Families of lines and the shift of origin

Where two lines meet

Two lines that are not parallel meet at exactly one point. Solve their two equations together to find it.

Example. The lines $2x + y - 6 = 0$ and $x - y = 0$. The second says $y = x$. Put that into the first: $3x - 6 = 0$, so $x = 2$ and the lines meet at $(2, 2)$.

The family through that point

Write $L_1 = 0$ and $L_2 = 0$ for the two equations. At their meeting point both $L_1$ and $L_2$ are $0$. So for any number $k$, the expression $L_1 + kL_2$ is $0$ there too.

is a line, and it passes through the meeting point for every $k$. Every line through that point, except $L_2$ itself, is one of them.

This saves work. You do not need the meeting point to write a line through it.

Example. Find the line through the meeting point of $2x + y - 6 = 0$ and $x - y = 0$ that also passes through $(0, 5)$.

Step 1. Write the family. $(2x + y - 6) + k(x - y) = 0$.

Step 2. Use the point. At $(0, 5)$: $(5 - 6) + k(0 - 5) = 0$, so $-1 - 5k = 0$ and $k = -\dfrac{1}{5}$.

Step 3. Put $k$ back. $(2x + y - 6) - \dfrac{1}{5}(x - y) = 0$. Multiply by $5$: $9x + 6y - 30 = 0$, which is $3x + 2y - 10 = 0$.

Check. $(2, 2)$ gives $6 + 4 - 10 = 0$. $(0, 5)$ gives $10 - 10 = 0$.

The line $L_2$ itself is never in the family. It would need $k$ to be infinitely large.

A fixed point in a family

Sometimes a line has a letter $k$ in its equation, and we are asked to show it passes through a fixed point. Collect the terms with $k$.

Example. Show that $(1 + 2k)x + (2 - k)y + (3 - 4k) = 0$ passes through a fixed point for every $k$.

Step 1. Separate the parts. $(x + 2y + 3) + k(2x - y - 4) = 0$.

Step 2. Solve both brackets together. $x + 2y + 3 = 0$ and $2x - y - 4 = 0$ give $y = 2x - 4$ and then $5x - 5 = 0$.

Answer: the fixed point is $(1, -2)$.

Other families

• Lines parallel to $Ax + By + C = 0$: $Ax + By + k = 0$.
• Lines perpendicular to it: $Bx - Ay + k = 0$.
• Lines through a point $(x_1, y_1)$: $y - y_1 = m(x - x_1)$, which leaves out only the vertical line.

Three lines through one point

Three lines are concurrent when they pass through one point. To test it, find the meeting point of two of them and put it into the third.

Example. Are $x + y - 4 = 0$, $2x - y - 2 = 0$ and $3x + 2y - 10 = 0$ concurrent?

Step 1. Meet the first two. Adding gives $3x - 6 = 0$, so $x = 2$ and $y = 2$.

Step 2. Test the third. $3(2) + 2(2) - 10 = 0$.

Answer: yes, all three pass through $(2, 2)$. If a constant is unknown, make the third line vanish at the meeting point and solve for it.

Shifting the origin

Sometimes an equation is simpler if the axes are moved, without turning them. Move the origin to the point $(h, k)$. A point with old coordinates $(x, y)$ then has new coordinates $(X, Y)$, where

Example 1. The origin moves to $(2, 3)$. The point $(5, 7)$ has new coordinates $X = 5 - 2 = 3$ and $Y = 7 - 3 = 4$.

Example 2. With the origin at $(2, 3)$, rewrite $3x + 2y = 12$.

Step 1. Replace $x$ and $y$. $3(X + 2) + 2(Y + 3) = 12$.

Step 2. Collect. $3X + 2Y + 12 = 12$, so $3X + 2Y = 0$.

The new equation has no constant term. The point $(2, 3)$ is on the line, so the new origin lies on the line. Shifting the origin onto a point of a line always removes the constant.

Shifting the axes changes only the labels of the points. Slopes, angles and distances stay the same.

Practise with new numbers

Try it in Python

The program meets two lines with exact fractions, builds members of the family, and checks that each one passes through the meeting point.

Common mistakes

• Forgetting the letter $k$. The family is $L_1 + kL_2 = 0$, and $k$ is found from one extra fact.
• Expecting $L_2$ in the family. $L_2$ itself is never a member, so test it on its own.
• Moving the origin the wrong way. The new coordinate is $X = x - h$, not $x + h$.
• Testing concurrency by looking only at slopes. Three lines can be pairwise non-parallel and still not meet at one point.
• Meeting two lines with wrong signs. Substitute the answer back into both equations.

Remember these

• $L_1 + kL_2 = 0$ gives every line through the meeting point of $L_1 = 0$ and $L_2 = 0$, except $L_2$.
• To find a fixed point, collect the terms with $k$ into $L_1 + kL_2$ and solve $L_1 = 0$, $L_2 = 0$.
• Concurrent lines: the meeting point of two lies on the third.
• Shift of origin to $(h, k)$: $X = x - h$, $Y = y - k$.
• A shift of axes does not change slopes or distances.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the fixed point of the family $(x + y - 3) + k(x - y + 1) = 0$.

2. Find the line through the meeting point of $x + y - 2 = 0$ and $x - y = 0$ that passes through $(3, 0)$.

3. Are $2x + y - 4 = 0$, $x - y - 2 = 0$ and $x + 3y - 2 = 0$ concurrent?

4. The origin moves to $(4, 3)$. What are the new coordinates of $(6, -1)$?

5. The origin moves to $(1, 2)$. Write $4x - y - 2 = 0$ in the new axes.