Question

10. Write the equation of the line that passes through the point P and is perpendicular to the line with the

given equation.
through: P(3,-2)
Perpendicular to line: y = 3x -4

Answer 1

The equation of the line perpendicular to the line y = 3 x - 4 is x + 3 y + 1 = 0

Explanation:

Given line equation as :

y = 3 x - 4

Now a line which is perpendicular to given line passes through point p ( 3 , - 2 )

The standard equation of line is given as

y = m x + c

where m is the slope of the line

Now, for line y = 3 x - 4

So By comparing the line, the slope of this line = m = 3

Now, when two lines are perpendicular then

The product of the slope of lines = - 1

Let the slope of other line = M

So, from property

m × M = - 1

∴ M =
`(-1)/(m)`

Or, M =
`(-1)/(3)`

∴ M =
`(-1)/(3)`

Now, The line having slope M =
`(-1)/(3)` , is passing through point p ( 3 , - 2 )

From standard equation of line

I.e y = m x + c

As the line y = m x + c passes through point ( 3 , - 2 ) having slope M =
`(-1)/(3)`

So, satisfying the points and slope in equation y = M x + c

I.e - 2 =
`(-1)/(3)` ( 3 ) + c

Or, - 2 =
`(- 3)/(3)` ( 3 ) + c

Or, - 2 = - 1 + c

∴ c = - 2 + 1

i.e c = - 1

So The equation of line can be written as

y =
`(-1)/(3)` x - 1

or, 3 y = - x - 1

or, x + 3 y + 1 = 0

So equation of new line is x + 3 y + 1 = 0

Hence The equation of the line perpendicular to the line y = 3 x - 4 is x + 3 y + 1 = 0 . Answer