Answer:
Option C
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
$\left(\frac{bg-fh}{h^{2}-ab},\frac{af-gh}{h^{2}-ab}\right)=\left(\frac{30-40}{25-24},\frac{-16+\frac{25}{2}}{25-24}\right)$
=$\left(-10,\frac{-7}{2}\right)$
