Just started ONLINE college Calculus, not seen any math in over 5 years!!!! Help.

[samistumbo]

Okay, I just started Calculus I (online) and there are a few problems with Analytic Geometry that I've no idea how to do. I've not seen any math in well over 5 years.... and now I'm in Calculus... and the highest math I took in College was College Algebra (over 5 years ago). I would love more so an explanation rather than just answers... or examples. I'd like to be able to do problems like these on my own and actually learn know to do them. I just need guidance. Here are the problems:
1.3.1
The x- and y- intercepts of a line L are respectively a and b. Show that an equation of L is x/a + y/b=1, if ab does not equal 0.

1.3.3
Find the coordinates of the point P (x,y) which is located so that the line L1, which passes through P and the origin, has slope equal to 2, and the line L2, which passes through the point P and the point A (-1, 0), has slope equal to 1.

1.3.4
(a)Find equation of the line L through A (-2, 2) and perpendicular to the line L' : 2x+y=4.
(b)Find the point B at which the lines L and L' of (a) intersect.
(c)Find the perpendicular distance from the point A (-2, 2) to the line L' whose equation is 2x + y= 4.

And I apologize... I would post what I've done, but I don't even know where to begin on these and I wanted to try and get help with these at once because they're all related.

 

[Deleted member 4993]

Okay, I just started Calculus I (online) and there are a few problems with Analytic Geometry that I've no idea how to do. I've not seen any math in well over 5 years.... and now I'm in Calculus... and the highest math I took in College was College Algebra (over 5 years ago). I would love more so an explanation rather than just answers... or examples. I'd like to be able to do problems like these on my own and actually learn know to do them. I just need guidance. Here are the problems:
1.3.1
The x- and y- intercepts of a line L are respectively a and b. Show that an equation of L is x/a + y/b=1, if ab does not equal 0.

1.3.3
Find the coordinates of the point P (x,y) which is located so that the line L1, which passes through P and the origin, has slope equal to 2, and the line L2, which passes through the point P and the point A (-1, 0), has slope equal to 1.

1.3.4
(a)Find equation of the line L through A (-2, 2) and perpendicular to the line L' : 2x+y=4.
(b)Find the point B at which the lines L and L' of (a) intersect.
(c)Find the perpendicular distance from the point A (-2, 2) to the line L' whose equation is 2x + y= 4.

And I apologize... I would post what I've done, but I don't even know where to begin on these and I wanted to try and get help with these at once because they're all related.

For a quick review, go to:

http://www.purplemath.com/modules/strtlneq.htm

 

[samistumbo]

Okay, I just started Calculus I (online) and there are a few problems with Analytic Geometry that I've no idea how to do. I've not seen any math in well over 5 years.... and now I'm in Calculus... and the highest math I took in College was College Algebra (over 5 years ago). I would love more so an explanation rather than just answers... or examples. I'd like to be able to do problems like these on my own and actually learn know to do them. I just need guidance. Here are the problems:
1.3.1
The x- and y- intercepts of a line L are respectively a and b. Show that an equation of L is x/a + y/b=1, if ab does not equal 0.

1.3.3
Find the coordinates of the point P (x,y) which is located so that the line L1, which passes through P and the origin, has slope equal to 2, and the line L2, which passes through the point P and the point A (-1, 0), has slope equal to 1.

1.3.4
(a)Find equation of the line L through A (-2, 2) and perpendicular to the line L' : 2x+y=4.
(b)Find the point B at which the lines L and L' of (a) intersect.
(c)Find the perpendicular distance from the point A (-2, 2) to the line L' whose equation is 2x + y= 4.

And I apologize... I would post what I've done, but I don't even know where to begin on these and I wanted to try and get help with these at once because they're all related.

I can do slope intercept when I have numbers and information. I get lost whenever I'm given a problem without any numbers or information. Referring to 1.3.1.

With 1.3.3, Can I figure out the slope intercept for L₂ with the information given? y= 1(-1) +0? where do I go from there? I'm lost from there.... I've not had any of this stuff since high school over ten years ago.

 

[samistumbo]

A couple of quick pointers about these boards

Please post one problem per thread; otherwise, things get confusing VERY fast.
Always explain (as you did on this post) that you are taking an online course and have not studied math in many years. It helps give us context.
Say whether you need a hint to get started OR show your work so we can see where you are stuck

Let's deal with 1.3.1 first OK

Two distinct points determine a unique line, right? This is simple plane geometry.

The x and y intercepts are distinct points unless there is an intercept at the origin. The problem has simply been tricky (careless ?) with regard to specifying the points numerically. When a curve or line intercepts the x-axis, the value of y is what? When a curve or line intercepts the y-axis, the value of x is what? So putting your two given points into (x, y) form, they are what?

If you are not able to answer those questions let me know.
If you can answer them, do you now see how to write the equation correcponding to the given line?

I just wanted to post them all here because I know they're related with how you do them.... I just don't know how.

y= (1) (x/a) + (y/b)? I don't know. I'm lost.

 

[galactus]

For 1.3.1:

Start with finding the slope. We are told that a and b are the x and y intercepts, respectively.

So, the line crosses the x-axis at some point a, then the coordinate is (a,0)

If it crosses the y axis at some point b, then the coordinate is (0,b)

Find the slope: \(\displaystyle m=\frac{0-b}{a-0}=\frac{-b}{a}\)

Plug it in the slope intercept form of a line, y=mx+b. Remember, the b in y=mx+b is where it crosses the y-axis.

\(\displaystyle y=\frac{-b}{a}x+b\)

Now, rearrange this so you have \(\displaystyle \frac{x}{a}+\frac{y}{b}=1\)

Can you do that?. Divide everything through by b and you should see it coming together. After all, what is b/b?. It is 1. Yes, you can do that. As long as you do it to everything.

1.3.3
Find the coordinates of the point P (x,y) which is located so that the line L1, which passes through P and the origin, has slope equal to 2, and the line L2, which passes through the point P and the point A (-1, 0), has slope equal to 1.

This one is actually pretty easy if you draw a diagram. The line L1 passes through the origin (0,0) and has slope 2. Start at the origin and go up 2 and across 1. Thus, the line passes through the point (1,2). Check this coordinate with the coordinate given for L2. Do you get a slope of 1?. \(\displaystyle \frac{2-0}{1-(-1)}\)

1.3.4
(a)Find equation of the line L through A (-2, 2) and perpendicular to the line L' : 2x+y=4.

Solve \(\displaystyle 2x+y=4\) for y. This gives \(\displaystyle y=-2x+4\)

See the slope?. It is the number in front of the x. -2

Thus, if the other line is perpendicular, then its slope is the negative reciprocal of -2. What is the negative reciprocal of -2?.

Plug this slope and the given coordinates (-2,2) into y=mx+b. Solve for b and that's it.

Another little tidbit: when a line equation is in the form \(\displaystyle ax+by=c\), you can find the slope without solving for y by noting that the slope is given by \(\displaystyle \frac{-a}{b}\)

 

[samistumbo]

This one is actually pretty easy if you draw a diagram. The line L1 passes through the origin (0,0) and has slope 2. Start at the origin and go up 2 and across 1. Thus, the line passes through the point (1,2). Check this coordinate with the coordinate given for L2. Do you get a slope of 1?. \(\displaystyle \frac{2-0}{1-(-1)}\)



Solve \(\displaystyle 2x+y=4\) for y. This gives \(\displaystyle y=-2x+4\)

See the slope?. It is the number in front of the x. -2

Thus, if the other line is perpendicular, then its slope is the negative reciprocal of -2. What is the negative reciprocal of -2?.

Plug this slope and the given coordinates (-2,2) into y=mx+b. Solve for b and that's it.

Another little tidbit: when a line equation is in the form \(\displaystyle ax+by=c\), you can find the slope without solving for y by noting that the slope is given by \(\displaystyle \frac{-a}{b}\)

What about the fact that it's L'? Will it still be done the same?

 

[galactus]

I do not understand. L' is the one they're giving you. 2x+y=4

The line you must find, L, passes through (-2,2) and is perp. to L': 2x+y=4

You can solve L' for y and get it into slope-intercept form. Solving L' for y gives us:

\(\displaystyle y=-2x+4\)

See the slope of L'?. It is -2.

Now, since the line we must find is perp. to this, the slope of L is the negative reciprocal of -2.

The negative reciprocal of -2 is 1/2

Thus, L has slope m=1/2 and passes through (-2,2). Plug x=-2, y=2, m=1/2 into y=mx+b

\(\displaystyle 2=1/2(-2)+b\)

\(\displaystyle b=3\)

The equation of L is \(\displaystyle y=\frac{1}{2}x+3\)

That's it.

Now, where do they intersect?. Set the two equations equal and solve for x.

\(\displaystyle -2x+4=\frac{1}{2}x+3\)

y follows from simply plugging the x value you just found back into either line equation.