Problems from Relative positions between straight lines

Given straight lines $r : 3 x - y + 2 = 0$ and $s : y = - x + 4$ , find a parallel straight line to $r$ that crosses point $P = (1, 1)$ .

Determine the relative position between $r$ and $s$ , and look for the intersection point between $s$ and the parallel of $r$ .

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Development:

To begin with, we remember that two straight lines are parallel if, and only if, their director vectors are so.

A vector director of the straight line $r$ is $\vec{v} = (1, 3)$ . Therefore, if we use the vector equation of the straight line that crosses $P = (1, 1)$ and it is parallel to $r$ , it is: $(x, y) = (1, 1) + k \cdot (1, 3)$ If we now look for the relative position between $r$ and $s$ :

$r : 3 x - y + 2 = 0 \to r : y = 3 x + 2$ (explicit equation)

$s : y = - x + 4$

Therefore, we have the slopes of $r$ and $s$ , $m_{r} = 3, m_{s} = - 1$ We observe that they are different and therefore the straight lines are neither parallel nor coincidental, and obviously they have to be secants.

Finally, we can calculate the intersection point easily by solving the system:

${\begin{matrix} y = - x + 4 \\ x = 1 + k \\ y = 1 + 3 k \end{matrix}$

equivalent to:

${\begin{matrix} y = - x + 4 \\ x - 1 = \frac{y - 1}{3} \end{matrix}$

and the solution is:

$3 x - 3 = - x + 4 - 1 \to 4 x = 6 \to x = 3 / 2$

$y = 5 / 2$

Solution:

The equation of the parallel straight line is: $(x, y) = (1, 1) + k \cdot (1, 3)$

The straight lines $r$ and $s$ are secant.

The straight line $s$ and the parallel to $r$ cross at point $P = (3 / 2, 5 / 2)$ .

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