\(\frac{9}{2}\)
\(\frac{3\sqrt17}{2}\)
\(\frac{3\sqrt17}{4}\)
9
Let point P be (h, k)
\((ℎ–1)^2+(k–2)^2+(ℎ+2)^2+(k–1)^2=14\)
\(2ℎ^2+2k^2+2ℎ–6k–4=0\)
Locus of point P : x2 + y2 + x – 3y – 2 = 0
Intersection with x-axis,
x2 + x – 2 = 0
x = –2, 1
Intersection with y-axis,
y2 – 3y – 2 = 0
\(y=\frac{3±\sqrt17}{2}\)
Area of the quadrilateral ACBD is \(=\frac{1}{2}(|x1|+|x2|)(|y1|+|y2|)\)
\(=\frac{1}{2}×3×\sqrt17=\frac{3\sqrt17}{2}\)
So, the correct option is (B): \(\frac{3\sqrt17}{2}\)