Q9. - Algebra 2

Question

Graph the line that satisfies each set of conditions.

Passes through $\left(0,3\right)$ parallel to graph of $\mathbf{6}\mathbf{y}-\mathbf{10}\mathbf{x}=\mathbf{30}.$

Step-by-Step Solution

Answer

The graph of the required line is:

Step 1- Given information

A point $(0,3)$ and an equation $6y-10x=30$.

Step 2- Calculation

From given information,

A point $(0,3)$ and $6y-10x=30$

Here, x-intercept is:

$\begin{array}{l}6\times 0-10x=30\\ \Rightarrow -10x=30\\ \Rightarrow x=-3\end{array}$

y- intercept is:

$\begin{array}{l}6y-10\times 0=30\\ \Rightarrow 6y=30\\ \Rightarrow y=5\end{array}$

Therefore, the line goes to 5 units up and 3 units right then.

So, the slope of the line is $\frac{5}{3}$

We get a new point $\left(3,8\right)$.

Now we draw the line which is contained both the points $(0,3)$ and $\left(3,8\right)$.

Step 3- Graph of the line

The graph of the required line is: