Parallel lines, perpendicular lines and meeting points

Every point on the $\displaystyle x$-axis has the form $\displaystyle (x,0)$, and every point on the $\displaystyle y$-axis the form $\displaystyle (0,y)$. So:

• where it crosses the $\displaystyle x$-axis: substitute $\displaystyle {y}={0}$ in the line’s equation;
• where it crosses the $\displaystyle y$-axis: substitute $\displaystyle {x}={0}$.

Example: the line $\displaystyle {2x-3y}={12}$.

• $\displaystyle {y}={0}$: $\displaystyle {2x}={12}$, so $\displaystyle {x}={6}$ – the point $\displaystyle A(6,0)$;
• $\displaystyle {x}={0}$: $\displaystyle {-3y}={12}$, so $\displaystyle {y}={-4}$ – the point $\displaystyle B(0,-4)$.

Mark the points where the line $\displaystyle {y}={-\frac{3}{4}x+3}$ crosses the axes.

Two lines that have slopes lie in one of three ways, and the slopes and the points where they cross the $\displaystyle y$-axis decide which:

• the same line – the same slope and the same $\displaystyle b$: one line, written twice. For example $\displaystyle {y}={2x+1}$ and $\displaystyle {4x-2y+2}={0}$;
• parallel – the same slope, a different $\displaystyle b$: the lines never meet. For example $\displaystyle {y}={2x+1}$ and $\displaystyle {y}={2x-3}$;
• crossing – different slopes: the lines have one point in common.

Parallel lines, and a line written twice, have the same slope:

Two lines are perpendicular when the product of their slopes is $\displaystyle -1$:

The drawing shows why: turn the line $\displaystyle {y}={2x}$ a quarter turn about the origin.

• The step from $\displaystyle (0,0)$ to $\displaystyle A(1,2)$: $\displaystyle 1$ to the right and $\displaystyle 2$ up;
• after a quarter turn it becomes the step from $\displaystyle (0,0)$ to $\displaystyle B(-2,1)$: $\displaystyle 2$ to the left and $\displaystyle 1$ up;
• the slope of the turned line: $\displaystyle {\frac{1}{-2}}={-\frac{1}{2}}$, and $\displaystyle {2\cdot\left(-\frac{1}{2}\right)}={-1}$.

So for any slope $\displaystyle m$: the step $\displaystyle (1,m)$ becomes $\displaystyle (-m,1)$, and the slope $\displaystyle -\frac{1}{m}$. A line parallel to the $\displaystyle x$-axis is perpendicular to a line parallel to the $\displaystyle y$-axis – the second has no slope.

How do the two lines lie: the same line, parallel, perpendicular, or crossing without being perpendicular?

1. $\displaystyle {y}={3x+1}$ and $\displaystyle {y}={3x-4}$
2. $\displaystyle {y}={-\frac{1}{2}x+3}$ and $\displaystyle {y}={2x-1}$
3. $\displaystyle {y}={4x}$ and $\displaystyle {y}={-4x+1}$
4. $\displaystyle {x-y+2}={0}$ and $\displaystyle {y}={x+2}$

Write the equation of the line through the point $\displaystyle P(2,5)$ parallel to the line $\displaystyle {y}={3x-4}$.

Write the equation of the line through the point $\displaystyle P(4,1)$ perpendicular to the line $\displaystyle {y}={2x+3}$.

The meeting point is on both lines, so its coordinates satisfy both equations: it is the solution of the system of equations.

Example: $\displaystyle {y}={2x-1}$ and $\displaystyle {y}={-x+5}$.

• $\displaystyle {2x-1}={-x+5}$, so $\displaystyle {3x}={6}$ and $\displaystyle {x}={2}$;
• $\displaystyle {y}={2\cdot2-1}={3}$;
• the lines meet at $\displaystyle M(2,3)$.

Parallel lines have no point in common, and their system has no solution.

The lines are $\displaystyle {y}={2x-3}$ and $\displaystyle {x+y}={6}$.

Mark on the board the point where the two lines meet.

The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(0,0)$, $\displaystyle B(4,2)$ and $\displaystyle C(3,4)$.

Find the slopes of $\displaystyle AB$ and of $\displaystyle BC$.

Is the triangle $\displaystyle ABC$ right-angled?
If it is – at which vertex is the right angle?
Explain.