Computing the Intersection of Two Lines

Two different straight lines that lie in a plane and are not parallel always have an intersection point.

Given are usually two linear functions $f$ and $g$ with the general equations of the straight line:

$f (x) = m_{1} \cdot x + t_{1}$ and
$g (x) = m_{2} \cdot x + t_{2}$ .

We are looking for the intersection point $S (a | b)$ .

For this point $S (a | b)$ , it holds that

f (a) = g (a) = b

Example

Consider the following linear functions:

$f (x) = 3 \cdot x + 2$
$g (x) = 2 \cdot x + 5$ .

Calculate the intersection point.

Set the functions equal and bring $x$ to one side of the equation:

\begin{array}{rcll} f (x) & = & g (x) \\ 3 \cdot x + 2 & = & 2 \cdot x + 5 & | - 2 \\ 3 \cdot x & = & 2 \cdot x + 3 & | - 2 \cdot x \\ x & = & 3 \end{array}

The function value of $f$ and $g$ are therefore equal at $x = 3$ . The point of intersection of the two lines is therefore at $S (3 | ?)$ .

You can now calculate the missing $y$ -value by plugging $x = 3$ into $f (x)$ or $g (x)$ . It doesn't make any difference which of both you choose, since according to the calculation above $f (3) = g (3)$ .

$x = 3$ plugged into $f (x)$ yields:

f (3) = 3 \cdot 3 + 2 = 9 + 2 = 11

The intersection point is therefore at $S (3 | 11)$ .

General procedure

You set the two functions equal and bring $x$ to one side of the equation.

\begin{array}{rcll} f (x) & = & g (x) \\ m_{1} \cdot x + t_{1} & = & m_{2} \cdot x + t_{2} & | - t_{1} \\ m_{1} \cdot x & = & m_{2} \cdot x + (t_{2} - t_{1}) & | - m_{2} \cdot x \\ m_{1} \cdot x - m_{2} \cdot x & = & t_{2} - t_{1} \\ (m_{1} - m_{2}) \cdot x & = & t_{2} - t_{1} & | : (m_{1} - m_{2}) \\ x & = & \frac{t_{2} - t_{1}}{m_{1} - m_{2}} \end{array}

The x-value of the intersection point is therefore at $x = \frac{t_{2} - t_{1}}{m_{1} - m_{2}}$ .

Now plug $x = \frac{t_{2} - t_{1}}{m_{1} - m_{2}}$ into $f (x)$ or $g (x)$ to get the $y$ -value of the intersection point:

f (x) = m_{1} \cdot x + t_{1}

\Rightarrow f (\frac{t_{2} - t_{1}}{m_{1} - m_{2}}) = \underset{y-value of the intersection}{\underset{⏟}{m_{1} \cdot \frac{t_{2} - t_{1}}{m_{1} - m_{2}} + t_{1}}}

The intersection point of the lines is therefore $S (\frac{t_{2} - t_{1}}{m_{1} - m_{2}} | m_{1} \cdot \frac{t_{2} - t_{1}}{m_{1} - m_{2}} + t_{1})$ .

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