The ratio in which the line x+y-1=0 divides t | vedclass

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(A) Let the equation of the pair of lines be $f(x, y) = 2x^2 - 13xy - 7y^2 + x + 23y - 6 = 0$. To find the point of intersection,we solve $\frac{\part

Vedclass · Accepted Answer

$15:11$

Vedclass · Answer

(A) Let the equation of the pair of lines be $f(x, y) = 2x^2 - 13xy - 7y^2 + x + 23y - 6 = 0$. To find the point of intersection,we solve $\frac{\partial f}{\partial x} = 4x - 13y + 1 = 0$ and $\frac{\partial f}{\partial y} = -13x - 14y + 23 = 0$. Solving these simultaneous equations,we multiply the first by $14$ and the second by $13$: $56x - 182y + 14 = 0$ $-169x - 182y + 299 = 0$ Subtracting the two gives $225x - 285 = 0$,so $x = \frac{285}{225} = \frac{19}{15}$. Substituting $x$ into $4x - 13y + 1 = 0$,we get $4(\frac{19}{15}) + 1 = 13y$,which simplifies to $\frac{76+15}{15} = 13y$,so $y = \frac{91}{15 	imes 13} = \frac{7}{15}$. The point of intersection is $P(\frac{19}{15}, \frac{7}{15})$. The line $L: x+y-1=0$ divides the segment joining $O(0,0)$ and $P(\frac{19}{15}, \frac{7}{15})$ in the ratio $k:1$. The ratio is given by $-\frac{L(O)}{L(P)} = -\frac{0+0-1}{\frac{19}{15} + \frac{7}{15} - 1} = -\frac{-1}{\frac{26}{15} - 1} = \frac{1}{\frac{11}{15}} = \frac{15}{11}$. Thus,the ratio is $15:11$.

The ratio in which the line $x+y-1=0$ divides the line segment joining the origin $(0,0)$ and the point of intersection of the lines represented by $2x^2-13xy-7y^2+x+23y-6=0$ is

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