Question

1. Find the equation of the line passing through the point (2,−4) that is parallel to the line y=3x+2 y= 2. Find the equation of the line passing through the point (1,−5) and perpendicular to y=18x+2 y=

Answer 1

Explanation:

1) Parallel lines have same slope

y = 3x + 2

m = 3

(2, -4) ; m = 3

equation: y - y1 = m (x - x1)

y - [-4] = 3(x - 2)

y + 4 = 3x - 6

y = 3x - 6 - 4

y = 3x - 10

2) y = 18x + 2

m1 = 18

Slope the line perpendicular to y = 18x + 2, m2 = -1/m1 = -1/18

m2 = -1/18

(1 , -5)

`y-[-5]=\frac{-1/18}(x-1)\\\\y+5=(-1)/(18)x + (1)/(18)\\\\y=(-1)/(18)x+(1)/(18)-5\\\\y=(-1)/(18)x+(1)/(18)-(5*18)/(1*18)\\\\y=(-1)/(18)x+(1)/(18)-(90)/(18)\\\\y=(-1)/(18)x-(89)/(18)\\\\`