99 Percentile Qs Bank for JEE MainMathematicsVector Algebra
A ( a → ) ,   B ( b → ) ,   C ( c → ) ,   D ( d → ) are four concyclic points, such that x a → + y b → + z c → + t d → = 0 → ,   x + y + z + t = 0 , where x ,   y ,   z ,   t are constants not all zero. If the chords A B and C D intersect at P , then
Options
- Ax y a → + c → 2 = z t b → + d → 2
- Bx y a → - b → 2 = z t c → - d → 2
- Cx t a → - c → 2 = y z b → - d → 2
- Dx z b → + d → 2 = y t a → + c → 2
Correct answer
B. x y a → - b → 2 = z t c → - d → 2
Step-by-step solution
We are given that A a → ,   B b → ,   C c →   &   D d → are four concyclic points such that x a → + y b → + z c → + t d → = 0 →     . . . i x + y + z + t = 0     . . . ii Now, we know that angle subtended by chord on same side of circle are equal therefore ∠ A = ∠ C ∠ B = ∠ D ∠ P = ∠ P Hence, by A A A △ A P D ~ ∆ C P B So, A P C P = P D P B ⇒ r - a → r - c