Question

The point of intersection of the straight lines represented by $6x^2+xy-40y^2-35x-83y+11=0$ is :

• A. $(0,\frac{-11}{5})$
• B. $(3,-1)$
• C. $(\frac{11}{2},0)$
• D. None of the above

Answer 1

The correct option is B $(3,-1)$

Point of intersection $=(\frac{hf-bg}{ab-h^2},\frac{hg-af}{ab-h^2})=(\frac{\frac{-83}{4}-\left(20\right)\left(35\right)}{\left(6\right)(-40)-{\left(\frac{1}{2}\right)}^2},\frac{\frac{-35}{4}+249}{\left(6\right)(-40)-{\left(\frac{1}{2}\right)}^2})$
$=(\frac{-2883}{-961},\frac{961}{-961})$
$\therefore$ point of intersection is $(3,-1)$