Question

Find the equation of the straight line passing through the point (0,5)

which is perpendicular to the line
y = 5 x + 2​

Answer 1

y = -x/5 + 5

Explanation:

Equation of a line L : y=mx+b perpendicular to another line L1 through a point P(p,q)

Given :

L1 : y = 5x+2

P : P(p,q) = P(0,5)

Solution :

Slope of L1 = 5. For L to be perpendicular, product of slopes = -1 =>

m*5=-1, or m = -1/5

Since L passes through P(0,5), using the point slope form of the line L :

L : (y-5) = -(x – 0) / 5

L : y = -x/5 + 5

Answer 2

y = -
`(1)/(5)` x + 5

Explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 5x + 2 ← is in slope- intercept form

with slope m = 5

Given a line with slope m then the slope of the line perpendicular to it is

`m_(perpendicular)` = -
`(1)/(m)` = -
`(1)/(5)`

The line crosses the y- axis at (0, 5 ) ⇒ c = 5

y = -
`(1)/(5)` x + 5 ← equation of perpendicular line