Question

Write the equation of the line passing through (−3, 4) and perpendicular to −2xx + 7yy = −3

Answer 1

`y=-3.5x-6.5`

Explanation:

Fist thing we need to do is find the slope of the original line, we have:

`-2x+7y=-3`

we want to clear for
`y` to get a slope-intercept equation (
`y=mx+b`, where
`m` is the slope and
`b` is the interception of the line with the y axis)

`7y=2x-3\\y=(2)/(7)x-(3)/(7)`

this way we can see that the number that represents the slope is
`(2)/(7)`. I will call this
`m_(1)=(2)/(7)` because it is the slope of the fisrt line.

Now to find the equation of the second line (the line perpendicular to
`-2x+7y=-3`), We need to apply the condition so that two lines are perpendicular:

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

we have
`m_(1)=(2)/(7)`, so the slope of the perpedicular line
`m_(2)` is:

`(2)/(7)m_(2)=-1\\ m_(2)=-(7)/(2)`

we already have the slope, and the problem mentions that the new line passes through the point (-3, 4), so we use the point-slope equation

`y=m(x-x_(0))+y_(0)`

where
`m` is the slope, and
`(x_(0),y_(0))` is the point, so
`x_(0)=-3`, and
`y_(0)=4`

thus:

`y=-3.5(x-(-3))+4 \\y=-3.5(x+3)+4\\y=-3.5x-10.5+4\\y=-3.5x-6.5`

the equation is
`y=-3.5x-6.5`