\right|=20\)
\(16-4k=20\) gives \(k=-1\)
\(16-4k=-20\) gives \(4k=36\), so \(k=9\)
The possible values are \(k=-1\) and \(k=9\); of the options, only \(9\) is one of them
Check with \(k=9\): \(\frac{1}{2}\left|16-36\right|=\frac{1}{2}\times20=10\) square units
