Can I get some help please?

[Michelle Roanoke]

I have two problems I need help with. How do I calculate the equation of a line when given the point (4,5) and knowing it is parallel to the line y=3x+2. The second one wants me to find the same thing, but the point is (-8,2) and the line is perpendicular to y=4/5x -5.

 

[MarkFL]

Let's look at the first question. What will the slope of the line we want be?

 

[Michelle Roanoke]

3?

 

[Michelle Roanoke]

y=3x-7? and for the other one y=5/4x +12?

 

[pka]

I have two problems I need help with. How do I calculate the equation of a line when given the point (4,5) and knowing it is parallel to the line y=3x+2. The second one wants me to find the same thing, but the point is (-8,2) and the line is perpendicular to y=4/5x -5.

Suppose we want a line parallel to \(\displaystyle \ell:y=mx+b\) & contains \(\displaystyle (p,q)\)?
Well that means that \(\displaystyle q=mp+b\) or \(\displaystyle b=q-mp\) so that
\(\displaystyle y=mx+q-mp \) is the line parallel to \(\displaystyle \ell\) that contains \(\displaystyle (p,q)~.\)

Now suppose that need a line perpendicular to \(\displaystyle \ell \) that contains \(\displaystyle (r,s)\).
We understand that the slope must be \(\displaystyle \frac{-1}{m}\).
That is \(\displaystyle (y-s)=\frac{-1}{m}(x-r)\)

 

[MarkFL]

3?

Yes, and so now armed with the slope, and a point on the line, we may use the point slope formula to write:

[MATH]y-5=3(x-4)[/MATH]
or:

[MATH]y=3x-7\quad\checkmark[/MATH]
Now, for the second one, we know the slope must be:

[MATH]m=-\frac{1}{\dfrac{4}{5}}=-\frac{5}{4}[/MATH]
Hence:

[MATH]y-2=-\frac{5}{4}(x+8)[/MATH]
or:

[MATH]y=-\frac{5}{4}x-8[/MATH]

 

[JeffM]

y=3x-7? and for the other one y=5/4x +12?

Yes and no. (By the way, our submission guidelines require that you post one problem per thread. Please do so in the future.)

If two lines are parallel, they have the same slope. So in the first problem, you know that the equation you are looking for has a slope of 3 because it is said to be parallel to the line described by y = 3x + 2. Now all you need is the intercept with the y-axis of a line with a slope of 3 and containing the point (4, 5).

[MATH]5 = b + 3 * 4 \implies 5 = b + 12 \implies b = -\ 7.[/MATH]
Well done.

In the second problem, the product of the slopes of two lines that are perpendicular is minus 1. In other words the slope of one is the additive inverse of the reciprocal of the other.

So the slope of the second line in the second problem is

[MATH]-\ \dfrac{1}{\dfrac{4}{5}} = -\ \dfrac{5}{4}.[/MATH]
This is where you made your mistake. You forgot that it is the additive inverse of the reciprocal rather than just the reciprocal. Now finish it up by finding the intercept with the y-axis given that you now know the slope and one point of the line.