Question

A point P moves so that the sum of squares of its distances from the points (1, 2) and (–2, 1) is 14. Let f(x, y) = 0 be the locus of P, which intersects the x-axis at the points A, B and the y-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to

• A. $\frac{9}{2}$
• B. $\frac{3\sqrt17}{2}$
• C. $\frac{3\sqrt17}{4}$
• D. 9

Answer 1

Answer: (B)

$\frac{3\sqrt17}{2}$

Answer 2

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): $\frac{3\sqrt17}{2}$