Question

Find the equation of the line with points (3,3) in y=mx+b where it is perpendicular to equation 5x+5-4y=0

Answer 1

y = -
`(4)/(5)` x +
`(27)/(5)` or

5y = -4x + 27

Explanation:

To find the equation of a line with points (3,3) and is perpendicular to 5x+5-4y=0

we will follow the steps below:

First find write the equation: 5x+5-4y=0 in standard form y= mx + c and find the slope

5x+5-4y=0

4y = 5x + 5

divide through by 4

4y /4= 5x/4 + 5/4

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

comparing the above with y = mx + c

m=
`(5)/(4)`

since the equations are perpendicular.

To find the new slope

`m_(1)`
`m_(2)` = -1

`(5)/(4)`
`m_(2)` = -1

multiply both-side of the equation by
`(4)/(5)`

`(4)/(5)`×
`(5)/(4)`
`m_(2)` = -1 ×
`(4)/(5)`

`m_(2)` = -
`(4)/(5)`

The slope of our new equation is -
`(4)/(5)`

The points are (3,3)

We can now go ahead and form our new equation

y -
`y_(1)` = m ( x -
`x_(1)`)

y - 3 = -
`(4)/(5)` ( x - 3)

y - 3 = -
`(4)/(5)` x +
`(12)/(5)`

y = -
`(4)/(5)` x + 3 +
`(12)/(5)`

y = -
`(4)/(5)` x +
`(27)/(5)`

5y = -4x + 27