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 ?

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

Answer:

$y=-4x+15$

Step-by-step explanation:

We are given that an equation

$y=\frac{x}{4}+5$

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

$\frac{dy}{dx}=\frac{1}{4}$

Using:$\frac{dx^n}{dx}=nx^{n-1}$

When two lines are perpendicular then

Slope of a line=$-\frac{1}{slope\;of\;other\;line}$

Slope of perpendicular line=$-\frac{1}{\frac{1}{4}}=-4$

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

$y=-4x+15$