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  • A straight line passing through the point <img alt="\mathrm{A \equiv(\lambda, 2 \lambda)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BA%20%5Cequiv%28%5Clambda%2C%202%20%5

A straight line passing through the point \mathrm{A \equiv(\lambda, 2 \lambda)} and making an angle of 60° with the positive direction of the x-axis intersects the lines \mathrm{2 x+y=5 \text { and } x+y=2} at the points B and C respectively. If A lies in midway between B and C, then \mathrm{\lambda=k \frac{7 \sqrt{3}+9}{7 \sqrt{3}+10}} where k =

Option: 1

-1


Option: 2

1


Option: 3

2


Option: 4

\frac{1}{2}


Answers (1)

best_answer

Let,

\mathrm{B \equiv\left(x_1, y_1\right) \& C \equiv\left(x_2, y_2\right)}

\therefore \frac{\mathrm{x}_1-\lambda}{\cos 60^{\circ}}=\frac{\mathrm{y}_1-2 \lambda}{\sin 60^{\circ}}=\mathrm{AB}=+\mathrm{r} \text { (say) }

\mathrm{\Rightarrow x_1=\frac{r}{2}+\lambda\: \& \: y_1=\frac{r \sqrt{3}}{2}+2 \lambda}

Also \mathrm{\frac{x_2-\lambda}{\cos 60^{\circ}}=\frac{y_2-2 \lambda}{\sin 60^{\circ}}=A C=-r[\text A B=AC]}

\mathrm{\Rightarrow x_2=\lambda-\frac{\frac{r}{2}}{2} \& y_2=2 \lambda-\frac{r \sqrt{3}}{2}}

Again \mathrm{2 x_1+y_1=5}

\mathrm{\Rightarrow 2{\left(\frac{r}{2}+\lambda\right)+\left(\frac{r \sqrt{3}}{2}+2 \lambda\right)}=5}

\mathrm{\Rightarrow r=\frac{5-4 \lambda}{1+\frac{\sqrt{3}}{2}}}                  .......(1)

and \mathrm{x_2+y_2=2 \Rightarrow\left(\lambda-\frac{r}{2}\right)+\left(2 \lambda-\frac{r \sqrt{3}}{2}\right)=2}

\mathrm{\Rightarrow r=\frac{3 \lambda-2}{\frac{1}{2}+\frac{\sqrt{3}}{2}}}                       .... (2)

Equating r from both the equations (1) & (2), we get

\mathrm{\frac{5-4 \lambda}{1+\frac{\sqrt{3}}{2}}=\frac{\frac{3 \lambda-2}{1+\sqrt{3}}}{2} \text { or }(5-4 \lambda)(1+\sqrt{3})=(3 \lambda-2)(2+\sqrt{3})}

Or \mathrm{9+7 \sqrt{3}=10 \lambda+7 \lambda \sqrt{3} \Rightarrow \lambda=\frac{7 \sqrt{3}+9}{7 \sqrt{3}+10}}

Posted by

avinash.dongre

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