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9.Straight Line
hard
ધારો કે ${ }^{ n } C _{ r -1}=28,{ }^{ n } C _{ r }=56$ અને ${ }^{ n } C _{ r +1}=70$. ધારો કે $A (4 \cos t, 4 \sin t ), B (2 \sin t ,-2 \cos t )$ અને $C$ $\left(3 r - n , r ^2- n -1\right)$ એ ત્રિકોણ $ABC$ ના શિરોબિંદુઓ છે, જ્યાં $t$ પ્રચલ છે. જો $(3 x -1)^2+(3 y )^2=\alpha$ એ ત્રિકોણ $ABC$ ના મધ્યકેન્દ્રનો બિંદુપથ હોય, તો $\alpha=$ __________.
A$20$
B$8$
C$6$
D$18$
(JEE MAIN-2025)
Solution
${ }^n C_{r-1}=28,{ }^n C_r=56$
$\frac{{ }^n C_{r-1}}{{ }^n C}=\frac{28}{56}$
$\frac{n!}{\frac{(r-1)!(n-r+1)!}{r!(n-r)!}}=\frac{1}{2}$
$\frac{r}{(n-r+1)}=\frac{1}{2}$
$3 r=n+1$
$\frac{{ }^{n} C_{r}}{{ }^{n} C_{r+1}}=\frac{56}{70}$
$\frac{(r+1)}{(n-r)}=\frac{56}{70} \Rightarrow 9 r=4 n-5 \text { (ii) }$
$By(i) \ (ii)$
$(r=3),(n=8)$
$A(4 \operatorname{cost}, 4 \sin t) B(2 \sin t,-2 \cos t) C\left(3 r-n, r^2-n-1\right)$
$A(4 \operatorname{cost}, 4 \sin t) B(2 \sin t,-2 \cos t) C(1,0)$
$(3 x-1)^2+(3 y)^2=(4 \operatorname{cost}+2 \operatorname{sint})^2+(4 \sin t-\operatorname{cost})^2$
$(3 x-1)^2+(3 y)^2=20 \therefore(1)$
$\frac{{ }^n C_{r-1}}{{ }^n C}=\frac{28}{56}$
$\frac{n!}{\frac{(r-1)!(n-r+1)!}{r!(n-r)!}}=\frac{1}{2}$
$\frac{r}{(n-r+1)}=\frac{1}{2}$
$3 r=n+1$
$\frac{{ }^{n} C_{r}}{{ }^{n} C_{r+1}}=\frac{56}{70}$
$\frac{(r+1)}{(n-r)}=\frac{56}{70} \Rightarrow 9 r=4 n-5 \text { (ii) }$
$By(i) \ (ii)$
$(r=3),(n=8)$
$A(4 \operatorname{cost}, 4 \sin t) B(2 \sin t,-2 \cos t) C\left(3 r-n, r^2-n-1\right)$
$A(4 \operatorname{cost}, 4 \sin t) B(2 \sin t,-2 \cos t) C(1,0)$
$(3 x-1)^2+(3 y)^2=(4 \operatorname{cost}+2 \operatorname{sint})^2+(4 \sin t-\operatorname{cost})^2$
$(3 x-1)^2+(3 y)^2=20 \therefore(1)$
Standard 11
Mathematics
