Question

Write an equation of a line through (15, 8) that is perpendicular to the line y =57 x −1.

Answer 1

Equation of a line through (15, 8) that is perpendicular to the line y =57 x −1 is
`\mathbf{y=-(1)/(57)x+(157)/(19)}`

Explanation:

We need to write an equation of a line through (15, 8) that is perpendicular to the line y =57 x −1.

The equation will be in slope-intercept form:
`\mathbf{y=mx+b}`

Where m is slope and b is y-intercept.

Finding Slope of new line

We are given equation
`y =57x-1` of line, which is perpendicular to the line whose equation we need to find.

If the lines are perpendicular there slopes are inverse of each other i.e
`m_1=-(1)/(m_2)`

Slope of given line is m= 57 ( compare the given equation with standard slope-intercept form the value of m is 57

Slope of new line is m=
`-(1)/(57)`

Finding y-intercept of new line

Now finding y-intercept (b) for new line. It can be found using point (15,8) and slope m =
`-(1)/(57)`

`y=mx+b\\8=-(1)/(57)(15)+b\\8=-(15)/(57)+b\\b=8+(15)/(57)\\b=(8*57+15)/(57)\\b=(471)/(57)\\b=(157)/(19)`

So, y-intercept b of new line is:
`b=(157)/(19)`

Equation of required line

Equation of line having slope m=
`-(1)/(57)` and y-intercept
`b=(157)/(19)` is:

`y=mx+b\\\mathbf{y=-(1)/(57)x+(157)/(19)}`

So, equation of a line through (15, 8) that is perpendicular to the line y =57 x −1 is
`\mathbf{y=-(1)/(57)x+(157)/(19)}`