Question

What is the equation of a line perpendicular to the line y=-10x+1 passing thru point (5,7)

Answer 1

To find a line perpendicular, use the opposite-reciprocal slope of the line given. Also, use the point-slope form of a line since you're given a point.

y-7 = 1/10 (x-5)

Simplify to y=mx+b (slope-int) form

y-7= 1/10x -1/2
y = 1/10x + 13/2

Answer 2

The equation of line is
`y=(1)/(10)x+(13)/(2)`

Explanation:

To find : What is the equation of a line perpendicular to the line
`y=-10x+1` passing through point (5,7) ?

Solution :

We know that,

When two lines are perpendicular then slope of one line is negative reciprocal of another line.

Let the required line
`y=mx+b`

A line perpendicular to the line
`y=-10x+1`

Slope of the line is
`m_1=-10`

Slope of the perpendicular line is
`m=-((1)/(m_1)=(1)/(10)`

The passing points of the equation is (5,7)

Substitute in the formula,

`7=(1)/(10)(5)+b`

`7=(1)/(2)+b`

`b=7-(1)/(2)`

`b=(13)/(2)`

Now, substitute
`m=(1)/(10)` and
`b=(13)/(2)` in required line.

`y=(1)/(10)x+(13)/(2)`

Therefore, The equation of line is
`y=(1)/(10)x+(13)/(2)`