In ΔABC,ab−r1r2r3≠

1r3ab−r1r2=ab−Δ2(s−a)(s−b)Δ(s−c)=[ab−s(s−c)](s−c)Δ=4ab−(a+b)2−c2(s−c)4Δ=s(s−a)(s−b)(s−c)sΔ=Δs=r Similarly, bc−r2r3r1=ac−r1r3r2=r

In ΔABC,ab−r1r2r3≠

1r3ab−r1r2=ab−Δ2(s−a)(s−b)Δ(s−c)=[ab−s(s−c)](s−c)Δ=4ab−(a+b)2−c2(s−c)4Δ=s(s−a)(s−b)(s−c)sΔ=Δs=r Similarly, bc−r2r3r1=ac−r1r3r2=r

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In $ΔABC, \frac{(ab - r_{1} r_{2})}{r_{3}} \neq$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{r_{1}} (b c - r_{2} r_{3})$

$\frac{1}{r_{2}} (a c - r_{1} r_{3})$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{matrix} \frac{1}{r_{3}} (a b - r_{1} r_{2}) = \frac{a b - (\frac{Δ^{2}}{(s - a) (s - b)})}{\frac{Δ}{(s - c)}} \\ = \frac{[a b - s (s - c)] (s - c)}{Δ} \\ = \frac{\{4 a b - [(a + b)^{2} - c^{2}]\} (s - c)}{4 Δ} \\ = \frac{s (s - a) (s - b) (s - c)}{s Δ} = \frac{Δ}{s} = r \\ Similarly, \frac{b c - r_{2} r_{3}}{r_{1}} = \frac{a c - r_{1} r_{3}}{r_{2}} = r \end{matrix}$