The equation of line ab is y = 5x + 1. write an equation of a line parallel to line ab in slope-intercept form that contains point (4, 5).

y = 5x − 15
y = 5x + 15
y = 1 over 5x + 21 over 5
y = 1 over 5x − 29 over 5

Y = 5x + 1the slope here is 5...a parallel line will have the same slope.

y = mx + b
slope(m) = 5
(4,5)x = 4 and y = 5
now we sub and find b, the y int
5 = 5(4) + b
5 = 20 +b
5 - 20 = b
-15 = b

so ur parallel line is : y = 5x - 15