Question

What is the equation of the line that is perpendicular to the like y=-4x + 7 and passes through the point (8,2)?

Answer 1

`y=(1)/(4)x`

Explanation:

Given equation of line:

`y=-4x+7`

To find the equation of line perpendicular to the line of the given equation and passes through point (8,2).

Applying slope relationship between perpendicular lines.

`m_1=-(1)/(m_2)`

where
`m_1` and
`m_2` are slopes of perpendicular lines.

For the given equation in the form
`y=mx+b` the slope
`m_2`can be found by comparing
`y=-4x+3` with standard form.

∴
`m_2=-4`

Thus slope of line perpendicular to this line
`m_1` would be given as:

`m_1=-(1)/(-4)`

∴
`m_1=(1)/(4)`

The line passes through point (8,2)

Using point slope form:

`y_-y_1=m(x_-x_1)`

Where
`(x_1,y_1)\rightarrow (8,2)` and
`m=m_1=(1)/(4)`

So,

`y-2=(1)/(4)(x-8)`

Using distribution.

`y-2=((1)/(4)x)-((1)/(4)* 8)`

`y-2=(1)/(4)x-2`

Adding 2 to both sides.

`y-2+2=(1)/(4)x-2+2`

`y=(1)/(4)x`

Thus the equation of line in standard form is given by:

`y=(1)/(4)x`