Question

What is the equation of the line perpendicular to y=5/4x-10 and passes through the point (11,6)?

Answer 1

`y=-(4)/(5)x+(74)/(5)`

Explanation:

Hi there!

What we need to know:

• Linear equations are typically organized in slope-intercept form:
  `y=mx+b` where m is the slope and b is the y-intercept (the value of y when x is equal to 0)
• Perpendicular lines always have slopes that are negative reciprocals (ex. 3 and -1/3, 5/6 and -6/5, etc.)

1) Determine the slope (m)

`y=(5)/(4) x-10`

From the given equation, we can identify clearly that the slope of this line is

`(5)/(4)`. The negative reciprocal of
`(5)/(4)` is
`-(4)/(5)`, so therefore, the slope of the line we're currently solving for is
`-(4)/(5)`. Plug this into
`y=mx+b`:

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

2) Determine the y-intercept (b)

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

Plug in the given point (11,6) and solve for b

`6=-(4)/(5)(11)+b\\6=-(44)/(5)+b`

Add
`-(44)/(5)` to both sides to isolate b

`6+(44)/(5)=-(44)/(5)+b+-(44)/(5)\\(74)/(5) =b`

Therefore, the y-intercept is
`(74)/(5)`. Plug this back into
`y=-(4)/(5)x+b`:

`y=-(4)/(5)x+(74)/(5)`

I hope this helps!