Question

Question:

1. Write an equation of the line that is parallel to y = -4x + 8 and passes through the point (5, 1).
2. Write an equation of the line that is perpendicular to y = -4x + 8 and passes through the point (5,1).

Answer 1

1. The equation of the line parallel to y=−4x+8 and passing through (5,1) is y=−4x+21.

2. The equation of the line perpendicular to y=−4x+8 and passing through the point (5, 1) is
`y=(1)/(4) x- (1)/(4)`.

To find the equation of a line parallel to y=−4x+8 and passing through the point (5,1), you can use the fact that parallel lines have the same slope.

Equation of the line parallel to

y=−4x+8:

Given the slope of the line

y=−4x+8 is -4, and parallel lines have the same slope, so the equation of the parallel line is:

Slope=−4

Point=(5,1)

Using the point-slope form of a line equation:

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

where

m is the slope and (
`x_1 ,y_1` ) is the given point:

y−1=−4(x−5)

y−1=−4x+20

y=−4x+21

Therefore, the equation of the line parallel to y=−4x+8 and passing through (5,1) is y=−4x+21.

To find the equation of a line perpendicular to y=−4x+8 and passing through the point (5, 1), note that perpendicular lines have slopes that are negative reciprocals. The given line has a slope of -4, so the perpendicular line will have a slope of
`(1)/(4)`.

Use the point-slope form again:

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

Substitute
`m = (1)/(4)` and (
`x_1, y_1` )=(5,1) into the equation:

`y - 1 = (1)/(4) (x -5)`

Expand and simplify:

`y - 1 = (1)/(4) x - (5)/(4)`

​
`y=(1)/(4) x- (1)/(4)`

So, the equation of the line perpendicular to y=−4x+8 and passing through the point (5, 1) is
`y=(1)/(4) x- (1)/(4)`.

Answer 2

Explanation is written below.

Explanation:

1)

Given the line

`y = -4x + 8`

Comparing with the slope-intercept form of the line equation

`y=mx+b`

Where m is the slope and b is the y-intercept

So, the slope of the line = m = -4

• We know that the parallel lines have equal slopes.

Thus, the equation of the line that is parallel to y = -4x + 8 and passes through the point (5, 1), using point slope form

`y-y_1=m\left(x-x_1\right)`

Here, m is the slope and (x₁, y₁) is the point

substituting the values m = -4 and the point (5, 1)

`y-y_1=m\left(x-x_1\right)`

y-1 = -4 (x - 5)

y-1 = -4x+20

y = -4x+20+1

y = -4x + 21

2)

Given the equation

`y = -4x + 8`

Comparing with the slope-intercept form of the line equation

`y=mx+b`

Where m is the slope and b is the y-intercept

So, the slope of the line = m = -4

As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line, so

The slope of the perpendicular line will be: 1/4

Thus, an equation of the line that is perpendicular to y = -4x + 8 and passes through the point (5,1), using point slope form

`y-y_1=m\left(x-x_1\right)`

Here, m is the slope and (x₁, y₁) is the point

substituting the values m = 1/4 and the point (5, 1)

`y-1\:=\:(1)/(4)\:\left(x\:-\:5\right)`

Add 1 to both sides

`y-1+1=(1)/(4)\left(x-5\right)+1`

`y=(1)/(4)x-(1)/(4)`