How do you write the standard form of a line given (6, -4) and is parallel to the line 5x-5y=2?

Question

Eva Adamson · Accepted Answer

#x-y=10# #Ax+By=C# => Equation of line in standard form, slope = -B/A: #5x-5y=2# #slope m=-(-5)/5 = 1# Start with point slope formula: in this case m = 1 , point(6, -4): #y-y_1=m(x-x_1)# #y-(-4)=1(x-6)# #y+4=x-6# #x-y=10# => the required line in standard form.

Julia Adams · Answer

undefined To write the standard form of a line parallel to $5x - 5y = 2$ passing through the point $(6, -4)$, follow these steps:

1. Find the slope of the given line by rearranging it into slope-intercept form $y = mx + b$.
2. Parallel lines have the same slope, so the slope of the line you're trying to find will be the same as the given line.
3. Once you have the slope, use the point-slope form $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the given point.
4. Plug in the values of the slope and the given point to find the equation of the line.
5. Rewrite the equation in standard form $Ax + By = C$ by rearranging the terms.

Following these steps, you'll find the standard form of the line parallel to $5x - 5y = 2$ passing through the point $(6, -4)$.