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 point C, D. Then the area of the quadrilateral ACBD is equal to

(A) \(\frac92\)

(B) \(\frac{3\sqrt{17}}{2}\)

(C) \(\frac{3\sqrt{17}}{4}\)

(D) 9

Answer 1

Correct option is (B) \(\frac{3\sqrt{17}}{2}\)  

Comments

THANKS

Answer 2

Correct option is (B) \(\frac{3\sqrt {17}}{2}\)

Area of quadrilateral ACBD = Area of △ACB + Area of △ABD

\(=\frac 12 \times AB \times OC + \frac12\times AB \times OD\)

\(= \frac 12 \times AB (OC + OD)\)

\(=\frac 12 \times AB\times CD\)

\(= \frac 12(1 - (-2)) \times \left(\frac{3 + \sqrt{17}}2 - \frac{3 - \sqrt{17}}2\right)\)

\(= \frac12 \times 3\times \sqrt{17}\)

\(= \frac{3\sqrt{17}}2\) square units