Question

Find an equation of a line parallel to the line $y=\frac{1}{2}x-3$that contains the point $(6,4)$. Write the equation in slope-intercept form.

Short answer

The equation is$y=\frac{1}{2}x+1$.

Step-by-step solution

Step 1. Find the slope of the given line

The given line is $y=\frac{1}{2}x-3$.

The line is already in the slope-intercept form.

Therefore the slope is:

$m=\frac{1}{2}$

Step 2. Find the slope of the parallel line and identify the point

As we know that parallel lines have the same slope.

Therefore the slope of the parallel line is also $\frac{1}{2}$.

The given point is $(6,4)$.

$(x_1,y_1)=(6,4)$

Step 3. Substitute the values into the point-slope form

The point-slope form is:

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-4=\frac{1}{2}(x-6) \\ y-4=\frac{1}{2}x-3 \end{gathered}$$

Step 4. Write the equation in slope-intercept form

The slope-intercept form of the equation is$y=\frac{1}{2}x+1$.