Question

A line Li passes through point (1, 2) and has a gradient of 5. Another line L2, is

perpendicular to L, and meets it at a point where x=4. Find the equation for Le in the
form y = mx +c.

Answer 1

The equation of line Le is 5 y + x + 89 = 0

Explanation:

Given as :

The line Li passes through point (1,2)

The slope of line Li (m) = 5

So, equation of line y = m x + c

or, 2 = 5 × 1 + c

Or, c = 2 - 5 = -3

∴ Eq of line Li is

y = 5 x - 3

Another line Le is perpendicular to line Li

Let the slope of line Le = M

The line Le meet at points where x = 4

So, for perpendicular property

Products of slopes = - 1

So, m × M = - 1

or, M = -
`(1)/(m)`

Or. M = -
`(1)/(5)`

Now equation of line Le is y = M x + c

∵ Line Le meet the line Li at x = 4

So, y = 5 × 4 - 3

I.e y = 20 - 3

Or, y = 17

Now, equation of line Le with slope -
`(1)/(5)` and passes through points ( 4 , 17 ) is

y = M x + c

or, 17 = -
`(1)/(5)` × 4 + c

or, 17 +
`(4)/(5)` = c

So, c =
`(85+4)/(5)`

Or, c = -
`(89)/(5)`

∴ Equation of line Le is y = -
`(1)/(5)` x -
`(89)/(5)`

ie. 5 y = - x - 89

or, 5 y + x + 89 = 0

Hence The equation of line Le is 5 y + x + 89 = 0 answer