Question

How do you find a standard form equation for the line that is perpendicular to `6x-2y+5=0` and has the same y-intercept as `x-y+8=0`?

Answer 1

When given an equation of a line of the form:

`ax-by+c=0`

The family of perpendicular lines are of the form:

`bx+ay+k=0`

Where k is an unknown constant.

Explanation:

Given: `6x-2y+5=0`

The family of perpendicular lines are:

`2x+6y+k=0" [1]"`

We must find the y intercept of the line `x-y+8=0`:

`0-y+8=0`

`y = 8`

Substitute the point `(0,8)` into equation [1] an then solve for k:

`2(0)+6(8)+k=0`

`k = -48`

Substitute this into equation [1]:

`2x+6y-48=0" [2]"`

Equation [2] is the answer.

Answer 2

`y=-1/3x+8`

This is the same as: `1/3x+y-8=0" "->" " x+3y-24=0`

Explanation:

First consider `x-y+8=0" "..................Equation(1)`

This can be written as: `y=x+8`

So the y-intercept for this equation is 8.

Thus the equation we are looking for passes through the point:
`(x,y)->(0,8)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider the given equation: `6x-2y+5=0" ".....Equation(2)`

Add `2y` to both sides

`2y=6x+5`

Divide both sides by 2

`y=3x+5/2" ".......Equation(2_a)`

Thus the gradient (slope) for `Equation(2_a)` is 3 from the `3x`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Compare `Equation(2_a)` to the standardised form of`" "y=mx+c`
where `m` is the gradient (slope).

A line that is perpendicular to it will have the gradient of:
`(-1)xx1/m = -1/m" "->" "(-1)xx1/3=-1/3`

Thus our standardised form for the new line is:

`y=-1/3x+c`

From `Equation(1)` we have determined that this line passes through the point `(x,y)->(0,8)`

So `y=-1/3x+c" "->" "8=-1/3(0)+c`

Thus `c=8`

So the equation of the perpendicular line is:
`y=-1/3x+8`

Tony B