Question

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

Answer 1

Y = 5x + 1....the 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

Answer 2

$k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ ||\ l\iff m_1=m_2\\\\AB:y=5x+1;\ CD:y=mx+b\ /the\ slope-intercept\ form/\\\\AB\ ||\ CD\ iff\ m=5\\\\therefore\ CD:y=5x+b\\\\The\ line\ CD\ \ contains\ point\ (4;\ 5)\to x=4\ and\ y=5\\\\subtitute\ to\ the\ equation\ of\ the\ line\ CD:\\\\5=5\cdot4+b\\5=20+b\ \ \ \ |subtract\ 20\ from\ both\ sides\\b=-15\\\\Answer:\boxed{y=5x-15}$