Question

What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y=−2/3x+5 ? Enter your answer in the box.

Answer 1

Answer: the equation is y'=
`(3)/(2)x'-11`

Explanation:

A line is given whose equation is given as
`y=-(2x)/(3)+5`------(1)

and we have to find the equation of another line which is perpendicular to this.

Let the equation of the line be y'=mx'+c------(2)

This line passes through a point (8,1) so we put the values of x & y in the equation.

⇒ 1 = m×8+c

⇒ 8m+c = 1------(3)

We know that if two lines are perpendicular to each other then multiplication of their slopes is equal to (-1).

Therefore m×
`(-(2)/(3))`=(-1)

⇒
`(2)/(3) m=1`

⇒m=
`(3)/(2)`

Now we put the value of m in equation (3) to get the value of c

8×
`((3)/(2))`+c = 1

12+c=1

c = (-11)

Therefore the equation will be y'=
`(3)/(2)x'+(-11)`

Answer 2

`y = (3)/(2) x-11`

Step-by-step explanation:

We are to find the equation a line that passes through the point (8, 1) and which is perpendicular to a line whose equation is
`y = - (2)/(3) x+5`.

We know that the slope of line which is perpendicular to another line is the negative reciprocal of the slope of the other line so it will be
`(3)/(2)`.

Then, we will find the y-intercept of the line using the standard equation of a line:

`y=mx+c`

`1=(3)/(2) (8)+c`

`c=-11`

Therefore, the equation of the line will be
`y = (3)/(2) x-11`.