# Line always passes through a fixed point for some parameters

My question goes like this

If 5a+4b+20c=t, then what is the value of t for which the line ax+by+c-1=0 always passes through a fixed point?

I tried but couldn't solve it so I looked at the solution. The solution says that the equation has 2 independent parameters. I get that. If we choose a and b, c automatically gets defined. Hence c is not really independent.

In the next line it says that it can pass through a fixed point if there is only one independent parameters. This is where I got stuck. I tried a lot but can't understand their logic.

Then they defined a relation between a/(c-1) and b/(c-1) to get the answer which is t=20.

Please help me because I feel understanding this would really clear my concepts and may prove very beneficial later.

Fix your point as $$(\alpha,\beta)$$. Then, since $$c = \frac{t - 5a - 4b}{20}$$, you need to find $$t$$ such that for any values of $$a$$ and $$b$$, $$a\alpha + b\beta + \frac{t- 5a - 4b}{20} - 1 = 0$$. In particular, with $$a = b = 0$$, we must have $$t = 20$$. But also, with $$a = 4, b = 0$$, we must have $$\alpha = \frac{1}{4}$$, and with $$a = 0, b = 5$$, we must have $$\beta = \frac{1}{5}$$.

Substituting these values in, we notice that, for any $$a$$ and $$b$$,

$$a\alpha + b\beta + \frac{t - 5a - 4b}{20} - 1 = \frac{a}{4} + \frac{b}{5} + \frac{20-5a-4b}{20} - 1 = \frac{5a+4b+20-5a-4b-20}{20} = 0,$$

so, indeed, this is a solution.

• Thank you, you made it absolutely clear. This method is definitely going in my armory. Jan 11 '19 at 13:43

A line in the plane normally has two degrees of freedom. You’re probably used to specifying them via slope and $$y$$-intercept (with an exception for vertical lines). For lines that go through a fixed point, you can still make an arbitrary choice of slope, but that choice determines the line’s $$y$$-intercept, so you’ve taken away a degree of freedom by requiring that the line pass through a certain point. To put it differently, there is a one-parameter family of lines through any given point. This means that, after substituting for $$c$$ in $$ax+by+c=1$$, the remaining unknown coefficients $$a$$ and $$b$$ are not independent, and this equation tells you how they’re related.