# Prove that for any scalar t, C is a conic that goes through the four intersection points

Given four lines in the plane: $L_k:a_k x+b_k y+c_k=0$, k=1…4

Prove that, for any scalar t, $C:L_1 L_2+tL_3 L_4=0$ is a conic that goes through the four intersection points $L_1∩L_3$, $L_2∩L_3$, $L_1∩L_4$, $L_2∩L_4$.

Here, $C=(a_1 x+b_1 y+c_1 )(a_2 x+b_2 y+c_2 )+t(a_3 x+b_3 y+c_3 )(a_4 x+b_4 y+c_4 )$.

Hints: $C$ is given by a quadratic equation in $x,y$ for any fixed $t$, so it is a conic by definition. Morevover, for each of the intersection points $L_{1,2} \cap L_{3,4}$ one term in each of the products will be $0$ so $C = 0 + t \cdot 0 = 0$.