How do you write an equation of a line through (3,3) and parallel to y= 2/3x -1?

Question

Ethan Adkins · Accepted Answer

#y=2/3x+1#

Our reference line is: #y=2/3x-1#

Hence the slope of the reference line is: #2/3#

Any straight line parallel to the reference line will have a slope of #2/3#

We are asked to find the equation of the line parallel to the reference line passing through the point #(3,3)#.

The equation of a straight line passing through point #(x_1, y_1)# is:

#(y-y_1) =m(x - x_1)# Where #m# is the slope of the line.

Hence the equation of our required line will be:

#(y-3) = 2/3(x-3)#

#3(y-3) = 2(x-3)#

#3y-9 = 2x-6#

#3y= 2x+3#

#y = 2/3x+1#

Ethan Adkins · Answer

undefined To write an equation of a line through the point (3,3) and parallel to the line $y = \frac{2}{3}x - 1$, you can use the fact that parallel lines have the same slope.

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

So, the equation of the line parallel to this line will also have a slope of $m = \frac{2}{3}$.

Using the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the given point (3,3), and $m$ is the slope, we can substitute the values:

$$y - 3 = \frac{2}{3}(x - 3)$$

Simplify:

$$y - 3 = \frac{2}{3}x - 2$$

$$y = \frac{2}{3}x + 1$$

Therefore, the equation of the line through (3,3) and parallel to $y = \frac{2}{3}x - 1$ is $y = \frac{2}{3}x + 1$.