Q.Show that lines 3x − 4y + 5 = 0, 7x − 8y + 5 = 0 and 4x + 5y − 45 = 0 are concurrent. Find their point of concurrence.

A family of lines is the set of all lines satisfying one stated condition, described by an equation containing an arbitrary constant (parameter) in place of the specific numbers that would pin down one particular line. Since a line's general equation $ax+by+c=0$ genuinely has only two independent constants, imposing one further linear condition on those constants leaves one free parameter — a one-parameter family; imposing no further condition (letting both constants vary freely, subject only to the line passing through a fixed point, say) gives a two-parameter family.

One-parameter families from $y=mx+b$:

• $m$ arbitrary, $b$ fixed — all lines with a common, fixed $y$-intercept $(0,b)$; different real $m$ sweep out every line through that one point except the vertical line $x=0$.
• $b$ arbitrary, $m$ fixed — all lines with a common, fixed slope, i.e. all lines parallel to one another. As a special case, the family of lines parallel to $ax+by+c=0$ is $ax+by+\lambda=0$, and the family of lines perpendicular to $ax+by+c=0$ is $bx-ay+\lambda=0$, for varying $\lambda$.

Two-parameter families:

• Both $m$ and $b$ vary in $y=mx+b$ — genuinely two independent parameters; a sub-case that is easy to visualise is $y-y_1=m(x-x_1)$ for a fixed point $(x_1,y_1)$ and varying slope $m$: this sweeps out every non-vertical line through $(x_1,y_1)$ (the vertical line $x=x_1$ through the same point is the one member point-slope form structurally cannot express, since its slope is undefined).
• Lines through the intersection of two given lines. If $L_1\equiv a_1x+b_1y+c_1=0$ and $L_2\equiv a_2x+b_2y+c_2=0$ meet at one point, then $L_1+\lambda L_2=0$ (for real $\lambda$) is the equation of every line through that intersection point except $L_2=0$ itself — a one-parameter family, even though it is built from two given lines, because it is really parametrising 'lines through one fixed point' just like the point-slope family above. …