Question

find the equation of the line which passes through the point (-5,5) and is parallel to the given line. express your answer in slope-intercept form. simplify your answer.

Answer 1

The given equation of a line is:

$7y-14=-3(5-x)$

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:

$y-y_1=m(x-x_1)$

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:

$\begin{gathered} 7y-14=-3(5-x) \\ Distribute\text{ -1 into the parentheses in the right-hand side:} \\ \Rightarrow7y-14=3(-5+x) \\ \text{Rewrite the expression in the parentheses using the co}mmutative\text{ property of addition:} \\ \Rightarrow7y-14=3(x-5) \\ \text{Divide both sides of the equation by 7:} \\ \Rightarrow y-2=\frac{3}{7}(x-5) \end{gathered}$

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:

$y-5=\frac{3}{7}(x-(-5))$

Rewrite the equation in the slope-intercept form and simplify as required:

$\begin{gathered} \Rightarrow y-5=\frac{3}{7}(x+5) \\ \Rightarrow y-5=\frac{3}{7}x+\frac{15}{7} \\ \Rightarrow y=\frac{3}{7}x+\frac{15}{7}+5 \\ \Rightarrow y=\frac{3}{7}x+\frac{50}{7} \end{gathered}$

The required equation of the line is y=3/7 x + 50/7.