NIMCET 2017 Question Paper with solutions

As we know that the pair of straight line
$\mathit{a}{\mathit{x}}^2$ + 2hxy + $\mathit{b}{\mathit{y}}^2$ + 2gx + 2fy + c = 0 intersect each other at
$\left(\frac{\mathit{h}\mathit{f}-\mathit{b}\mathit{g}}{\mathit{a}\mathit{b}-{\mathit{h}}^2},\frac{\mathit{h}\mathit{g}-\mathit{a}\mathit{f}}{\mathit{a}\mathit{b}-{\mathit{h}}^2}\right)$
Therefore, point of intersection of pair of straight line
${\mathit{x}}^2$ + 3xy + $2{\mathit{y}}^2$ - x - 4y - 6 = 0 is
= $(\frac{\frac{3}{2}\times \left(-2\right)-\left(2\right)\left(-\frac{1}{2}\right)}{\left(1\times 2\right)-{\left(\frac{3}{2}\right)}^2}$ , $\frac{\frac{3}{2}\times \left(-\frac{1}{2}\right)-1\times \left(-2\right)}{1\times 2-{\left(\frac{3}{2}\right)}^2})$
= $\frac{-3+1}{2-\frac{9}{4}}$ , $\frac{-\frac{3}{4}+2}{2-\frac{9}{4}}$
= $\left(\frac{-2}{-\frac{1}{4}},\frac{\frac{5}{4}}{-\frac{1}{4}}\right)$ = (8 , - 5)