The distance from the origin to the ortho-cetre of the triangle formed by the lines x+y-1=0 and $6 x^{2} - 13 x y + 5 y^{2} = 0 i s$

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$\frac{11 \sqrt{2}}{24}$

$\frac{11 \sqrt{2}}{2}$

$11$

$13$

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n s 6 x^{2} - 13 x y + 5 y^{2} = 0 \\ c o m p a r i g w i t h a x^{2} + 2 h x y + b y^{2} = 0 \\ a = 6, h = \frac{- 13}{2}, b = 5 \\ a n d x + y - 1 = 0 w i t h l x + m y + n = 0 \\ l = 1, m = 1, n = - 1 \\ K = \frac{- n (a + b)}{a m^{2} - 2 h l m + b l^{2}} \\ \Rightarrow \frac{1 (6 + 5)}{6 (1) + 13 (1) (1) + 5 {(1)}^{2}} \Rightarrow \frac{11}{24} \\ P = o r t h o c e n t r e = (k l, k m) = (\frac{11}{24} (1), \frac{11}{24} (1)) \Rightarrow (\frac{11}{24}, \frac{11}{24}) \\ a n d (0, 0) \\ O P = \sqrt{{(\frac{11}{24})}^{2} + {(\frac{11}{24})}^{2}} = \frac{11 \sqrt{2}}{24} \end{array}$