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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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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
User Kuldeep Dubey
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2 votes

Answer:


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

User Spencer Rose
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8.5k points

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