Question

What is the equation of the line that passes through the point (1,0) and is perpendicular to the line x+5y=30

Answer 1

The equation of line passing through points (1 , 0) is x - 5 y - 1 = 0

Explanation:

Given equation of line as

x + 5 y = 30

Now, equation of line in standard form is y = m x + c

where m is the slope

So, x + 5 y = 30

Or, 5 y = - x + 30

Or, y = -
`(1)/(5)` x + 6

So, Slope of this line m = -
`(1)/(5)`

Again , let the slope of other line passing through point (1 , 0) is M

And Both lines are perpendicular , So , products of line = - 1

i.e m × M = - 1

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

Or, M = - 1 × -
`(1)/((1)/(5))` =
`(1)/(5)`

So, equation of line with slope M and points (1, 0) is

y -
`y_1` = M × (x -
`x_1`)

Or, y - ( 0 ) =
`(1)/(5)` × ( x - 1 )

Or, y =
`(1)/(5)` x -
`(1)/(5)` × 1

Or, y =
`(1)/(5)` x -
`(1)/(5)`

or, y +
`(1)/(5)` =
`(1)/(5)` x

Or, 5×y + 1 = x

∴ 5 y + 1 = x

I.e x - 5 y - 1 = 0

Hence The equation of line passing through points (1 , 0) is x - 5 y - 1 = 0 Answer