Answer:
The equation of the fourth line is;
[tex]\begin{gathered} y-15=2(x-3)\text{ }\rightarrow\text{ point-slope form} \\ y=2x+9\text{ }\rightarrow\text{ Slope-intercept form} \end{gathered}[/tex]
Explanation:
Given that the fourth line would also be parallel to the other three lines.
[tex]\text{slope m = 2}[/tex]
And passes through the point;
[tex](x_1,y_1)=(3,15)[/tex]
Since, we have the value of the slope and we have a point on the fourth line;
Then we can use the point-slope form;
[tex]y-y_{1_{}}=m(x-x_1)[/tex]
substituting the given values;
[tex]\begin{gathered} y-15=2(x-3) \\ y-15=2x-6 \\ y=2x-6+15 \\ y=2x+9 \end{gathered}[/tex]
Therefore, the equation of the fourth line is;
[tex]\begin{gathered} y-15=2(x-3)\text{ }\rightarrow\text{ point-slope form} \\ y=2x+9\text{ }\rightarrow\text{ Slope-intercept form} \end{gathered}[/tex]
To derive the equation of the fourth line, the best and easiest equation to use is the point-slope form. because we were given the slope and a point on the slope.
