Chapter 4 · Geometry and Equations · Concept 2
4-14 Line Passing Through a Fixed Point
Change one number in an equation and you get a different line — different slope, different position. And yet a whole family of them can be pinned through a single point that never moves. This lesson finds it.
Guiding question
What can stay the same while everything else changes?
Learning outcome. Find the fixed point a family of lines passes through.
Make k a spectator
When a line “passes through a fixed point for all values of k”:
1 · Rearrange with k as the subject
(ax + by + c)k + (a′x + b′y + c′) = 0
Gather every term that has a k in it into the first bracket, and everything else into the second.
2 · Both brackets must be zero
ax + by + c = 0
a′x + b′y + c′ = 0
Solve the two together. That crossing point is the answer.
Why both brackets have to be zero.
The statement has to hold for every k — including k = 0, which wipes out the first bracket entirely and leaves the second one equalling zero on its own. And once the second is zero, the first must be too, since (something) × k = 0 for every k only when the something is 0. That is what the book means by treating it as an identity in k.
This is not solving for k.
k is never found and never needs to be — the answer is a point, in x and y. If your working ends with a value of k, the question has been misread.
Drag k first, on its own, and watch the line swing about. Then press the button and see every line at once.
(k − 4)x + (k + 2)y + k + 5 = 0
Every k gives a different line — different slope, different position. Drag it and watch what does not move.
The line (k − 4)x + (k + 2)y + k + 5 = 0 passes through a fixed point whatever k is. Find it.
- 1
Open every bracket first, so the k terms can be seen.
kx − 4x + ky + 2y + k + 5 = 0Nothing clever yet — just multiplying out so the sorting can happen.
Work with a partner. Once you have the point, each of you picks a different value of k, writes out that particular line, and checks the point satisfies it. Two different k values agreeing is convincing in a way the algebra alone is not.
Question 1 has an easy second bracket.
2x − 6 = 0 gives x = 3 straight away, with no simultaneous solving at all. Look at both brackets before choosing which to start with — sometimes one hands you a coordinate for free.
−(k − 2)x + (k − 3)y = −3k + 5
This passes through a fixed point for all values of k. Find it.
Solving for k.
k is never found. The answer is a point (x, y), and every trace of k disappears on the way there.
Not moving everything to one side first.
The method needs "… = 0". The Challenge is written with a right-hand side on purpose.
Losing a sign when opening brackets.
−(k − 2)x is −kx + 2x, and −(k + 2)y is −ky − 2y. Both terms inside change sign, not just the first.
Missing a k term when sorting.
Every term with a k in it goes in the first bracket. Count them before and after — the totals must match.
Setting only one bracket to zero.
Both must be. One equation cannot pin down two coordinates.
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