AP EAMCET · Maths · Straight Lines
The distance from the origin to the orthocentre of the triangle formed by the lines \(x+y-1=0\) and \(6 x^2-13 x y+5 y^2=0\) is
- A \(\frac{11 \sqrt{2}}{2}\)
- B 13
- C 11
- D \(\frac{11 \sqrt{2}}{24}\)
Answer & Solution
Correct Answer
(D) \(\frac{11 \sqrt{2}}{24}\)
Step-by-step Solution
Detailed explanation
Given lines are \(x+y-1=0\) and \(6 x^2-13 x y+5 y^2=0\) \[ \begin{array}{lc} \Rightarrow & 6 x^2-10 x y-3 x y+5 y^2=0 \\ \Rightarrow & 2 x(3 x-5 y)-y(3 x-5 y)=0 \\ \Rightarrow & (2 x-y)(3 x-5 y)=0 \\ \Rightarrow & 2 x-y=0 \\ \text { or } & 3 x-5 y=0 \end{array} \] Let…
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