Relative position of two lines: intersecting, parallel, coincident

Decision table

Let $(d_1) : y = m_1 x + p_1$ and $(d_2) : y = m_2 x + p_2$.

Coefficients	Position
$m_1 \neq m_2$	Intersecting (1 common point)
$m_1 = m_2$ and $p_1 \neq p_2$	Strictly parallel (0 common points)
$m_1 = m_2$ and $p_1 = p_2$	Coincident (infinitely many points)

3-step method

1. Put both equations in the form $y = mx + p$.
2. Compare slopes $m_1$ and $m_2$.
3. If equal, compare y-intercepts.

Example

$(d_1) : y = 2x + 3$ and $(d_2) : 4x - 2y + 6 = 0$.

Isolate $y$ in $(d_2)$: $y = 2x + 3$.

$m_1 = m_2 = 2$ and $p_1 = p_2 = 3$ ⇒ coincident.

Intersection point (intersecting)

Solve the system: $\begin{cases} y = m_1 x + p_1 \\ y = m_2 x + p_2 \end{cases}$

By equality: $m_1 x + p_1 = m_2 x + p_2$ ⇒ $x = \dfrac{p_2 - p_1}{m_1 - m_2}$.

Vertical line case

$x = a$ has no slope (infinite slope).

• $x = a$ and $x = b$: parallel if $a \neq b$, coincident if $a = b$.
• $x = a$ and $y = mx + p$: always intersecting at $(a, ma + p)$.

Common mistakes

1. Comparing $p$ before $m$ (order matters).
2. Forgetting to put in the form $y = mx + p$.
3. Confusing parallel and coincident: parallel = 0 points, coincident = all points.

?À toi de jouer

(d1) : y = 3x − 1 and (d2) : y = 3x + 2. Position?

A.Strictly parallelB.IntersectingC.CoincidentD.Perpendicular

Valider