Question

Write an equation in slope-intercept form for the line that passes through (5,0) and is perpendicular to the line described by y=-5/2x+6

Answer 1

For this case we have to;

We have that an equation in slope-intercept form is given by:

`y = mx + b`

Where:

m is the slope

b is the cut point with the y axis

Also, by definition, two lines are perpendicular when the product of their slopes is -1. That is:
`m_ {1} * m_ {2} = - 1`

We have the line as data:
`y_ {1} = \frac {-5} {2} x + 6`

Then
`m_ {1} = \frac {-5} {2}`

We found
`m_ {2}`:

`m_ {1} * m_ {2} = - 1`

`\frac {-5} {2} * m_ {2} = - 1`

`m_ {2} = \frac {-1} {(\frac {-5} {2})}`

`m_ {2} = \frac {(2) (- 1)} {(- 5) (1)}`

`m_ {2} = \frac {2} {5}`

Thus,
`y_ {2} = \frac {2} {5} x_ {2} + b_ {2}`

We must find
`b_ {2}`:

We know that
`y_ {2}` passes through the point
`(x_ {2}, y_ {2}) = (5,0)`

We substitute the point in the equation of
`y_ {2}`:

`0 = \frac {2} {5} (5) + b_ {2}\\0 = 2 + b_ {2}\\b_ {2} = - 2`

Thus,
`y_ {2} = \frac {2} {5} x_ {2} -2`

Then the equation in slope-intercept for the line that passes through (5,0) and is perpendicular to the line described by
`y_ {1} = \frac {-5} {2} x_(1) + 6` is:
`y_ {2} = \frac {2} {5} x_ {2} -2`

Answer:

`y_ {2} = \frac {2} {5} x_ {2} -2`