Finding the equation of a line though points

[Brenten]

Ok, how about this one:

Find the equation of a line through the point (3 , -1) with slope 3.

y-(-1) = 3 (x-3)

y+1 = 3x -9

y +1 = 3(0) -9
add -1 to both sides
y = -10

Then
y(0) +1 =3x -9
add -3x to both sides
add -1 to both sides
-3x = -10
Divide both sides by -3
x = 3 1/3

So I get y -10 = 3 (x +1/3)

But it doesn't look right.

 

[Mrspi]

Ok, how about this one:

Find the equation of a line through the point (3 , -1) with slope 3.

y-(-1) = 3 (x-3)

y+1 = 3x -9<--------correct!!

y +1 = 3(0) -9<-------why do you substitute 0 for x?? You HAVE the equation
add -1 to both sides
y = -10

Then
y(0) +1 =3x -9
add -3x to both sides
add -1 to both sides
-3x = -10
Divide both sides by -3
x = 3 1/3

So I get y -10 = 3 (x +1/3)

But it doesn't look right.

You have (CORRECTLY) gotten the equation

y + 1 = 3x - 9

Are you trying to get this in slope-intercept form, y = mx + b? If you are, simply add -1 to both sides of your equation to get y by itself:

y + 1 + (-1) = 3x - 9 + (-1)
y = 3x - 10

There you go! Slope-intercept form.

 

[Brenten]

I should have added that it also needs to be graphed, which is why I did all the extra
parts to get y = -10 and x = 3 1/3, which are the points I need to plot right?

(0 , -10) and (3 1/3 , 0)

My brain hurts =/

 

[Mrspi]

I should have added that it also needs to be graphed, which is why I did all the extra
parts to get y = -10 and x = 3 1/3, which are the points I need to plot right?

(0 , -10) and (3 1/3 , 0)

My brain hurts =/

Yes, (0, -10) and (3 1/3, 0) are points on the line which you could use to graph the line.

But the equation

y = 3x - 10

has all of the information you need to graph the line. You know the y-intercept is -10. Plot (0, -10). The slope is 3, or 3/1. Starting at your point (0, -10), move UP three units and to the RIGHT one unit. Mark that point. Use a straightedge to draw the line through those two points. That's your graph.

Remember that there are often SEVERAL ways to accomplish something....don't lock yourself into just one.

 

[Brenten]

How about if you want to find the equation of the line that passes through the point (-1,3) and (-1,5)?

I started with m = y2 - y1 / x2 - x1
which gave me m = 5 - 3 / -1 - (-1)
Which becomes m = 2 / -1 +1
So M = 0 ??

Therefore, y + 5 = 0 (x -1) ?

 

[Mrspi]

How about if you want to find the equation of the line that passes through the point (-1,3) and (-1,5)?

I started with m = y2 - y1 / x2 - x1
which gave me m = 5 - 3 / -1 - (-1)
Which becomes m = 2 / -1 +1
So M = 0 ??

Therefore, y + 5 = 0 (x -1) ?

I don't think so...-1 + 1 is 0 all right, so you have 2 / 0. Think a bit. Is division by 0 defined??

It also might be informative for you if you graph the two points, and draw the line through them. What kind of line is it? Does that kind of line have a "typical" equation form?

 

[Brenten]

If I graph the points (-1 , 3) and -1 , 5) I get a straight line. So then my rise/run is 2/0, going up 2pts at a time but not moving anywhere on the x-axis.

I'm lost on it being a 'typical' equation form but something CAN = 0 right?

 

[Mrspi]

If I graph the points (-1 , 3) and -1 , 5) I get a straight line. So then my rise/run is 2/0, going up 2pts at a time but not moving anywhere on the x-axis.

I'm lost on it being a 'typical' equation form but something CAN = 0 right?

Any line is a "straight" line.

Do the words "vertical" and "horizontal" ring a bell with regard to lines?

Yes, "something can = 0"....but you CANNOT divide anything by 0. How many times does 0 go into 5???? 1 time? 17 times? 10000 times? I could argue that any of those answers would be correct...

Once you have decided what "special" kind of line these two points determine, I strongly suggest that you review what your textbook has to say about that kind of line and a typical equation for that kind of line. HINT: use the glossary in the back of your textbook to find where information about your special kind of line is located.

 

[Brenten]

Gotcha, I was looking at it too simplisitcally. The line is a vertical line and anything divided by 0 will always be 0. Since this is a vertical line , x = a and there is no slope nor is there a y-intercept.

 

[Mrspi]

Gotcha, I was looking at it too simplisitcally. The line is a vertical line and anything divided by 0 will always be 0. Since this is a vertical line , x = a and there is no slope nor is there a y-intercept.

I agree with everything you said except this "anything divided by 0 will always be 0." You can't divide by 0. And I concede that I made an error in my previous post with regard to division by 0. Consider the division 5/0. What might the answer be? Remember how you learned to check division in grade school? Multiply the divisor by the quotient (answer) and you should get the number you were dividing into. For example, if you have 15 / 3 = 5, you check by multiplication. The divisor is 3....the quotient is 5. And if you multiply the divisor by the quotient, you get the dividend (the number you were dividing INTO): 3*5 = 15. It checks.

Now, if you have 5/0 = "something", you should be able to check that the same way. Divisor * quotient = dividend. 0 * "something" = 5. Can that EVER be true? It can't, because anything multiplied by 0 is 0. So...mathematicians agreed a long time ago that division by 0 is NOT DEFINED.

And a line with an "undefined" slope is a vertical line.