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If \mathrm{S=x^2+m x y-2 y^2+3 y-1=0} represents two straight lines, they intersect at the point
 

Option: 1

\mathrm{\left(\frac{1}{3}, \frac{2}{3}\right)}
 


Option: 2

\mathrm{\left(-\frac{1}{3},-\frac{2}{3}\right)}
 


Option: 3

\mathrm{\left(\frac{1}{3},-\frac{2}{3}\right)}
 


Option: 4

\mathrm{\left(\frac{1}{3}, \frac{1}{3}\right)}


Answers (1)

best_answer

In standard form, the condition for a pair of straight lines is

\mathrm{ a b c+2 f g h-a f^2-b g^2-c h^2=0 }

\mathrm{Or \: 1(-2)(-1)+2\left(\frac{3}{2}\right) 0-1\left(\frac{3}{2}\right)^2-(-2) 0-(-1)\left(\frac{m}{2}\right)^2=0 }

\mathrm{Or \: \: m^2=1 \Rightarrow m= \pm 1 }

The point of intersection of the lines is given by the solution of

\mathrm{ \frac{\partial s}{\partial x} \equiv 2 x+m y=0, \frac{\partial s}{\partial y}=m x-4 y+3=0 }

When \mathrm{ m=1 }, the point of intersection is \mathrm{ x=-\frac{1}{3}, y=\frac{2}{3} }
When \mathrm{ m=-1 }, the point of intersection is \mathrm{ x=\frac{1}{3}, y=\frac{2}{3} }.

Hence option 1 is correct.

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