Q.Find the separate equations of the following pair of straight lines

Each part asks us to split a degree-2 equation into its two linear factors, using whichever technique fits the equation's shape.

Step 1 (i). Solve $3x^2+2xy-y^2=0$ as a quadratic in $x$. Treating $y$ as constant: $3x^2+2yx-y^2=0$, so

This gives $x=\tfrac{2y}{6}=\tfrac{y}{3}$ or $x=\tfrac{-6y}{6}=-y$, i.e. $3x-y=0$ or $x+y=0$.

Step 2 (ii). Reduce $6(x-1)^2+5(x-1)(y-2)-4(y-2)^2=0$ by substitution. Let $X=x-1,\ Y=y-2$; the equation becomes $6X^2+5XY-4Y^2=0$. Solve for $X$ in terms of $Y$:

This gives $X=\tfrac{6Y}{12}=\tfrac{Y}{2}$ (i.e. $2X-Y=0$) or $X=\tfrac{-16Y}{12}=-\tfrac{4Y}{3}$ (i.e. $3X+4Y=0$).

Step 3 (ii cont.). Re-substitute $X=x-1,Y=y-2$.

$3X+4Y=0\Rightarrow 3(x-1)+4(y-2)=0\Rightarrow 3x+4y-11=0.$ …

1. quadratic-in-$x$;
2. shift variables then quadratic; …

• Forgetting to re-substitute $X=x-1,Y=y-2$ back to $x,y$ in part (ii)
• Sign errors while comparing coefficients $p,q$ in part (iii) …