Discussion Forum

If $2x^{2}-10xy+2\lambda y^{2}+5x-16y-3=0$ represents a pair of straight  lines, then the point of intersection of those lines is 

Answer:

Explanation:

 On comparing  the given equations with

 $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ , we get

 $a=2, b=2 \lambda,h=-5, g=\frac{5}{2}$

     f=-8,c=-3

 Since , the given equations  represents a pair of straight  lines, therefore

$\begin{bmatrix}a & h&g \\h &b&f\\g&f&c \end{bmatrix}=0$

  $\Rightarrow\begin{bmatrix}2 & -5&5/2 \\-5 &2\lambda&-8\\5/2&-8&-3 \end{bmatrix}=0$

$\Rightarrow$      $2(-6 \lambda-64)+5(15+20)+\frac{5}{2}(40-5\lambda)=0$

$\Rightarrow$   $-12 \lambda-128+175+100-\frac{25}{2} \lambda=0$

$\Rightarrow$     $-\frac{49}{2} \lambda=-147$

$\Rightarrow$   $\lambda= \frac{147 \times 2}{49}=6$

Now, the point of intersection of given lines is given by

    $\left(\frac{bg-fh}{h^{2}-ab},\frac{af-gh}{h^{2}-ab}\right)=\left(\frac{30-40}{25-24},\frac{-16+\frac{25}{2}}{25-24}\right)$

              =$\left(-10,\frac{-7}{2}\right)$