The straight line $5 x+4 y=0$ passes through the point of intersection of the straight lines $x+2 y-10=0$ and $2 x+y+5=0$.

The vertex of an equilateral triangle is $(2,3)$ and the equation of the opposite side is $x+y=2$. Then, the other two sides are $y-3=(2 \pm \sqrt{3})(x-2)$.

The equation of the line joining the point $(3,5)$ to the point of intersection of the lines $4 x+y-1=0$ and $7 x-3 y-35=0$ is equidistant from the points $(0,0)$ and $(8,34)$.

The line $\frac{x}{a}+\frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$, where $c$ is a constant. The locus of the foot of the perpendicular from the origin on the given line is $x^2+y^2=c^2$.

The lines $a x+2 y+1=0, \quad b x+3 y+1=0$ and $c x+4 y+1=0$ are concurrent, if $a, b$ and $c$ are in GP.