Helloaaa
Badges: 9
Rep:
?
#1
Report Thread starter 11 months ago
#1
Can someone check if I have done this correctly
Answer must be given in the form ax + by + c = 0Name:  15741969157537250224537406477457.jpg
Views: 30
Size:  53.7 KB

I got x=3 but answer is x-3= 0 .... is it because they rearranged it to give it in the form ax+ by + c = 0 ???
Last edited by Helloaaa; 11 months ago
0
reply
MarkFromWales
Badges: 10
Rep:
?
#2
Report 11 months ago
#2
You really don't need to use such a complicated method for this question. If you have two points both with an x-coordinate of 3 then obviously the equation of the line is x = 3 or, in the format requested, x - 3 = 0.
Last edited by MarkFromWales; 11 months ago
0
reply
Helloaaa
Badges: 9
Rep:
?
#3
Report Thread starter 11 months ago
#3
(Original post by MarkFromWales)
You really don't need to use such a complicated method for this question. If you have two points both with an x-coordinate of 3 then obviously the equation of the line is x = 3 or, in the format requested, x - 3 = 0.
That is certainly true .... perhaps I am a little bit extra.
If we have a vertical line that passes through the point (0, 2/3) how can we find the equation cor this line in the form ax + by + c = 0 or in the form y = mx + c (which I guess we can rearrange to write it in a different format)

I haven't been given anything apart from these two coordinates so confused on how to do this? Also wouldn't it be a horizontal line as we have a y coordinate and not an x ?
0
reply
MarkFromWales
Badges: 10
Rep:
?
#4
Report 11 months ago
#4
(Original post by Helloaaa)
If we have a vertical line that passes through the point (0, 2/3) how can we find the equation cor this line in the form ax + by + c = 0 or in the form y = mx + c (which I guess we can rearrange to write it in a different format)

I haven't been given anything apart from these two coordinates so confused on how to do this? Also wouldn't it be a horizontal line as we have a y coordinate and not an x ?
(0, 2/3) is on the y-axis. I think that's what you mean by a vertical line. So the equation is x = 0. All points on the y-axis have an x-coordinate of 0. You can't write it in the form y = mx + c, that's only for slanting lines or lines parallel to the x-axis. It's already in the form ax + by + c = 0 ... that's with a = 1, b = 0, c = 0.
0
reply
Helloaaa
Badges: 9
Rep:
?
#5
Report Thread starter 11 months ago
#5
(Original post by MarkFromWales)
(0, 2/3) is on the y-axis. I think that's what you mean by a vertical line. So the equation is x = 0. All points on the y-axis have an x-coordinate of 0. You can't write it in the form y = mx + c, that's only for slanting lines or lines parallel to the x-axis. It's already in the form ax + by + c = 0 ... that's with a = 1, b = 0, c = 0.
The answer is x =0 ... why is the equation in terms of x and not y? And why is a=1?

Thank you for your time
0
reply
MarkFromWales
Badges: 10
Rep:
?
#6
Report 11 months ago
#6
Some teachers say that you can write the equation of any straight line in the form y = mx + c, but it's not true. It doesn't work for lines that are parallel to the y-axis. The equation of such a line is x = constant, for example, x = 0 or x = -5.

The equation of any straight line can be written in the form ax + by + c = 0 and I guess that's why some writers prefer it to y=mx+c.
If you replace a with 1 and b and c with 0 you get 1x + 0 + 0 = 0 which is simply x = 0, the equation of the y-axis.
Last edited by MarkFromWales; 11 months ago
0
reply
Helloaaa
Badges: 9
Rep:
?
#7
Report Thread starter 11 months ago
#7
(Original post by MarkFromWales)
Some teachers say that you can write the equation of any straight line in the form y = mx + c, but it's not true. It doesn't work for lines that are parallel to the y-axis. The equation of such a line is x = constant, for example, x = 0 or x = -5.

The equation of any straight line can be written in the form ax + by + c = 0 and I guess that's why some writers prefer it to y=mx+c.
If you replace a with 1 and b and c with 0 you get 1x + 0 + 0 = 0 which is simply x = 0, the equation of the y-axis.
Thank you so much!

Name:  15742053024167009153937374031038.jpg
Views: 34
Size:  166.1 KB

Do you know how to rearrange to radius equation to find c?
Attached files
0
reply
mqb2766
Badges: 18
Rep:
?
#8
Report 11 months ago
#8
(Original post by Helloaaa)
Thank you so much!

Name:  15742053024167009153937374031038.jpg
Views: 34
Size:  166.1 KB

Do you know how to rearrange to radius equation to find c?
Do you want to form the equation of a circle from this quadratic? If so, complete the square for x and y.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (163)
14.54%
I'm not sure (51)
4.55%
No, I'm going to stick it out for now (330)
29.44%
I have already dropped out (33)
2.94%
I'm not a current university student (544)
48.53%

Watched Threads

View All