Question

What is the equation of a line that passes through the point (2, 7) and is perpendicular to the line whose equation is y=x4+5 ?

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Answer 1

gradient of the line= -1/4
equation of line: y=-1/4x + c
substitute (2,7)
c= 15/2
hence the equation is: y= -1/4x + 15/2

Answer 2

`y=-4x+15`

Explanation:

We are given that an equation

`y=(x)/(4)+5`

We have to find the equation of a line that passes through the point (2,7) and perpendicular to given line.

Differentiate w.r.t x

`(dy)/(dx)=(1)/(4)`

Using:
`(dx^n)/(dx)=nx^(n-1)`

When two lines are perpendicular then

Slope of a line=
`-(1)/(slope\;of\;other\;line)`

Slope of perpendicular line=
`-(1)/((1)/(4))=-4`

The equation of a line passing through the point (
`x_1,y_1)`

with slope m is given by

`y-y_1=m(x-x_1)`

Using this formula

The equation of perpendicular line with slope -4 and passing through the point (2,7) is given by

`y-7=-4(x-2)`

`y-7=-4x+8`

`y=-4x+8+7=-4x+15`

Hence,the equation of the line which is passing through the point (2,7) and perpendicular to the given line is given by

`y=-4x+15`