Find the value of \(k\) so that the area of the triangle with vertices \(A(k+1,\,1)\), \(B(4,\,-3)\) and \(C(7,\,-k)\) is 6 square units.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3
The area of a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\) is \(\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|\). Setting it equal to 6 gives an equation in \(k\).
The expression inside the bars (with \(A(k+1,\,1)\), \(B(4,\,-3)\), \(C(7,\,-k)\)):
Putting in the coordinates: \((k+1)(-3+k)+4(-k-1)+7(1+3)\)
This comes to \(k^2-2k-3-4k-4+28=k^2-6k+21\)
Solving for \(k\) (the area is 6, so the expression is \(\pm12\)):
\(k^2-6k+21=12\) gives \(k^2-6k+9=0\), that is \((k-3)^2=0\), so \(k=3\)
\(k^2-6k+21=-12\) gives \(k^2-6k+33=0\), which has no real root (\(36-132\lt0\))