Question

What is the standard form of the equation of the line passing through the point (4,-4) and the perpendicular to the line 4x-7y=7x+4y

Answer 1

The slope-intercept form:

`y=mx+b`

m - slope

b - y-intercept

Convert the equation of the line 4x - 7y = 7x + 4y to the slope-intercept form:

`4x-7y=7x+4y` subtract 4x from both sides

`-7y=3x+4y` subtract 4y from both sides

`-11y=3x` divide both sides by (-11)

`y=-(3)/(11)x`

Let
`k:y=m_1x+b_1` and
`l:y=m_2x+b_2` then

`l\ \perp\ k\iff m_1m_2=-1\to m_2=-(1)/(m_1)`

We have
`m_1=-(3)/(11)` therefore the slope of the line perpendicular is

`m_2=-(1)/(-(3)/(11))=(11)/(3)`

Therefore we have:

`y=(11)/(3)x+b`

Put the coordinates of the point (4, -4) to the equation of a line:

`-4=(11)/(3)(4)+b`

`-4=(44)/(3)+b`

`-(12)/(3)=(44)/(3)+b` subtract
`(44)/(3)` from both sides

`-(56)/(3)=b`

Answer:
`\boxed{y=(11)/(3)x-(56)/(3)}`