A line L1 passes through point (1,2) and has a gradient of 5.Another line L2,is perpendicular to L1 and meets it at a point where X=4.Find the equation for L2 in the form y=mx+c

Answer 1

equation of first line:
y-2 = 5(x-1)
y = 5x - 5 + 2 ---> y = 5x - 3

The slope of the perpendicular must be -1/5
it meets the old line when x = 4
which means at y = 5(4)-3 or 17 and the point (4,17) must be common to both lines
new line:
y = (-1/5)x + b, but (4,17) lies on it, so ...
17 = (-1/5)(4) + b
b = 89/5

new line: y = (-1/5)x + 89/5

proof:
https://www.wolframalpha.com/input/?i=solve+y+%3D+%28-1%2F5%29x+%2B+89%2F5%2C+y+%3D+5x-3