JEE Mains · Maths · STD 11 - 9. straight line
Let the line \(x+y=1\) meet the axes of \(x\) and \(y\) at A and B, respectively. A right angled triangle AMN is inscribed in the triangle OAB , where O is the origin and the points M and N lie on the lines \(O B\) and \(A B\), respectively. If the area of the triangle \(A M N\) is \(\frac{4}{9}\) of the area of the triangle \(O A B\) and AN : NB \(=\lambda: 1\), then the sum of all possible value(s) of is \(\lambda\) :
- A 2
- B \(\frac{5}{2}\)
- C \(\frac{1}{2}\)
- D \(\frac{13}{6}\)
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text {Area of } \triangle \mathrm{AOB}=\frac{1}{2} \\ & \text {Area of } \triangle \mathrm{AMN}=\frac{4}{9} \times \frac{1}{2}=\frac{2}{9} \end{aligned}\) Equation of AB is \(\mathrm{x}+\mathrm{y}=1\)…
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