Chapter B, Problem 41E

To determine

To show: The lines $2x-y=4$ and $6x-2y=10$ are not parallel and also find their point of intersection.

Answer to Problem 41E

The point of the intersection is $\left(1,-2\right)$.

Explanation of Solution

Formula used:

The non-vertical lines are parallel if and only if they have the same slope. That is, $m_1=m_2$.

Calculation:

It is enough to show that $m_1=m_2$, in order to say the two lines are parallel.

Find the slope of the line $2x-y=4$ as follows.

$-y=4-2x$

$y=2x-4$. (1)

Therefore, the slope is $m_1=2$.

Find the slope of the line $6x-2y=10$ as follows.

$\begin{array}{l}-2y=10-6x\\ -y=5-3x\end{array}$

$y=3x-5$. (2)

Therefore, the slope is $m_2=3$.

Here, $m_1\ne m_2$.

From the given formula, the given two lines are not parallel.

Hence, it is proved that the lines $2x-y=4$ and $6x-2y=10$ are not parallel.

In order to find the intersection of their points, equate the equations (1) and (2) as follows.

$\begin{array}{l}2x-4=3x-5\\ 2x-3x=-5+4\\ -x=-1\\ x=1\end{array}$

Now substitute $x=1$ in either $2x-y=4$ or $6x-2y=10$.

$\begin{array}{l}2\left(1\right)-y=4\\ 2-y=4\\ -y=4-2\\ y=-2\end{array}$

Therefore, the point of the intersection is $\left(1,-2\right)$.