Question

Find the slope-intercept form of the equation of the line that passes through the point (5,1) and is perpendicular to the line 2x + 5y = 10

Answer 1

The slope-intercept form:

`y=mx+b`

m - slope

b - y-intercept

Convert 2x + 5y = 10 to the slope0intercept form:

`2x+5y=10` subtract 2x from both sides

`5y=-2x+10` divide both sides by 5

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

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

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

We have
`m_1=-(2)/(5)`

Therefore

`m_2=-\frac{-(2)/(5)}=\dfac{5}{2}`

We have the equation of a line:

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

Put the coordinates of the point (5, 1) to the equation of a line:

`1=(5)/(2)(5)+b`

`1=(25)/(2)+b` subtract
`(25)/(2)` from both sides

`-(23)/(2)=b\to b=(23)/(2)`

Answer:
`\boxed{y=(5)/(2)x+(23)/(2)}`