Math Question

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VowOFSilence

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Ok so i've been falling behind the rest of the class for a while now, I think its time to catchup so I need help understanding these questions

A line is drawn through (−1, 2), perpendicular to the line 4x + 3y − 6 = 0. Find its equation


in general form.








A line is drawn through the point (−1, −1), parallel to the line 2x − 3y + 9 = 0. Where will
it cross the x-axis?






A line is drawn parallel to 4x−3y+1=0, through the points(1,3)and(6,a).Whatisthe


value of a?





The answers are in the book but I don't understand how to get them. Any help is appreciated :D. I know a bit of background info on perpendicular lines and the equation MX + b = y. However i've been trying to get the answers in the book through different ways, and it never seems to work.
 
"A little behind"? Frankly it sounds like you have never attended class nor even opened a text book! These are problems in three dimensions and one of the first things you should have learned is that lines, in three dimensions, cannot be written in terms of "y= mx+ b". That only works in two dimensions. In three dimensions you need to include "z" and you need either two linear equations in x, y, and z, or parametric equation. Parametric equations would be simpler. Do you know what "parametric equations" are? If not look them up in your text book and then we can continue.
 
HallsofIvy said:
"A little behind"? Frankly it sounds like you have never attended class nor even opened a text book! These are problems in three dimensions and one of the first things you should have learned is that lines, in three dimensions, cannot be written in terms of "y= mx+ b". That only works in two dimensions. In three dimensions you need to include "z" and you need either two linear equations in x, y, and z, or parametric equation. Parametric equations would be simpler. Do you know what "parametric equations" are? If not look them up in your text book and then we can continue.
Click to expand...

Yeah, believe me we don't need to learn about parametric equations. We also don't need z's. The answer in the book just has the equation in general form.
 
My error- these are NOT three dimensional problems, only two. Yes, the equation of a line in two dimensions can be written as y= mx+ b.

Further, the slope of a line perpendicular to y= mx+ b (which has slope m) is -1/m.

So the line 4x+ 3y- 6= 0 can be written 3y= 6- 4x (I added 6 to and subtracted 4x from both sides) so y= 2- (4/3)x (I divided both sides by 6). The line has slope -4/3 and any line perpendicular to it has slope -3/4 and equation y= (-3/4)x+ b for some number b. If we require that the line also go through (-1, 2), we must have y= 2 when x= -1 so 2= (-3/4)(-1)+ b, 2= (3/4)+ b, b= 2- (3/4)= 8/4- 3/4= 5/4. y= (-3/4)x+ 5/4. Multiply both side by 4 to get 4y= -3x+ 5 and add 3x to both sides to get 3x+ 4y= 5.

The others can be done the same way- the crucial point is that "the slope of a line perpendicular to y= mx+ b (which has slope m) is -1/m."

Sorry about my misunderstanding.
 
HallsofIvy said:
My error- these are NOT three dimensional problems, only two. Yes, the equation of a line in two dimensions can be written as y= mx+ b.

Further, the slope of a line perpendicular to y= mx+ b (which has slope m) is -1/m.

So the line 4x+ 3y- 6= 0 can be written 3y= 6- 4x (I added 6 to and subtracted 4x from both sides) so y= 2- (4/3)x (I divided both sides by 6). The line has slope -4/3 and any line perpendicular to it has slope -3/4 and equation y= (-3/4)x+ b for some number b. If we require that the line also go through (-1, 2), we must have y= 2 when x= -1 so 2= (-3/4)(-1)+ b, 2= (3/4)+ b, b= 2- (3/4)= 8/4- 3/4= 5/4. y= (-3/4)x+ 5/4. Multiply both side by 4 to get 4y= -3x+ 5 and add 3x to both sides to get 3x+ 4y= 5.

The others can be done the same way- the crucial point is that "the slope of a line perpendicular to y= mx+ b (which has slope m) is -1/m."

Sorry about my misunderstanding.
Click to expand...

Thanks for taking the time out to help me Ivy :)

But this is the answer in the notebook 3x−4y+11=0
 
VowOFSilence said:
Ok so i've been falling behind the rest of the class for a while now, I think its time to catchup so I need help understanding these questions

A line is drawn through (−1, 2), perpendicular to the line 4x + 3y − 6 = 0. Find its equation


in general form.

Equation of a line passing through (x1,y1) and slope m1 is:

(y - y1) = m1 * (x - x1)

here

m1 = -1/(-4/3) = 3/4, x1 = -1 and y1 = 2

so the equation of the line is:

(y - 2) =3/4* (x - (-1))

4y - 8 = 3x + 3

3x - 4y + 11 = 0



A line is drawn through the point (−1, −1), parallel to the line 2x − 3y + 9 = 0. Where will
it cross the x-axis?



A line is drawn parallel to 4x−3y+1=0, through the points(1,3)and(6,a).Whatisthe


value of a?





The answers are in the book but I don't understand how to get them. Any help is appreciated :D. I know a bit of background info on perpendicular lines and the equation MX + b = y. However i've been trying to get the answers in the book through different ways, and it never seems to work.
Click to expand...

Now you try the rest and show us your work if you are still stuck.
 
I got that and the secnod one correct, then the third question I don't even get close to the answer? Thanks for all the help above btw
 
Denis said:
That's the 3rd question:
"A line is drawn parallel to 4x−3y+1=0, through the points (1,3) and (6,a).
What is the value of a?

Step 1: get slope of new line
4x - 3y + 1 = 0
3y = 4x + 1
y = (4/3)x + 1/3
So slope = 4/3

Were you able to get that far?
Where did you get stuck?
What is the given solution?
Click to expand...

I got that far, but then I tried the formula khan provided

(y - 2) =3/4* (x - (-1))
given solution = 29/3 = A
 
Denis said:
WHY do you show x = -1; From (1,3), x = 1 .... OK?

So:
y - 3 = 4/3(x - 1)
3y - 9 = 4x + 4
4x - 3y + 5 = 0
OK?
Now substitute (6,a) in there, and solve for a : OK? ;)
Click to expand...

You my friend, are a saint! I got 3y-4x-5=0 then I substitued 3 x a and 4 x 6 - 5 so its 29/3 = a
Screen Shot 2014-10-19 at 18.35.26.jpg
Can you help me get started here, then i'll do the rest :) I just dont get it.
 
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