Question

Find the equation of the line that contains the point (-4, -4) and is perpendicular to the line 4x+5y=5

Answer 1

`5x-4y+4=0`

Explanation:

Given: A point which is perpendicular to the line .

To find: The equation of the line which passes through
`(-4, -4)` and is perpendicular to the line
`4x+5y=5`.

Solution:

We have,
`4x+5y=5`.

Slope of the line
`4x+5y=5` is
`-(4)/(5)`.

The line which passes through
`(-4, -4)` is perpendicular to the line
`4x+5y=5`.

Now, the product of the slopes of two perpendicular lines is
`-1`.

Therefore,
`m*-(4)/(5)=-1`

So, its slope is
`(5)/(4)`.

Now, the equation of the line which passes through
`(-4, -4)` and slope
`(5)/(4)` is:

`y-(-4)=(5)/(4) [x-(-4)]`

`\Rightarrow y+4=(5)/(4) (x+4)`

`\Rightarrow 4(y+4)=5(x+4)`

`\Rightarrow 4y+16=5x+20`

`\Rightarrow 5x-4y+20-16=0`

`\Rightarrow 5x-4y+4=0`

Hence, the equation of the line that contains the point
`(-4, -4)` and is perpendicular to the line
`4x+5y=5` is
`5x-4y+4=0`.