Prove that the pair of lines $6x^2+5xy-4y^2+7x+13y-3=0$ form a parallelogram with the pair of lines $6x^2+5xy-4y^2=0$. Find its area.
My attempt/ I factorized the second equation to get the two lines represented by it: $$6x^2+5xy-4y^2=0$$ $$6x^2+8xy-3xy-4y^2=0$$ $$(3x+4y)(2x-y)=0$$.
No what should I should next to complete.?
The other pair would simplify to $(3x+4y+k)(2x-y-l)=0; $ Now area of the parallelogram is $$k*l\over{\sqrt{(3^2+4^2)(2^2+1^2)}}$$
Just find k and l from substitution.
Distance between $(3x+4y+k=0) and (3x+4y=0)$ = $k\over{\sqrt{3^2+4^2}}$
Distance between $(2x-y+l=0) and (2x-y=0)$ = $l\over{\sqrt{2^2+1^2}}$
Area of parallelogram = Product of the distances.