The given equation of a line is:
[tex]7y-14=-3(5-x)[/tex]It is required to find the equation of a line that is parallel to this line and passes through the point (-5,5).
Recall that the equation of a line in point-slope form is given as:
[tex]y-y_1=m(x-x_1)[/tex]Where m is the slope of the line, and the line passes through the point (x₁,y₁).
Rewrite the given equation in the point-slope form:
[tex]\begin{gathered} 7y-14=-3(5-x) \\ Distribute\text{ -1 into the parentheses in the right-hand side:} \\ \Rightarrow7y-14=3(-5+x) \\ \text{Rewrite the expression in the parentheses using the co}mmutative\text{ property of addition:} \\ \Rightarrow7y-14=3(x-5) \\ \text{Divide both sides of the equation by 7:} \\ \Rightarrow y-2=\frac{3}{7}(x-5) \end{gathered}[/tex]Compare this equation with the standard point-slope form written above, it can be seen that the slope of the line is 3/7.
Recall that the slopes of parallel lines are the same or equal.
It follows that the slope of the required parallel line is also 3/7.
Substitute m=3/7 and the point (x₁,y₁)=(-5,5) into the point-slope form of the equation of a line:
[tex]y-5=\frac{3}{7}(x-(-5))[/tex]Rewrite the equation in the slope-intercept form and simplify as required:
[tex]\begin{gathered} \Rightarrow y-5=\frac{3}{7}(x+5) \\ \Rightarrow y-5=\frac{3}{7}x+\frac{15}{7} \\ \Rightarrow y=\frac{3}{7}x+\frac{15}{7}+5 \\ \Rightarrow y=\frac{3}{7}x+\frac{50}{7} \end{gathered}[/tex]The required equation of the line is y=3/7 x + 50/7.
