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Find an equation of the line that is parallel to `y=3/4 x -1` and passes through the...

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gturner49 | (Level 1) Honors

Posted November 8, 2013 at 2:58 AM via web

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Find an equation of the line that is parallel to

`y=3/4 x -1`

and passes through the point (4,0).

Find an equation of the line that is perpendicular to

`y=x+4`

and through the point (-7, 1).

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted November 8, 2013 at 3:23 AM (Answer #1)

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If two lines are parallel, they have the same slope.

The slope of  `y=3/4 x -1`   is    `3/4`

So we want a line with slope   `3/4` , that goes through the point  `(4,0)`

We know the slope, we want to find the y-intercept:

`y=3/4 x + b`

`0 = 3/4 (4) + b`

`0=3+b`

`-3=b`

So:  `y=3/4 x - 3`

This is the equation of the line that is parallel to `y=3/4 x -1` that passes through (4,0)

Two lines are perpendicular if you multiply their slopes and get -1.  In other words, you want the "negative reciprocal"

`y=x+4`

This line has slope = 1.

So a perpendicular line would have slope = -1.

`y=-1 x + b=-x + b`

Now we need to find the y-intercept.

Plug in (-7,1)

`1=-(-7)+b`

`1=7+b`

`-6=b`

so:

`y=-x-6`

This is the line that is perpendicular to `y=x+4` and passes through the point (-7,1)

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baxthum8 | High School Teacher | (Level 3) Associate Educator

Posted November 8, 2013 at 9:45 AM (Answer #2)

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An equation paralle to `y= 3/4x-1` and passes through the point (4,0).  Since the equation is written in slope intercept form the line has a slope of `3/4` .

In order to be parallel the line must havet he same slope, `3/4.`

Use point slope form `y - y_(1)= m(x-x_(1))` , to write the equation of the line.

`y - 0 = 3/4(x-4) rArr y = 3/4x - 3`

A line perpendicular to `y = x+4` , must have a slope of `-1` as this is the opposite reciprocal of the line's slope of 1.

Again use point slope form to write the equation as it passes through the point (-7, 1)

`y-1=-1(x-(-7))`

`y - 1 = -1(x+7)`

`y = -1x - 7 + 1`

`y = -x -6`

The solution to part 1 is `y = 3/4x - 3` and the solution to part 2 is` y = -x-6` .

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ksri | High School Teacher | (Level 3) eNoter

Posted November 8, 2013 at 6:26 PM (Answer #3)

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y = 3/4 x - 1

Any straight line parallel to the given line will be of the form

y = 3/4 x + k

When this line passes through the point ( 4 , 0 ) we have

0 = 3/4 * 4 + k

Hence k + 3 = 0 .

k = - 3 .

Hence y= 3/4 x - 3  is the required equation.

__________________________________________

y = x + 4

That is x - y + 4 =0  is the given line .

Any line perpendicular to this line will be of the form

x + y +k =0 .

Since this passes through the point ( - 7 , 1 )

1 - 7 + k = 0 .   ......  That is k = 6

Hence the required equation is x + y + 6 = 0 ..

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