TS EAMCET · Maths · Straight Lines
The line \(2 x+y-3=0\) divides the line segment joining the points \(A(1,2)\) and \(B(-2,1)\) in the ratio \(a: b\) at the point \(C\). If the point \(C\) divides the line segment joining the points \(P\left(\frac{b}{3 a},-3\right)\) and \(Q\left(-3,-\frac{b}{3 a}\right)\) in the ratio \(p: q\), then \(\frac{p}{q}+\frac{q}{p}=\)
- A \(\frac{29}{10}\)
- B \(\frac{17}{10}\)
- C 6
- D 5
Answer & Solution
Correct Answer
(A) \(\frac{29}{10}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & C \equiv\left(\frac{-2 a+b}{a+b}, \frac{a+2 b}{a+b}\right), C \text { lies on } 2 x+y-3=0 \\ & \frac{-4 a+2 b}{a+b}+\frac{a+2 b}{a+b}-3=0 \\ & \Rightarrow-4 a+2 b+a+2 b-3 a-1 b=0 \Rightarrow 6 a=b \\ & C \equiv\left(\frac{4}{7}, \frac{13}{7}\right) \Rightarrow…
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