Question

Write the equation of a line that passes

through point (-5,3) and is perpendicular to
y=3/5x +2.

Answer 1

`y=-(5)/(3)x-(16)/(3)`

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 the line crosses the y-axis)
• Perpendicular lines always have slopes that are negative reciprocals (ex. 2 and -1/2, 4/3 and -3/4, etc.)

1) Determine the slope (m)

`y=(3)/(5) x +2`

From the given equation, we can see that
`(3)/(5)` is the slope of the line since it's in the place of m. Because perpendicular lines always have slopes that are negative reciprocals, the line we're currently solving for would have a slope of
`-(5)/(3)`. Plug this into
`y=mx+b`:

`y=-(5)/(3)x+b`

2) Determine the y-intercept (b)

`y=-(5)/(3)x+b`

Plug in the given point (-5,3)

`3=-(5)/(3)(-5)+b\\3=(25)/(3)+b`

Subtract
`(25)/(3)` from both sides to isolate b

`3-(25)/(3)=(25)/(3)+b-(25)/(3)\\-(16)/(3) = b`

Therefore, the y-intercept is
`-(16)/(3)`. Plug this back into
`y=-(5)/(3)x+b`

`y=-(5)/(3)x+(-(16)/(3))\\y=-(5)/(3)x-(16)/(3)`

I hope this helps!